Chapter 11 Three Dimensional Geometry

We are given the angles that a line makes with the positive direction of the coordinate axes. We need to find the direction cosines of this line.

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Given:

Angles made by the line with the positive x, y, and z axes are $\alpha = 90^\circ$, $\beta = 60^\circ$, and $\gamma = 30^\circ$, respectively.

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To Find:

The direction cosines of the line.

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Solution:

The direction cosines of a line that makes angles $\alpha$, $\beta$, and $\gamma$ with the positive direction of the x, y, and z axes are given by $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

Substitute the given angles:

$l = \cos(90^\circ)$

$m = \cos(60^\circ)$

$n = \cos(30^\circ)$

Calculate the cosine of each angle:

$\cos(90^\circ) = 0$

$\cos(60^\circ) = \frac{1}{2}$

$\cos(30^\circ) = \frac{\sqrt{3}}{2}$

So, the direction cosines are:

$l = 0$

$m = \frac{1}{2}$

$n = \frac{\sqrt{3}}{2}$

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Verification:

A fundamental property of direction cosines $(l, m, n)$ is that the sum of their squares is equal to 1: $l^2 + m^2 + n^2 = 1$.

Let's check this property with our calculated direction cosines:

$(0)^2 + (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 0 + \frac{1}{4} + \frac{3}{4} = \frac{1 + 3}{4} = \frac{4}{4} = 1$

Since the sum of the squares is 1, our direction cosines are correct.

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Result:

The direction cosines of the line are $\mathbf{\left( 0 \;,\; \frac{1}{2} \;,\; \frac{\sqrt{3}}{2} \right)}$.

We are given the direction ratios of a line and need to find its direction cosines. Direction ratios are numbers proportional to the direction cosines.

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Given:

The direction ratios of the line are $a = 2$, $b = -1$, and $c = -2$.

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To Find:

The direction cosines of the line.

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Solution:

Let the direction ratios of a line be $a$, $b$, and $c$. The direction cosines $l$, $m$, and $n$ are related to the direction ratios by the formulas:

$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$

$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$

$n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$

First, calculate the value of $\sqrt{a^2 + b^2 + c^2}$ using the given direction ratios $a=2$, $b=-1$, $c=-2$.

$\sqrt{a^2 + b^2 + c^2} = \sqrt{(2)^2 + (-1)^2 + (-2)^2}$

$= \sqrt{4 + 1 + 4}$

$= \sqrt{9}$

$= 3$

Now, substitute the values of $a$, $b$, $c$, and $\sqrt{a^2 + b^2 + c^2}$ into the formulas for the direction cosines:

$l = \frac{2}{3}$

$m = \frac{-1}{3}$

$n = \frac{-2}{3}$

So, the direction cosines are $\left( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \right)$.

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Verification:

The sum of the squares of the direction cosines must be 1 ($l^2 + m^2 + n^2 = 1$).

$(\frac{2}{3})^2 + (-\frac{1}{3})^2 + (-\frac{2}{3})^2 = \frac{4}{9} + \frac{1}{9} + \frac{4}{9} = \frac{4 + 1 + 4}{9} = \frac{9}{9} = 1$

The property holds, so the direction cosines are correct.

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Result:

The direction cosines of the line are $\mathbf{\left( \frac{2}{3} \;,\; -\frac{1}{3} \;,\; -\frac{2}{3} \right)}$.

We are given the coordinates of two points that a line passes through. We need to find the direction cosines of this line. The direction ratios of the line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are proportional to $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$. Once we have the direction ratios, we can find the direction cosines.

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Given:

Point P with coordinates $(x_1, y_1, z_1) = (-2, 4, -5)$.

Point Q with coordinates $(x_2, y_2, z_2) = (1, 2, 3)$.

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To Find:

The direction cosines of the line passing through P and Q.

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Solution:

The direction ratios of the line passing through points P$(x_1, y_1, z_1)$ and Q$(x_2, y_2, z_2)$ are proportional to the differences in their coordinates:

$a = x_2 - x_1$

$b = y_2 - y_1$

$c = z_2 - z_1$

Substitute the given coordinates:

$a = 1 - (-2) = 1 + 2 = 3$

$b = 2 - 4 = -2$

$c = 3 - (-5) = 3 + 5 = 8$

So, the direction ratios of the line are $(3, -2, 8)$.

Now, we use the direction ratios $(a, b, c)$ to find the direction cosines $(l, m, n)$ using the formulas:

$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$

$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$

$n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$

First, calculate $\sqrt{a^2 + b^2 + c^2}$:

$\sqrt{a^2 + b^2 + c^2} = \sqrt{(3)^2 + (-2)^2 + (8)^2}$

$= \sqrt{9 + 4 + 64}$

$= \sqrt{77}$

Now, substitute the values into the formulas for the direction cosines:

$l = \frac{3}{\sqrt{77}}$

$m = \frac{-2}{\sqrt{77}}$

$n = \frac{8}{\sqrt{77}}$

So, the direction cosines of the line are $\left( \frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}} \right)$.

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Verification:

Check if the sum of the squares of the direction cosines is 1:

$(\frac{3}{\sqrt{77}})^2 + (-\frac{2}{\sqrt{77}})^2 + (\frac{8}{\sqrt{77}})^2 = \frac{9}{77} + \frac{4}{77} + \frac{64}{77}$

$= \frac{9 + 4 + 64}{77} = \frac{13 + 64}{77} = \frac{77}{77} = 1$

The property holds, confirming the direction cosines are correct.

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Result:

The direction cosines of the line passing through the points (-2, 4, -5) and (1, 2, 3) are $\mathbf{\left( \frac{3}{\sqrt{77}} \;,\; -\frac{2}{\sqrt{77}} \;,\; \frac{8}{\sqrt{77}} \right)}$.

We need to find the direction cosines for each of the standard coordinate axes (x, y, and z) in a 3D Cartesian system. The direction cosines of a line represent the cosines of the angles that the line makes with the positive x, y, and z axes.

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Given:

The coordinate axes: x-axis, y-axis, and z-axis.

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To Find:

The direction cosines for each axis.

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Solution:

Let $(l, m, n)$ be the direction cosines of a line, where $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$. Here, $\alpha$, $\beta$, and $\gamma$ are the angles made by the line with the positive x, y, and z axes, respectively.

For the x-axis:

The positive x-axis makes an angle of $0^\circ$ with the positive x-axis.

It is perpendicular to the yz-plane, so it makes an angle of $90^\circ$ with the positive y-axis and the positive z-axis.

So, for the x-axis: $\alpha = 0^\circ$, $\beta = 90^\circ$, $\gamma = 90^\circ$.

The direction cosines are:

$l = \cos(0^\circ) = 1$

$m = \cos(90^\circ) = 0$

$n = \cos(90^\circ) = 0$

The direction cosines of the x-axis are $(1, 0, 0)$.

For the y-axis:

The positive y-axis is perpendicular to the xz-plane, so it makes an angle of $90^\circ$ with the positive x-axis.

It makes an angle of $0^\circ$ with the positive y-axis.

It is perpendicular to the xy-plane, so it makes an angle of $90^\circ$ with the positive z-axis.

So, for the y-axis: $\alpha = 90^\circ$, $\beta = 0^\circ$, $\gamma = 90^\circ$.

The direction cosines are:

$l = \cos(90^\circ) = 0$

$m = \cos(0^\circ) = 1$

$n = \cos(90^\circ) = 0$

The direction cosines of the y-axis are $(0, 1, 0)$.

For the z-axis:

The positive z-axis is perpendicular to the xy-plane, so it makes an angle of $90^\circ$ with the positive x-axis and the positive y-axis.

It makes an angle of $0^\circ$ with the positive z-axis.

So, for the z-axis: $\alpha = 90^\circ$, $\beta = 90^\circ$, $\gamma = 0^\circ$.

The direction cosines are:

$l = \cos(90^\circ) = 0$

$m = \cos(90^\circ) = 0$

$n = \cos(0^\circ) = 1$

The direction cosines of the z-axis are $(0, 0, 1)$.

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Verification:

Check the property $l^2 + m^2 + n^2 = 1$ for each set of direction cosines:

For x-axis: $(1)^2 + (0)^2 + (0)^2 = 1 + 0 + 0 = 1$. (Correct)

For y-axis: $(0)^2 + (1)^2 + (0)^2 = 0 + 1 + 0 = 1$. (Correct)

For z-axis: $(0)^2 + (0)^2 + (1)^2 = 0 + 0 + 1 = 1$. (Correct)

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Result:

The direction cosines of the x-axis are $\mathbf{(1, 0, 0)}$.

The direction cosines of the y-axis are $\mathbf{(0, 1, 0)}$.

The direction cosines of the z-axis are $\mathbf{(0, 0, 1)}$.

We need to show that the three given points A, B, and C in 3D space are collinear. Points are collinear if they lie on the same straight line. One way to show this is to find the direction ratios of the lines formed by pairs of these points and demonstrate that these direction ratios are proportional.

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Given:

Point A (2, 3, -4)

Point B (1, -2, 3)

Point C (3, 8, -11)

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To Show:

Points A, B, and C are collinear.

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Solution:

Find the direction ratios of the line segment AB.

The direction ratios of the line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are proportional to $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$.

For line AB, using A(2, 3, -4) and B(1, -2, 3):

$a_{AB} = 1 - 2 = -1$

$b_{AB} = -2 - 3 = -5$

$c_{AB} = 3 - (-4) = 3 + 4 = 7$

The direction ratios of AB are $(-1, -5, 7)$.

Find the direction ratios of the line segment BC.

For line BC, using B(1, -2, 3) and C(3, 8, -11):

$a_{BC} = 3 - 1 = 2$

$b_{BC} = 8 - (-2) = 8 + 2 = 10$

$c_{BC} = -11 - 3 = -14$

The direction ratios of BC are $(2, 10, -14)$.

To show that the points are collinear, the direction ratios of AB and BC must be proportional. This means that $(a_{BC}, b_{BC}, c_{BC}) = k(a_{AB}, b_{AB}, c_{AB})$ for some scalar $k$.

Compare the ratios of the corresponding direction ratios:

$\frac{a_{BC}}{a_{AB}} = \frac{2}{-1} = -2$

$\frac{b_{BC}}{b_{AB}} = \frac{10}{-5} = -2$

$\frac{c_{BC}}{c_{AB}} = \frac{-14}{7} = -2$

Since the ratios of the corresponding direction ratios are equal to a constant value $(-2)$, the direction ratios of AB and BC are proportional. This means that the line segment AB is parallel to the line segment BC.

Since the line segments AB and BC are parallel and share the common point B, the points A, B, and C must lie on the same straight line.

Therefore, the points A, B, and C are collinear.

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Alternative Method using Vectors:

Let $\overrightarrow{AB}$ and $\overrightarrow{BC}$ be the vectors joining the points.

$\overrightarrow{AB} = (1 - 2)\hat{i} + (-2 - 3)\hat{j} + (3 - (-4))\hat{k} = -\hat{i} - 5\hat{j} + 7\hat{k}$

$\overrightarrow{BC} = (3 - 1)\hat{i} + (8 - (-2))\hat{j} + (-11 - 3)\hat{k} = 2\hat{i} + 10\hat{j} - 14\hat{k}$

Observe the relationship between $\overrightarrow{AB}$ and $\overrightarrow{BC}$.

$\overrightarrow{BC} = 2\hat{i} + 10\hat{j} - 14\hat{k} = -2(-\hat{i} - 5\hat{j} + 7\hat{k}) = -2\overrightarrow{AB}$

Since $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$ (with scalar $k = -2$), the vectors are parallel. As they share the common point B, the points A, B, and C are collinear.

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Conclusion:

By showing that the direction ratios of AB and BC are proportional, or equivalently that $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$, we have proven that the points A, B, and C are collinear.

We are given the angles that a line makes with the positive directions of the x, y, and z axes. We need to find the direction cosines of this line.

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Given:

The angles made by the line with the positive x, y, and z axes are $\alpha = 90^\circ$, $\beta = 135^\circ$, and $\gamma = 45^\circ$, respectively.

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To Find:

The direction cosines of the line.

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Solution:

The direction cosines of a line that makes angles $\alpha$, $\beta$, and $\gamma$ with the positive x, y, and z axes are given by $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

Substitute the given angles:

$l = \cos(90^\circ)$

$m = \cos(135^\circ)$

$n = \cos(45^\circ)$

Calculate the cosine of each angle:

$\cos(90^\circ) = 0$

$\cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ) = -\frac{1}{\sqrt{2}}$

$\cos(45^\circ) = \frac{1}{\sqrt{2}}$

So, the direction cosines are:

$l = 0$

$m = -\frac{1}{\sqrt{2}}$

$n = \frac{1}{\sqrt{2}}$

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Verification:

Check if the sum of the squares of the direction cosines is 1 ($l^2 + m^2 + n^2 = 1$).

$(0)^2 + (-\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$

The property holds, confirming the direction cosines are correct.

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Result:

The direction cosines of the line are $\mathbf{\left( 0 \;,\; -\frac{1}{\sqrt{2}} \;,\; \frac{1}{\sqrt{2}} \right)}$.

We need to find the direction cosines of a line that makes equal angles with the coordinate axes. Let the angle made by the line with the positive x, y, and z axes be $\theta$. Since the angles are equal, $\alpha = \beta = \gamma = \theta$.

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Given:

A line makes equal angles with the coordinate axes.

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To Find:

The direction cosines of the line.

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Solution:

Let the equal angle made by the line with the positive x, y, and z axes be $\theta$. So, $\alpha = \theta$, $\beta = \theta$, and $\gamma = \theta$.

The direction cosines of the line are $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

In this case, the direction cosines are:

$l = \cos\theta$

$m = \cos\theta$

$n = \cos\theta$

We know that the sum of the squares of the direction cosines of any line is equal to 1:

$l^2 + m^2 + n^2 = 1$

Substitute the expressions for $l$, $m$, and $n$:

$(\cos\theta)^2 + (\cos\theta)^2 + (\cos\theta)^2 = 1$

$\cos^2\theta + \cos^2\theta + \cos^2\theta = 1$

$3\cos^2\theta = 1$

Solve for $\cos^2\theta$:

$\cos^2\theta = \frac{1}{3}$

Take the square root of both sides to find $\cos\theta$:

$\cos\theta = \pm\sqrt{\frac{1}{3}} = \pm\frac{1}{\sqrt{3}}$

Since the direction cosines are $l = \cos\theta$, $m = \cos\theta$, and $n = \cos\theta$, they are all equal to $\pm\frac{1}{\sqrt{3}}$.

Thus, the direction cosines of the line are either $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ or $\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$.

Both sets represent lines making equal angles with the axes, just in opposite directions.

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Result:

The direction cosines of a line which makes equal angles with the coordinate axes are $\mathbf{± \left( \frac{1}{\sqrt{3}} \;,\; \frac{1}{\sqrt{3}} \;,\; \frac{1}{\sqrt{3}} \right)}$.

We are given the direction ratios of a line and need to find its direction cosines. Direction ratios are numbers proportional to the direction cosines. We use the formula relating direction ratios and direction cosines.

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Given:

The direction ratios of the line are $a = -18$, $b = 12$, and $c = -4$.

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To Find:

The direction cosines of the line.

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Solution:

Let the direction ratios of a line be $a$, $b$, and $c$. The direction cosines $l$, $m$, and $n$ are given by:

$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$

$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$

$n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$

First, calculate the normalizing factor $\sqrt{a^2 + b^2 + c^2}$ using the given direction ratios $a=-18$, $b=12$, $c=-4$.

$\sqrt{a^2 + b^2 + c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$= \sqrt{324 + 144 + 16}$

$= \sqrt{484}$

To find the square root of 484, we can look for factors or use prime factorization:

$\begin{array}{c|cc} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$484 = 2^2 \times 11^2 = (2 \times 11)^2 = 22^2$

So, $\sqrt{484} = 22$.

Now, substitute the values of $a$, $b$, $c$, and $\sqrt{a^2 + b^2 + c^2} = 22$ into the formulas for the direction cosines:

$l = \frac{-18}{22} = \frac{-9}{11}$

$m = \frac{12}{22} = \frac{6}{11}$

$n = \frac{-4}{22} = \frac{-2}{11}$

So, the direction cosines are $\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$.

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Verification:

Check if the sum of the squares of the direction cosines is 1 ($l^2 + m^2 + n^2 = 1$).

$(-\frac{9}{11})^2 + (\frac{6}{11})^2 + (-\frac{2}{11})^2 = \frac{81}{121} + \frac{36}{121} + \frac{4}{121}$

$= \frac{81 + 36 + 4}{121} = \frac{117 + 4}{121} = \frac{121}{121} = 1$

The property holds, confirming the direction cosines are correct.

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Result:

The direction cosines of the line are $\mathbf{\left( -\frac{9}{11} \;,\; \frac{6}{11} \;,\; -\frac{2}{11} \right)}$.

We need to show that the three given points are collinear. We can do this by finding the direction ratios of the line segments formed by pairs of these points and checking if they are proportional. Alternatively, we can use vectors and check if one vector is a scalar multiple of another, sharing a common point.

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Given:

Point A (2, 3, 4)

Point B (-1, -2, 1)

Point C (5, 8, 7)

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To Show:

Points A, B, and C are collinear.

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Solution using Direction Ratios:

Find the direction ratios of the line segment AB.

Using A(2, 3, 4) and B(-1, -2, 1):

$a_{AB} = -1 - 2 = -3$

$b_{AB} = -2 - 3 = -5$

$c_{AB} = 1 - 4 = -3$

The direction ratios of AB are $(-3, -5, -3)$.

Find the direction ratios of the line segment BC.

Using B(-1, -2, 1) and C(5, 8, 7):

$a_{BC} = 5 - (-1) = 5 + 1 = 6$

$b_{BC} = 8 - (-2) = 8 + 2 = 10$

$c_{BC} = 7 - 1 = 6$

The direction ratios of BC are $(6, 10, 6)$.

To check for collinearity, check if the direction ratios are proportional. That is, if $\frac{a_{BC}}{a_{AB}} = \frac{b_{BC}}{b_{AB}} = \frac{c_{BC}}{c_{AB}} = k$ for some scalar $k$.

$\frac{a_{BC}}{a_{AB}} = \frac{6}{-3} = -2$

$\frac{b_{BC}}{b_{AB}} = \frac{10}{-5} = -2$

$\frac{c_{BC}}{c_{AB}} = \frac{6}{-3} = -2$

Since the ratios of the corresponding direction ratios are equal to a constant value $(-2)$, the direction ratios of AB and BC are proportional. This implies that the line segment AB is parallel to the line segment BC. As they share the common point B, the points A, B, and C must lie on the same line.

Therefore, the points A, B, and C are collinear.

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Alternative Method using Vectors:

Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$.

$\overrightarrow{AB} = (-1 - 2)\hat{i} + (-2 - 3)\hat{j} + (1 - 4)\hat{k} = -3\hat{i} - 5\hat{j} - 3\hat{k}$

$\overrightarrow{BC} = (5 - (-1))\hat{i} + (8 - (-2))\hat{j} + (7 - 1)\hat{k} = 6\hat{i} + 10\hat{j} + 6\hat{k}$

Check if $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$.

$\overrightarrow{BC} = 6\hat{i} + 10\hat{j} + 6\hat{k} = -2(-3\hat{i} - 5\hat{j} - 3\hat{k}) = -2\overrightarrow{AB}$

Since $\overrightarrow{BC} = -2\overrightarrow{AB}$, the vectors are parallel. Because they share the common point B, the points A, B, and C are collinear.

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Conclusion:

We have shown that the direction ratios of AB and BC are proportional, or that $\overrightarrow{BC}$ is a scalar multiple of $\overrightarrow{AB}$. This proves that the points (2, 3, 4), (-1, -2, 1), and (5, 8, 7) are collinear.

We are given the vertices of a triangle and need to find the direction cosines of each of its sides. A side of a triangle is a line segment joining two vertices. The direction cosines of a line segment are the direction cosines of the line containing the segment.

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Given:

Vertices of the triangle are A(3, 5, -4), B(-1, 1, 2), and C(-5, -5, -2).

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To Find:

The direction cosines of the sides AB, BC, and CA.

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Solution:

The direction ratios of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$. If the direction ratios are $(a, b, c)$, the direction cosines are $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$.

Side AB:

Points are A(3, 5, -4) and B(-1, 1, 2).

Direction ratios of AB: $a = -1 - 3 = -4$, $b = 1 - 5 = -4$, $c = 2 - (-4) = 2 + 4 = 6$.

Direction ratios of AB are $(-4, -4, 6)$.

Calculate $\sqrt{(-4)^2 + (-4)^2 + (6)^2} = \sqrt{16 + 16 + 36} = \sqrt{68}$.

$\sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$.

Direction cosines of AB are $\left(\frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}}\right) = \left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right)$.

Side BC:

Points are B(-1, 1, 2) and C(-5, -5, -2).

Direction ratios of BC: $a = -5 - (-1) = -5 + 1 = -4$, $b = -5 - 1 = -6$, $c = -2 - 2 = -4$.

Direction ratios of BC are $(-4, -6, -4)$.

Calculate $\sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68}$.

$\sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$.

Direction cosines of BC are $\left(\frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}\right) = \left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right)$.

Side CA:

Points are C(-5, -5, -2) and A(3, 5, -4).

Direction ratios of CA: $a = 3 - (-5) = 3 + 5 = 8$, $b = 5 - (-5) = 5 + 5 = 10$, $c = -4 - (-2) = -4 + 2 = -2$.

Direction ratios of CA are $(8, 10, -2)$.

Calculate $\sqrt{(8)^2 + (10)^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168}$.

$\sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}$.

Direction cosines of CA are $\left(\frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}}\right) = \left(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}\right)$.

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Verification (Optional):

Check the sum of squares of direction cosines for each side.

For AB: $(-\frac{2}{\sqrt{17}})^2 + (-\frac{2}{\sqrt{17}})^2 + (\frac{3}{\sqrt{17}})^2 = \frac{4}{17} + \frac{4}{17} + \frac{9}{17} = \frac{4+4+9}{17} = \frac{17}{17} = 1$. (Correct)

For BC: $(-\frac{2}{\sqrt{17}})^2 + (-\frac{3}{\sqrt{17}})^2 + (-\frac{2}{\sqrt{17}})^2 = \frac{4}{17} + \frac{9}{17} + \frac{4}{17} = \frac{4+9+4}{17} = \frac{17}{17} = 1$. (Correct)

For CA: $(\frac{4}{\sqrt{42}})^2 + (\frac{5}{\sqrt{42}})^2 + (-\frac{1}{\sqrt{42}})^2 = \frac{16}{42} + \frac{25}{42} + \frac{1}{42} = \frac{16+25+1}{42} = \frac{42}{42} = 1$. (Correct)

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Result:

The direction cosines of the sides of the triangle are:

For side AB: $\mathbf{\left( -\frac{2}{\sqrt{17}} \;,\; -\frac{2}{\sqrt{17}} \;,\; \frac{3}{\sqrt{17}} \right)}$

For side BC: $\mathbf{\left( -\frac{2}{\sqrt{17}} \;,\; -\frac{3}{\sqrt{17}} \;,\; -\frac{2}{\sqrt{17}} \right)}$

For side CA: $\mathbf{\left( \frac{4}{\sqrt{42}} \;,\; \frac{5}{\sqrt{42}} \;,\; -\frac{1}{\sqrt{42}} \right)}$

We need to find the vector and Cartesian equations of a line given a point it passes through and a vector it is parallel to.

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Given:

The line passes through the point $(x_1, y_1, z_1) = (5, 2, -4)$.

The line is parallel to the vector $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$.

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To Find:

The vector equation of the line.

The Cartesian equation of the line.

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Solution:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda\vec{b}$

where $\vec{r}$ is the position vector of any point $(x, y, z)$ on the line, and $\lambda$ is a scalar parameter.

The position vector of the point $(5, 2, -4)$ is $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$.

The vector parallel to the line is $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$.

Substitute $\vec{a}$ and $\vec{b}$ into the vector equation formula:

$\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$

$\vec{r} = (5 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (-4 - 8\lambda)\hat{k}$

This is the vector equation of the line.

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To find the Cartesian equation, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Equating the components of the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = (5 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (-4 - 8\lambda)\hat{k}$

Equating the corresponding components:

$x = 5 + 3\lambda$

$y = 2 + 2\lambda$

$z = -4 - 8\lambda$

From these equations, we can express $\lambda$ in terms of $x$, $y$, and $z$:

$3\lambda = x - 5 \implies \lambda = \frac{x - 5}{3}$

$2\lambda = y - 2 \implies \lambda = \frac{y - 2}{2}$

$-8\lambda = z + 4 \implies \lambda = \frac{z + 4}{-8}$

Since all these expressions are equal to $\lambda$, we can equate them to get the Cartesian equation of the line:

$\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$

This is the Cartesian equation of the line in symmetric form. The denominators (3, 2, -8) are the direction ratios of the line, which are the same as the components of the parallel vector $\vec{b}$. The numerators $(x-5)$, $(y-2)$, $(z+4)$ indicate that the line passes through the point $(5, 2, -4)$.

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Result:

The vector equation of the line is $\mathbf{\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})}$.

The Cartesian equation of the line is $\mathbf{\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}}$.

We need to find the vector equation of a line passing through two given points. The vector equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$, where $\vec{r}$ is the position vector of any point on the line and $\lambda$ is a scalar parameter.

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Given:

Point A with coordinates $(-1, 0, 2)$.

Point B with coordinates $(3, 4, 6)$.

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To Find:

The vector equation of the line passing through points A and B.

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Solution:

Let the position vector of point A be $\vec{a}$ and the position vector of point B be $\vec{b}$.

$\vec{a} = -1\hat{i} + 0\hat{j} + 2\hat{k} = -\hat{i} + 2\hat{k}$

$\vec{b} = 3\hat{i} + 4\hat{j} + 6\hat{k}$

The vector representing the direction of the line is the vector from A to B, $\overrightarrow{AB}$, which is given by $\vec{b} - \vec{a}$.

$\vec{b} - \vec{a} = (3\hat{i} + 4\hat{j} + 6\hat{k}) - (-\hat{i} + 2\hat{k})$

$\vec{b} - \vec{a} = (3 - (-1))\hat{i} + (4 - 0)\hat{j} + (6 - 2)\hat{k}$

$\vec{b} - \vec{a} = (3 + 1)\hat{i} + 4\hat{j} + 4\hat{k}$

$\vec{b} - \vec{a} = 4\hat{i} + 4\hat{j} + 4\hat{k}$

The vector equation of the line passing through $\vec{a}$ and parallel to $\vec{b} - \vec{a}$ is $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.

Substitute the position vector of A and the direction vector:

$\vec{r} = (-\hat{i} + 2\hat{k}) + \lambda(4\hat{i} + 4\hat{j} + 4\hat{k})$

This is one form of the vector equation.

Alternatively, we can use the position vector of B as the starting point:

$\vec{r} = \vec{b} + \mu(\vec{a} - \vec{b})$

Here, $\vec{a} - \vec{b} = -(\vec{b} - \vec{a}) = -(4\hat{i} + 4\hat{j} + 4\hat{k}) = -4\hat{i} - 4\hat{j} - 4\hat{k}$.

$\vec{r} = (3\hat{i} + 4\hat{j} + 6\hat{k}) + \mu(-4\hat{i} - 4\hat{j} - 4\hat{k})$

This is an equivalent vector equation.

Another common form of the two-point vector equation is $\vec{r} = (1-\lambda)\vec{a} + \lambda\vec{b}$.

$\vec{r} = (1-\lambda)(-\hat{i} + 2\hat{k}) + \lambda(3\hat{i} + 4\hat{j} + 6\hat{k})$

$\vec{r} = (-1 + \lambda)\hat{i} + (2 - 2\lambda)\hat{k} + 3\lambda\hat{i} + 4\lambda\hat{j} + 6\lambda\hat{k}$

$\vec{r} = (-1 + \lambda + 3\lambda)\hat{i} + (4\lambda)\hat{j} + (2 - 2\lambda + 6\lambda)\hat{k}$

$\vec{r} = (-1 + 4\lambda)\hat{i} + 4\lambda\hat{j} + (2 + 4\lambda)\hat{k}$

Let's compare the first form with this one:

$\vec{r} = (-\hat{i} + 2\hat{k}) + \lambda'(4\hat{i} + 4\hat{j} + 4\hat{k})$

$\vec{r} = -\hat{i} + 4\lambda'\hat{i} + 4\lambda'\hat{j} + 2\hat{k} + 4\lambda'\hat{k}$

$\vec{r} = (-1 + 4\lambda')\hat{i} + 4\lambda'\hat{j} + (2 + 4\lambda')\hat{k}$

Letting $\lambda' = \lambda$ shows these two forms are equivalent.

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Result:

The vector equation for the line passing through the points (-1, 0, 2) and (3, 4, 6) is $\mathbf{\vec{r} = (-\hat{i} + 2\hat{k}) + \lambda(4\hat{i} + 4\hat{j} + 4\hat{k})}$.

An alternative form is $\mathbf{\vec{r} = (1-\lambda)(-\hat{i} + 2\hat{k}) + \lambda(3\hat{i} + 4\hat{j} + 6\hat{k})}$.

We are given the Cartesian equation of a line in symmetric form and need to find its vector equation. The Cartesian equation in the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ represents a line passing through the point $(x_1, y_1, z_1)$ and parallel to the vector $a\hat{i} + b\hat{j} + c\hat{k}$.

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Given:

The Cartesian equation of the line is $\frac{x + 3}{2} = \frac{y − 5}{4} = \frac{z + 6}{2}$.

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To Find:

The vector equation for the line.

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Solution:

The given Cartesian equation is $\frac{x - (-3)}{2} = \frac{y - 5}{4} = \frac{z - (-6)}{2}$.

Comparing this with the standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, we can identify the coordinates of a point on the line and the direction ratios.

A point on the line is $(x_1, y_1, z_1) = (-3, 5, -6)$.

The direction ratios of the line are $(a, b, c) = (2, 4, 2)$.

The position vector of the point $(-3, 5, -6)$ is $\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}$.

The vector parallel to the line (using the direction ratios as components) is $\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda\vec{b}$

where $\vec{r}$ is the position vector of any point on the line, and $\lambda$ is a scalar parameter.

Substitute the identified $\vec{a}$ and $\vec{b}$ into the vector equation formula:

$\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})$

This is the vector equation of the line.

We can factor out a 2 from the direction vector $(2, 4, 2)$, giving direction ratios proportional to $(1, 2, 1)$. The parallel vector can also be taken as $\hat{i} + 2\hat{j} + \hat{k}$.

In that case, the vector equation would be $\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k})$. This is also a valid vector equation for the same line, just with a different parameter $\mu$ where $\mu = 2\lambda$. The form using the original denominators is standard.

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Result:

The vector equation for the line is $\mathbf{\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})}$.

We are given the vector equations of two lines and need to find the angle between them. The angle between two lines is the angle between their direction vectors. If the lines are parallel to vectors $\vec{b}_1$ and $\vec{b}_2$, the angle $\theta$ between the lines can be found using the dot product formula: $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$. We use the absolute value in the numerator because the angle between lines is typically taken as the acute angle ($0 \leq \theta \leq \frac{\pi}{2}$).

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Given:

Line 1: $\vec{r} = 3\hat{i} + 2\hat{j} − 4\hat{k} + λ(\hat{i} + 2\hat{j} + 2\hat{k})$

Line 2: $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$

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To Find:

The angle between the two lines.

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Solution:

The direction vector of the first line is the vector multiplied by the parameter $\lambda$. Let $\vec{b}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$.

The direction vector of the second line is the vector multiplied by the parameter $\mu$. Let $\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$.

The angle $\theta$ between the lines is the angle between their direction vectors $\vec{b}_1$ and $\vec{b}_2$. We use the formula:

$\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

First, calculate the dot product $\vec{b}_1 \cdot \vec{b}_2$:

$\vec{b}_1 \cdot \vec{b}_2 = (\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (3\hat{i} + 2\hat{j} + 6\hat{k})$

$\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (2)(2) + (2)(6)$

$\vec{b}_1 \cdot \vec{b}_2 = 3 + 4 + 12$

$\vec{b}_1 \cdot \vec{b}_2 = 19$

The absolute value $|\vec{b}_1 \cdot \vec{b}_2| = |19| = 19$.

Next, calculate the magnitudes of the direction vectors $|\vec{b}_1|$ and $|\vec{b}_2|$.

$|\vec{b}_1| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

$|\vec{b}_2| = \sqrt{(3)^2 + (2)^2 + (6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

Now, substitute these values into the formula for $\cos\theta$:

$\cos\theta = \frac{|19|}{(3)(7)}$

$\cos\theta = \frac{19}{21}$

To find the angle $\theta$, take the inverse cosine:

$\theta = \cos^{-1}\left(\frac{19}{21}\right)$

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Result:

The angle between the given pair of lines is $\mathbf{\cos^{-1}\left(\frac{19}{21}\right)}$.

We know that the angle $\theta$ between the lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by the formula:

$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

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The given lines are in the symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.

The direction vector of a line in this form is $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

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For the first line, $\frac{x + 3}{3} = \frac{y − 1}{5} = \frac{z + 3}{4}$, the direction vector is $\vec{b}_1 = 3\hat{i} + 5\hat{j} + 4\hat{k}$.

For the second line, $\frac{x + 1}{1} = \frac{y − 4}{1} = \frac{z − 5}{2}$, the direction vector is $\vec{b}_2 = 1\hat{i} + 1\hat{j} + 2\hat{k}$.

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Now we calculate the dot product of the direction vectors:

$\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (5)(1) + (4)(2)$

$\vec{b}_1 \cdot \vec{b}_2 = 3 + 5 + 8$

$\vec{b}_1 \cdot \vec{b}_2 = 16$

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Next, we calculate the magnitudes of the direction vectors:

$|\vec{b}_1| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50}$

$|\vec{b}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

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Substitute these values into the formula for $\cos \theta$:

$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

$\cos \theta = \frac{|16|}{\sqrt{50} \sqrt{6}}$

$\cos \theta = \frac{16}{\sqrt{50 \times 6}}$

$\cos \theta = \frac{16}{\sqrt{300}}$

$\cos \theta = \frac{16}{\sqrt{100 \times 3}}$

$\cos \theta = \frac{16}{10\sqrt{3}}$

$\cos \theta = \frac{8}{5\sqrt{3}}$

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We can rationalize the denominator:

$\cos \theta = \frac{8}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{5 \times 3} = \frac{8\sqrt{3}}{15}$

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Therefore, the angle $\theta$ between the lines is:

$\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$

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The angle between the pair of lines is $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

Given:

Equation of line l1:

Equation of line l2:

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To Find:

The shortest distance between lines l1 and l2.

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Solution:

The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$.

From equation (1), we have:

From equation (2), we have:

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The formula for the shortest distance ($d$) between two skew lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

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First, calculate $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (2 \hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j})$

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = \hat{i} + 0\hat{j} - \hat{k}$

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Next, calculate $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-1)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3)$

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Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} - 7\hat{k})$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(3) + (0)(-1) + (-1)(-7)$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 3 + 0 + 7$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |3\hat{i} - \hat{j} - 7\hat{k}|$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-7)^2}$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{9 + 1 + 49}$

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Finally, substitute the values into the shortest distance formula:

$d = \left| \frac{10}{\sqrt{59}} \right|$

$d = \frac{10}{\sqrt{59}}$

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The shortest distance between the given lines is $\frac{10}{\sqrt{59}}$.

We can rationalize the denominator if required, but it's not explicitly asked:

$d = \frac{10}{\sqrt{59}} \times \frac{\sqrt{59}}{\sqrt{59}} = \frac{10\sqrt{59}}{59}$

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The shortest distance between the lines is $\frac{10}{\sqrt{59}}$ units or $\frac{10\sqrt{59}}{59}$ units.

Given:

Equation of line l₁:

$\vec{r} = \hat{i} +2\hat{j} − 4\hat{k} + λ (2\hat{i} + 3\hat{j} + 6\hat{k})$

Equation of line l₂:

$\vec{r} = 3\hat{i} +3\hat{j} −5\hat{k} + µ (2\hat{i} + 3\hat{j} +6\hat{k})$

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To Find:

The distance between the lines l₁ and l₂.

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Solution:

The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}$.

From the equation of line l₁, we have:

$\vec{a}_1 = \hat{i} + 2\hat{j} − 4\hat{k}$

and the direction vector is:

$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

From the equation of line l₂, we have:

$\vec{a}_2 = 3\hat{i} + 3\hat{j} − 5\hat{k}$

and the direction vector is:

$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

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Since the direction vectors of the two lines are the same ($2\hat{i} + 3\hat{j} + 6\hat{k}$), the lines are parallel.

The formula for the shortest distance ($d$) between two parallel lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}$ is given by:

$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$

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First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (3\hat{i} + 3\hat{j} − 5\hat{k}) - (\hat{i} + 2\hat{j} − 4\hat{k})$

$\vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k}$

$\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k}$

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Next, calculate the cross product $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$:

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = (2\hat{i} + \hat{j} - \hat{k}) \times (2\hat{i} + 3\hat{j} + 6\hat{k})$

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \hat{i}((1)(6) - (-1)(3)) - \hat{j}((2)(6) - (-1)(2)) + \hat{k}((2)(3) - (1)(2))$

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \hat{i}(6 + 3) - \hat{j}(12 + 2) + \hat{k}(6 - 2)$

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = 9\hat{i} - 14\hat{j} + 4\hat{k}$

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Calculate the magnitude of the cross product $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|$:

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = |9\hat{i} - 14\hat{j} + 4\hat{k}|$

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2}$

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{81 + 196 + 16}$

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{293}$

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Calculate the magnitude of the direction vector $|\vec{b}|$:

$|\vec{b}| = |2\hat{i} + 3\hat{j} + 6\hat{k}|$

$|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2}$

$|\vec{b}| = \sqrt{4 + 9 + 36}$

$|\vec{b}| = \sqrt{49}$

$|\vec{b}| = 7$

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Substitute the values into the shortest distance formula:

$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$

$d = \frac{\sqrt{293}}{7}$

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The distance between the parallel lines is $\frac{\sqrt{293}}{7}$ units.

Given:

Direction cosines of three lines:

Line 1: $(l_1, m_1, n_1) = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$

Line 2: $(l_2, m_2, n_2) = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$

Line 3: $(l_3, m_3, n_3) = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$

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To Show:

The three lines are mutually perpendicular.

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Solution:

Two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if and only if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

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Consider Line 1 and Line 2:

We need to check if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

$l_1 l_2 + m_1 m_2 + n_1 n_2 = \left(\frac{12}{13}\right)\left(\frac{4}{13}\right) + \left(\frac{-3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$

$= \frac{48}{169} - \frac{36}{169} - \frac{12}{169}$

$= \frac{48 - 36 - 12}{169}$

$= \frac{0}{169} = 0$

Since $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$, Line 1 is perpendicular to Line 2.

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Consider Line 1 and Line 3:

We need to check if $l_1 l_3 + m_1 m_3 + n_1 n_3 = 0$.

$l_1 l_3 + m_1 m_3 + n_1 n_3 = \left(\frac{12}{13}\right)\left(\frac{3}{13}\right) + \left(\frac{-3}{13}\right)\left(\frac{-4}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{12}{13}\right)$

$= \frac{36}{169} + \frac{12}{169} - \frac{48}{169}$

$= \frac{36 + 12 - 48}{169}$

$= \frac{0}{169} = 0$

Since $l_1 l_3 + m_1 m_3 + n_1 n_3 = 0$, Line 1 is perpendicular to Line 3.

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Consider Line 2 and Line 3:

We need to check if $l_2 l_3 + m_2 m_3 + n_2 n_3 = 0$.

$l_2 l_3 + m_2 m_3 + n_2 n_3 = \left(\frac{4}{13}\right)\left(\frac{3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right) + \left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$

$= \frac{12}{169} - \frac{48}{169} + \frac{36}{169}$

$= \frac{12 - 48 + 36}{169}$

$= \frac{0}{169} = 0$

Since $l_2 l_3 + m_2 m_3 + n_2 n_3 = 0$, Line 2 is perpendicular to Line 3.

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Since all three pairs of lines are perpendicular, the three lines are mutually perpendicular.

Given:

Line 1 passes through points A(1, – 1, 2) and B(3, 4, – 2).

Line 2 passes through points C(0, 3, 2) and D(3, 5, 6).

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To Show:

Line 1 is perpendicular to Line 2.

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Solution:

The direction ratios (DRs) of a line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

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Direction ratios of Line 1 (through A(1, – 1, 2) and B(3, 4, – 2)):

$a_1 = 3 - 1 = 2$

$b_1 = 4 - (-1) = 4 + 1 = 5$

$c_1 = -2 - 2 = -4$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (2, 5, -4)$.

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Direction ratios of Line 2 (through C(0, 3, 2) and D(3, 5, 6)):

$a_2 = 3 - 0 = 3$

$b_2 = 5 - 3 = 2$

$c_2 = 6 - 2 = 4$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (3, 2, 4)$.

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Let's calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$ for the given lines:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(3) + (5)(2) + (-4)(4)$

$= 6 + 10 - 16$

$= 16 - 16$

$= 0$

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Since the sum of the products of their corresponding direction ratios is zero, i.e., $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, the two lines are perpendicular.

Hence, the line through points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through points (0, 3, 2) and (3, 5, 6).

Given:

Line 1 passes through points A(4, 7, 8) and B(2, 3, 4).

Line 2 passes through points C(– 1, – 2, 1) and D(1, 2, 5).

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To Show:

Line 1 is parallel to Line 2.

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Solution:

The direction ratios (DRs) of a line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

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Direction ratios of Line 1 (through A(4, 7, 8) and B(2, 3, 4)):

$a_1 = 2 - 4 = -2$

$b_1 = 3 - 7 = -4$

$c_1 = 4 - 8 = -4$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (-2, -4, -4)$.

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Direction ratios of Line 2 (through C(– 1, – 2, 1) and D(1, 2, 5)):

$a_2 = 1 - (-1) = 1 + 1 = 2$

$b_2 = 2 - (-2) = 2 + 2 = 4$

$c_2 = 5 - 1 = 4$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (2, 4, 4)$.

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel if their direction ratios are proportional, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Let's check the ratios for the given lines:

$\frac{a_1}{a_2} = \frac{-2}{2} = -1$

$\frac{b_1}{b_2} = \frac{-4}{4} = -1$

$\frac{c_1}{c_2} = \frac{-4}{4} = -1$

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Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -1$, the direction ratios are proportional.

Therefore, the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

Given:

The line passes through the point A(1, 2, 3).

The line is parallel to the vector $\vec{b} = 3\hat{i} + 2\hat{j} −2\hat{k}$.

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To Find:

The equation of the line in vector and Cartesian form.

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Solution (Vector Form):

The position vector of the given point A(1, 2, 3) is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The line is parallel to the vector $\vec{b} = 3\hat{i} + 2\hat{j} −2\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

where $\lambda$ is a scalar parameter.

Substituting the given values of $\vec{a}$ and $\vec{b}$, we get:

This is the vector equation of the line.

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Solution (Cartesian Form):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

From the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} −2\hat{k})$

$x\hat{i} + y\hat{j} + z\hat{k} = (1 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 - 2\lambda)\hat{k}$

Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$:

$x = 1 + 3\lambda \implies x - 1 = 3\lambda \implies \lambda = \frac{x-1}{3}$

$y = 2 + 2\lambda \implies y - 2 = 2\lambda \implies \lambda = \frac{y-2}{2}$

$z = 3 - 2\lambda \implies z - 3 = -2\lambda \implies \lambda = \frac{z-3}{-2}$

Since all these expressions are equal to $\lambda$, we have the Cartesian equation of the line:

Alternatively, this can be written as:

This is the Cartesian equation of the line passing through (1, 2, 3) with direction ratios (3, 2, -2).

Given:

The line passes through the point with position vector $\vec{a} = 2\hat{i} − \hat{j} + 4\hat{k}$.

The line is in the direction of the vector $\vec{b} = \hat{i} + 2\hat{j} − \hat{k}$.

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To Find:

The equation of the line in vector and Cartesian form.

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Solution (Vector Form):

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

where $\lambda$ is a scalar parameter.

Substituting the given values of $\vec{a}$ and $\vec{b}$, we get:

This is the vector equation of the line.

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Solution (Cartesian Form):

Let the coordinates of a point on the line be $(x, y, z)$. Its position vector is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

The line passes through the point $(2, -1, 4)$, so $(x_1, y_1, z_1) = (2, -1, 4)$.

The direction vector is $\vec{b} = \hat{i} + 2\hat{j} − \hat{k}$, so the direction ratios are $(a, b, c) = (1, 2, -1)$.

The Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Substituting the values, we get:

$\frac{x - 2}{1} = \frac{y - (-1)}{2} = \frac{z - 4}{-1}$

This is the Cartesian equation of the line.

Given:

The line passes through the point P(– 2, 4, – 5).

The line is parallel to the line given by $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$.

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To Find:

The Cartesian equation of the required line.

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Solution:

The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is given by:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

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The required line passes through the point (– 2, 4, – 5).

So, $(x_1, y_1, z_1) = (-2, 4, -5)$.

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The required line is parallel to the given line $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$.

The given line is in the standard Cartesian form, $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$, where $(a, b, c)$ are the direction ratios.

The direction ratios of the given line are (3, 5, 6).

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Since parallel lines have proportional direction ratios, the direction ratios of the required line can also be taken as (3, 5, 6).

So, $(a, b, c) = (3, 5, 6)$.

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Substitute the values of $(x_1, y_1, z_1)$ and $(a, b, c)$ into the Cartesian equation formula:

$\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$

Simplifying the equation:

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Thus, the Cartesian equation of the line passing through (– 2, 4, – 5) and parallel to the given line is $\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$.

Given:

The Cartesian equation of the line is $\frac{x − 5}{3} = \frac{ y + 4}{7} = \frac{z − 6}{2}$.

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To Find:

The vector form of the equation of the line.

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Solution:

The general Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is given by:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

Comparing the given equation $\frac{x − 5}{3} = \frac{ y + 4}{7} = \frac{z − 6}{2}$ with the general form, we can identify:

The point $(x_1, y_1, z_1)$ on the line is $(5, -4, 6)$.

The direction ratios $(a, b, c)$ of the line are $(3, 7, 2)$.

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The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

where $\lambda$ is a scalar parameter.

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The position vector of the point (5, -4, 6) is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$.

The direction vector of the line with direction ratios (3, 7, 2) is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

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Substituting these into the vector equation formula, we get:

This is the required vector form of the equation of the line.

Given:

The line passes through two points:

Point 1: Origin O(0, 0, 0)

Point 2: A(5, – 2, 3)

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To Find:

The vector and Cartesian equations of the line.

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Solution (Vector Form):

Let $\vec{a}$ be the position vector of the first point (Origin) and $\vec{b}$ be the position vector of the second point (5, – 2, 3).

$\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$

$\vec{b} = 5\hat{i} - 2\hat{j} + 3\hat{k}$

The vector equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})$

where $\lambda$ is a scalar parameter.

First, calculate the vector $(\vec{b} - \vec{a})$:

$\vec{b} - \vec{a} = (5\hat{i} - 2\hat{j} + 3\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k})$

$\vec{b} - \vec{a} = 5\hat{i} - 2\hat{j} + 3\hat{k}$

Now, substitute $\vec{a}$ and $(\vec{b} - \vec{a})$ into the vector equation formula:

$\vec{r} = \vec{0} + \lambda (5\hat{i} - 2\hat{j} + 3\hat{k})$

This is the vector equation of the line passing through the origin and (5, – 2, 3).

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Solution (Cartesian Form):

Let the two points be $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (5, -2, 3)$.

The Cartesian equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of the two points:

$\frac{x - 0}{5 - 0} = \frac{y - 0}{-2 - 0} = \frac{z - 0}{3 - 0}$

Simplifying the equation:

This is the Cartesian equation of the line passing through the origin and (5, – 2, 3).

Given:

The line passes through two points:

Point 1: A(3, – 2, – 5)

Point 2: B(3, – 2, 6)

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To Find:

The vector and Cartesian equations of the line.

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Solution (Vector Form):

Let $\vec{a}$ be the position vector of the first point A and $\vec{b}$ be the position vector of the second point B.

$\vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k}$

$\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}$

The vector equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})$

where $\lambda$ is a scalar parameter.

First, calculate the difference vector $(\vec{b} - \vec{a})$:

$\vec{b} - \vec{a} = (3\hat{i} - 2\hat{j} + 6\hat{k}) - (3\hat{i} - 2\hat{j} - 5\hat{k})$

$\vec{b} - \vec{a} = (3-3)\hat{i} + (-2 - (-2))\hat{j} + (6 - (-5))\hat{k}$

$\vec{b} - \vec{a} = 0\hat{i} + 0\hat{j} + 11\hat{k}$

Now, substitute $\vec{a}$ and $(\vec{b} - \vec{a})$ into the vector equation formula:

$\vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda (0\hat{i} + 0\hat{j} + 11\hat{k})$

This is the vector equation of the line.

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Solution (Cartesian Form):

Let the two points be $(x_1, y_1, z_1) = (3, -2, -5)$ and $(x_2, y_2, z_2) = (3, -2, 6)$.

The Cartesian equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of the two points:

$\frac{x - 3}{3 - 3} = \frac{y - (-2)}{-2 - (-2)} = \frac{z - (-5)}{6 - (-5)}$

$\frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}$

When the denominator of the Cartesian equation is zero, it implies that the numerator must also be zero (unless the denominator is also zero, which indicates indeterminacy, but here it means the coordinate is constant). This indicates that the line is parallel to the coordinate axis corresponding to the zero denominator.

From $\frac{x - 3}{0}$, we get $x - 3 = 0$, which implies $x = 3$.

From $\frac{y + 2}{0}$, we get $y + 2 = 0$, which implies $y = -2$.

The third part is $\frac{z + 5}{11}$.

So, the Cartesian equations of the line are $x = 3$, $y = -2$, and $\frac{z + 5}{11}$. This means the line lies in the plane $x=3$ and the plane $y=-2$, and its $z$-coordinate varies along the line. The equation $\frac{z+5}{11}$ describes how the $z$-coordinate changes parametrically, proportional to the other parts (which are constant 0 in this case, effectively meaning $z$ varies along the line where $x$ and $y$ are fixed).

The Cartesian equations can be written as:

along with the ratio for $z$, or simply stating $x=3, y=-2$ defines a line parallel to the z-axis through the point $(3, -2, 0)$. The points $(3, -2, -5)$ and $(3, -2, 6)$ both lie on this line.

We know that the angle $\theta$ between the lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by the formula:

$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

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(i) For the first pair of lines:

Line 1: $\vec{r} =2\hat{i} − 5\hat{j} + \hat{k} + λ (3\hat{i} + 2\hat{j} + 6\hat{k})$

The direction vector is $\vec{b}_1 = 3\hat{i} + 2\hat{j} + 6\hat{k}$.

Line 2: $\vec{r} = 7\hat{i} − 6\hat{k} + µ (\hat{i} + 2\hat{j} + 2\hat{k})$

The direction vector is $\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$.

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Calculate the dot product $\vec{b}_1 \cdot \vec{b}_2$:

$\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19$

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Calculate the magnitudes $|\vec{b}_1|$ and $|\vec{b}_2|$:

$|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

$|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|19|}{(7)(3)} = \frac{19}{21}$

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The angle $\theta$ between the first pair of lines is $\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

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(ii) For the second pair of lines:

Line 1: $\vec{r} = 3\hat{i} + \hat{j} − 2\hat{k} + λ (\hat{i} − \hat{j} − 2\hat{k})$

The direction vector is $\vec{b}_1 = \hat{i} − \hat{j} − 2\hat{k}$.

Line 2: $\vec{r} = 2\hat{i} − \hat{j} − 56\hat{k} + µ (3\hat{i} − 5\hat{j} − 4\hat{k})$

The direction vector is $\vec{b}_2 = 3\hat{i} − 5\hat{j} − 4\hat{k}$.

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Calculate the dot product $\vec{b}_1 \cdot \vec{b}_2$:

$\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$

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Calculate the magnitudes $|\vec{b}_1|$ and $|\vec{b}_2|$:

$|\vec{b}_1| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

$|\vec{b}_2| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|16|}{(\sqrt{6})(5\sqrt{2})}$

$\cos \theta = \frac{16}{5\sqrt{12}} = \frac{16}{5 \times 2\sqrt{3}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$

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Rationalize the denominator:

$\cos \theta = \frac{8}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{5 \times 3} = \frac{8\sqrt{3}}{15}$

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The angle $\theta$ between the second pair of lines is $\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

We know that the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:

$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$

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(i) For the first pair of lines:

Line 1: $\frac{x − 2}{2} = \frac{y − 1}{5} = \frac{z + 3}{−3}$

The direction ratios are $(a_1, b_1, c_1) = (2, 5, -3)$.

Line 2: $\frac{x + 2}{−1} = \frac{y − 4}{8} = \frac{z − 5}{4}$

The direction ratios are $(a_2, b_2, c_2) = (-1, 8, 4)$.

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Calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(-1) + (5)(8) + (-3)(4)$

$= -2 + 40 - 12 = 26$

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Calculate the magnitudes of the direction vectors:

$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$

$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|26|}{\sqrt{38} \times 9} = \frac{26}{9\sqrt{38}}$

The angle $\theta$ is $\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)$.

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(ii) For the second pair of lines:

Line 1: $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$

The direction ratios are $(a_1, b_1, c_1) = (2, 2, 1)$.

Line 2: $\frac{x − 5}{4} = \frac{y − 2}{1} = \frac{z − 3}{8}$

The direction ratios are $(a_2, b_2, c_2) = (4, 1, 8)$.

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Calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(4) + (2)(1) + (1)(8)$

$= 8 + 2 + 8 = 18$

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Calculate the magnitudes of the direction vectors:

$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|18|}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$

The angle $\theta$ is $\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

Given:

Line 1: $\frac{1 − x}{3} = \frac{7y − 14}{2p} = \frac{z − 3}{2}$

Line 2: $\frac{7 − 7x}{3p} = \frac{y − 5}{1} = \frac{6 − z}{5}$

The lines are at right angles.

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To Find:

The value of p.

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Solution:

First, we need to write the equations of the lines in the standard Cartesian form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ to find their direction ratios.

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For Line 1: $\frac{1 − x}{3} = \frac{7y − 14}{2p} = \frac{z − 3}{2}$

Rewrite the terms:

$\frac{-(x - 1)}{3} = \frac{7(y - 2)}{2p} = \frac{z - 3}{2}$

$\frac{x - 1}{-3} = \frac{y - 2}{\frac{2p}{7}} = \frac{z - 3}{2}$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = \left(-3, \frac{2p}{7}, 2\right)$.

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For Line 2: $\frac{7 − 7x}{3p} = \frac{y − 5}{1} = \frac{6 − z}{5}$

Rewrite the terms:

$\frac{-7(x - 1)}{3p} = \frac{y - 5}{1} = \frac{-(z - 6)}{5}$

$\frac{x - 1}{\frac{-3p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{-5}$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = \left(\frac{-3p}{7}, 1, -5\right)$.

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are at right angles if the sum of the products of their corresponding direction ratios is zero:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

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Substitute the direction ratios we found:

$(-3)\left(\frac{-3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0$

$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

Combine the terms with $p$:

$\frac{9p + 2p}{7} - 10 = 0$

$\frac{11p}{7} - 10 = 0$

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Solve for $p$:

$\frac{11p}{7} = 10$

$11p = 10 \times 7$

$11p = 70$

$p = \frac{70}{11}$

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Thus, the value of p for which the given lines are at right angles is $\frac{70}{11}$.

Given:

Line 1: $\frac{x − 5}{7} = \frac{y + 2}{−5} = \frac{z}{1}$

Line 2: $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

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To Show:

The lines are perpendicular to each other.

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Solution:

The Cartesian equation of a line in the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ has direction ratios $(a, b, c)$.

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For Line 1: $\frac{x − 5}{7} = \frac{y + 2}{−5} = \frac{z}{1}$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (7, -5, 1)$.

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For Line 2: $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (1, 2, 3)$.

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if the sum of the products of their corresponding direction ratios is zero, i.e., $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

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Let's calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$ for the given lines:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (7)(1) + (-5)(2) + (1)(3)$

$= 7 - 10 + 3$

$= 10 - 10$

$= 0$

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Since $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, the two lines are perpendicular to each other.

Given:

Equation of Line 1:

$\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + λ (\hat{i} − \hat{j} + \hat{k})$

Equation of Line 2:

$\vec{r} = 2\hat{i} − \hat{j} - \hat{k} + µ (2\hat{i} + \hat{j} + 2\hat{k})$

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To Find:

The shortest distance between the two lines.

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Solution:

The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$.

From the equation of Line 1, we have:

$\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$

$\vec{b}_1 = \hat{i} − \hat{j} + \hat{k}$

From the equation of Line 2, we have:

$\vec{a}_2 = 2\hat{i} − \hat{j} - \hat{k}$

$\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$

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Since $\vec{b}_1$ is not proportional to $\vec{b}_2$ (i.e., $(1, -1, 1)$ is not proportional to $(2, 1, 2)$), the lines are skew lines.

The formula for the shortest distance ($d$) between two skew lines is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

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First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (2\hat{i} − \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k})$

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}$

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Next, calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-1)(2) - (1)(1)) - \hat{j}((1)(2) - (1)(2)) + \hat{k}((1)(1) - (-1)(2))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2)$

$\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 0\hat{j} + 3\hat{k}$

$\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 3\hat{k}$

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Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k})$

$= (1)(-3) + (-3)(0) + (-2)(3)$

$= -3 + 0 - 6 = -9$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |-3\hat{i} + 3\hat{k}|$

$= \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18}$

$= \sqrt{9 \times 2} = 3\sqrt{2}$

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Substitute the values into the shortest distance formula:

$d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{|-9|}{|3\sqrt{2}|} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}$

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To rationalize the denominator:

$d = \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

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The shortest distance between the lines is $\frac{3}{\sqrt{2}}$ units or $\frac{3\sqrt{2}}{2}$ units.

Given:

Line 1: $\frac{x + 1}{7} = \frac{y + 1}{−6} = \frac{z + 1}{1}$

Line 2: $\frac{x − 3}{1} = \frac{y − 5}{−2} = \frac{z − 7}{1}$

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To Find:

The shortest distance between the two lines.

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Solution:

The given lines are in the Cartesian form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. We can convert them into vector form $\vec{r} = \vec{a} + \lambda \vec{b}$.

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For Line 1: $\frac{x - (-1)}{7} = \frac{y - (-1)}{-6} = \frac{z - (-1)}{1}$

A point on Line 1 is $(-1, -1, -1)$, so $\vec{a}_1 = -\hat{i} - \hat{j} - \hat{k}$.

The direction ratios are $(7, -6, 1)$, so the direction vector is $\vec{b}_1 = 7\hat{i} - 6\hat{j} + \hat{k}$.

Vector equation of Line 1: $\vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda (7\hat{i} - 6\hat{j} + \hat{k})$

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For Line 2: $\frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1}$

A point on Line 2 is $(3, 5, 7)$, so $\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}$.

The direction ratios are $(1, -2, 1)$, so the direction vector is $\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$.

Vector equation of Line 2: $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu (\hat{i} - 2\hat{j} + \hat{k})$

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Since $\vec{b}_1$ is not proportional to $\vec{b}_2$ (as $\frac{7}{1} \neq \frac{-6}{-2}$), the lines are skew lines.

The formula for the shortest distance ($d$) between two skew lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

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First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (3\hat{i} + 5\hat{j} + 7\hat{k}) - (-\hat{i} - \hat{j} - \hat{k})$

$\vec{a}_2 - \vec{a}_1 = (3 - (-1))\hat{i} + (5 - (-1))\hat{j} + (7 - (-1))\hat{k}$

$\vec{a}_2 - \vec{a}_1 = 4\hat{i} + 6\hat{j} + 8\hat{k}$

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Next, calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = (7\hat{i} - 6\hat{j} + \hat{k}) \times (\hat{i} - 2\hat{j} + \hat{k})$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-6)(1) - (1)(-2)) - \hat{j}((7)(1) - (1)(1)) + \hat{k}((7)(-2) - (-6)(1))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-6 + 2) - \hat{j}(7 - 1) + \hat{k}(-14 + 6)$

$\vec{b}_1 \times \vec{b}_2 = -4\hat{i} - 6\hat{j} - 8\hat{k}$

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Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (4\hat{i} + 6\hat{j} + 8\hat{k}) \cdot (-4\hat{i} - 6\hat{j} - 8\hat{k})$

$= (4)(-4) + (6)(-6) + (8)(-8)$

$= -16 - 36 - 64 = -116$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |-4\hat{i} - 6\hat{j} - 8\hat{k}|$

$= \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116}$

$= \sqrt{4 \times 29} = 2\sqrt{29}$

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Substitute the values into the shortest distance formula:

$d = \left| \frac{-116}{2\sqrt{29}} \right| = \frac{|-116|}{|2\sqrt{29}|} = \frac{116}{2\sqrt{29}}$

$d = \frac{58}{\sqrt{29}}$

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To rationalize the denominator:

$d = \frac{58}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{58\sqrt{29}}{29}$

$d = 2\sqrt{29}$

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The shortest distance between the lines is $2\sqrt{29}$ units.

Given:

Equation of Line 1:

$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + λ (\hat{i} − 3\hat{j} + 2\hat{k})$

Equation of Line 2:

$\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + µ (2\hat{i} + 3\hat{j} + \hat{k})$

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To Find:

The shortest distance between the two lines.

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Solution:

The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$.

From the equation of Line 1, we have:

$\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{b}_1 = \hat{i} − 3\hat{j} + 2\hat{k}$

From the equation of Line 2, we have:

$\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$

$\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$

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Since $\vec{b}_1$ is not parallel to $\vec{b}_2$ (as $(1, -3, 2)$ is not proportional to $(2, 3, 1)$), the lines are skew lines.

The formula for the shortest distance ($d$) between two skew lines is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

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First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}$

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Next, calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = (\hat{i} − 3\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} + \hat{k})$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6)$

$\vec{b}_1 \times \vec{b}_2 = -9\hat{i} + 3\hat{j} + 9\hat{k}$

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Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k})$

$= (3)(-9) + (3)(3) + (3)(9)$

$= -27 + 9 + 27 = 9$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |-9\hat{i} + 3\hat{j} + 9\hat{k}|$

$= \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171}$

To simplify $\sqrt{171}$: $\sqrt{171} = \sqrt{9 \times 19} = \sqrt{9} \times \sqrt{19} = 3\sqrt{19}$.

$|\vec{b}_1 \times \vec{b}_2| = 3\sqrt{19}$

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Substitute the values into the shortest distance formula:

$d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{|9|}{|3\sqrt{19}|} = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$

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To rationalize the denominator:

$d = \frac{3}{\sqrt{19}} \times \frac{\sqrt{19}}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$

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The shortest distance between the lines is $\frac{3}{\sqrt{19}}$ units or $\frac{3\sqrt{19}}{19}$ units.

Given:

Equation of Line 1:

$\vec{r} = (1−t)\hat{i} + (t−2)\hat{j} + (3−2t)\hat{k}$

Equation of Line 2:

$\vec{r} = (s+1)\hat{i} + (2s−1)\hat{j} − (2s+1)\hat{k}$

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To Find:

The shortest distance between the two lines.

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Solution:

First, we need to rewrite the given vector equations in the standard form $\vec{r} = \vec{a} + \lambda \vec{b}$.

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For Line 1: $\vec{r} = (1−t)\hat{i} + (t−2)\hat{j} + (3−2t)\hat{k}$

$\vec{r} = (1)\hat{i} - t\hat{i} + t\hat{j} - 2\hat{j} + 3\hat{k} - 2t\hat{k}$

Group terms independent of $t$ and terms dependent on $t$:

$\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})$

This is in the form $\vec{r} = \vec{a}_1 + t \vec{b}_1$, where:

$\vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k}$

$\vec{b}_1 = -\hat{i} + \hat{j} - 2\hat{k}$

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For Line 2: $\vec{r} = (s+1)\hat{i} + (2s−1)\hat{j} − (2s+1)\hat{k}$

$\vec{r} = s\hat{i} + \hat{i} + 2s\hat{j} - \hat{j} - 2s\hat{k} - \hat{k}$

Group terms independent of $s$ and terms dependent on $s$:

$\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$

This is in the form $\vec{r} = \vec{a}_2 + s \vec{b}_2$, where:

$\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$

$\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$

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Since $\vec{b}_1$ is not parallel to $\vec{b}_2$ (i.e., $(-1, 1, -2)$ is not proportional to $(1, 2, -2)$), the lines are skew lines.

The formula for the shortest distance ($d$) between two skew lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

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First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$

$\vec{a}_2 - \vec{a}_1 = (1-1)\hat{i} + (-1 - (-2))\hat{j} + (-1 - 3)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = 0\hat{i} + \hat{j} - 4\hat{k}$

$\vec{a}_2 - \vec{a}_1 = \hat{j} - 4\hat{k}$

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Next, calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = (-\hat{i} + \hat{j} - 2\hat{k}) \times (\hat{i} + 2\hat{j} - 2\hat{k})$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((1)(-2) - (-2)(2)) - \hat{j}((-1)(-2) - (-2)(1)) + \hat{k}((-1)(2) - (1)(1))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1)$

$\vec{b}_1 \times \vec{b}_2 = 2\hat{i} - 4\hat{j} - 3\hat{k}$

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Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k})$

$= (0)(2) + (1)(-4) + (-4)(-3)$

$= 0 - 4 + 12 = 8$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |2\hat{i} - 4\hat{j} - 3\hat{k}|$

$= \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$

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Substitute the values into the shortest distance formula:

$d = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}}$

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To rationalize the denominator:

$d = \frac{8}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{8\sqrt{29}}{29}$

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The shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units or $\frac{8\sqrt{29}}{29}$ units.

Given:

Distance of the plane from the origin, $d = \frac{6}{\sqrt{29}}$.

Normal vector from the origin to the plane, $\vec{N} = 2\hat{i} − 3\hat{j} + 4\hat{k}$.

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To Find:

The vector equation and the Cartesian equation of the plane.

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Solution (Vector Form):

The vector equation of a plane at a distance $d$ from the origin and having a normal vector $\vec{N}$ is given by:

$\vec{r} \cdot \hat{n} = d$

where $\hat{n}$ is the unit normal vector to the plane.

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First, find the unit normal vector $\hat{n}$ from the given normal vector $\vec{N}$.

$|\vec{N}| = |2\hat{i} − 3\hat{j} + 4\hat{k}|$

$|\vec{N}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

$\hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2\hat{i} − 3\hat{j} + 4\hat{k}}{\sqrt{29}}$

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Now, substitute $\hat{n}$ and $d$ into the vector equation formula:

$\vec{r} \cdot \left(\frac{2\hat{i} − 3\hat{j} + 4\hat{k}}{\sqrt{29}}\right) = \frac{6}{\sqrt{29}}$

Multiply both sides by $\sqrt{29}$:

This is the required vector equation of the plane.

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Solution (Cartesian Form):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} − 3\hat{j} + 4\hat{k}) = 6$

Perform the dot product:

$(x)(2) + (y)(-3) + (z)(4) = 6$

This is the required Cartesian equation of the plane.

Given:

The equation of the plane is $\vec{r} \;.\; (6\hat{i} − 3\hat{j} − 2\hat{k}) + 1 = 0$.

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To Find:

The direction cosines of the unit vector perpendicular to the plane from the origin.

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Solution:

The given equation of the plane is $\vec{r} \;.\; (6\hat{i} − 3\hat{j} − 2\hat{k}) + 1 = 0$.

We can rewrite this equation as:

$\vec{r} \;.\; (6\hat{i} − 3\hat{j} − 2\hat{k}) = -1$

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The standard form of the vector equation of a plane in normal form is $\vec{r} \cdot \hat{n} = d$, where $d$ is the distance of the plane from the origin and $\hat{n}$ is the unit normal vector from the origin to the plane.

In the equation $\vec{r} \cdot \vec{N} = D$, if $D$ is negative, we multiply by -1 to make it positive, which changes the direction of the normal vector $\vec{N}$ to point away from the origin.

Multiplying the equation by -1, we get:

$\vec{r} \;.\; (-6\hat{i} + 3\hat{j} + 2\hat{k}) = 1$

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Now, the normal vector from the origin is $\vec{N}' = -6\hat{i} + 3\hat{j} + 2\hat{k}$.

The distance from the origin is $d = 1$.

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To find the unit normal vector $\hat{n}$, we divide the normal vector $\vec{N}'$ by its magnitude:

$|\vec{N}'| = |-6\hat{i} + 3\hat{j} + 2\hat{k}| = \sqrt{(-6)^2 + (3)^2 + (2)^2}$

$|\vec{N}'| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$

The unit normal vector is:

$\hat{n} = \frac{\vec{N}'}{|\vec{N}'|} = \frac{-6\hat{i} + 3\hat{j} + 2\hat{k}}{7}$

$\hat{n} = -\frac{6}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{2}{7}\hat{k}$

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The direction cosines of the unit normal vector $\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$ are $(l, m, n)$.

Comparing the unit normal vector with $l\hat{i} + m\hat{j} + n\hat{k}$, we find the direction cosines:

$l = -\frac{6}{7}$

$m = \frac{3}{7}$

$n = \frac{2}{7}$

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The direction cosines of the unit vector perpendicular to the plane from the origin are $\left(-\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\right)$.

Given:

The equation of the plane is $2x - 3y + 4z - 6 = 0$.

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To Find:

The distance of the plane from the origin.

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Solution:

The equation of the plane is in the general form $Ax + By + Cz + D = 0$, where:

$A = 2$

$B = -3$

$C = 4$

$D = -6$

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The distance of a plane $Ax + By + Cz + D = 0$ from the origin $(0, 0, 0)$ is given by the formula:

Distance $= \frac{|A(0) + B(0) + C(0) + D|}{\sqrt{A^2 + B^2 + C^2}}$

Distance $= \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$

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Substitute the values of A, B, C, and D into the formula:

Distance $= \frac{|-6|}{\sqrt{2^2 + (-3)^2 + 4^2}}$

Distance $= \frac{6}{\sqrt{4 + 9 + 16}}$

Distance $= \frac{6}{\sqrt{29}}$

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The distance of the plane $2x - 3y + 4z - 6 = 0$ from the origin is $\frac{6}{\sqrt{29}}$ units.

Given:

The equation of the plane is $2x - 3y + 4z - 6 = 0$.

The point from which the perpendicular is drawn is the Origin (0, 0, 0).

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To Find:

The coordinates of the foot of the perpendicular from the origin to the plane.

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Solution:

The equation of the plane is $2x - 3y + 4z - 6 = 0$. Comparing this with the general form $Ax + By + Cz + D = 0$, we get the normal vector to the plane $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

So, the normal vector to the given plane is $\vec{n} = 2\hat{i} - 3\hat{j} + 4\hat{k}$.

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The line drawn from the origin perpendicular to the plane will be parallel to the normal vector of the plane.

The direction ratios of this line are the components of the normal vector, which are (2, -3, 4).

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The equation of a line passing through the origin $(0, 0, 0)$ with direction ratios $(a, b, c) = (2, -3, 4)$ is given by:

$\frac{x - 0}{2} = \frac{y - 0}{-3} = \frac{z - 0}{4}$

$\frac{x}{2} = \frac{y}{-3} = \frac{z}{4}$

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Let the coordinates of the foot of the perpendicular (the point where the line intersects the plane) be $(x, y, z)$. This point lies on the line.

Let $\frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = \lambda$ (where $\lambda$ is a scalar parameter).

From this, we get the coordinates of any point on the line in terms of $\lambda$:

$x = 2\lambda$

$y = -3\lambda$

$z = 4\lambda$

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Since the foot of the perpendicular $(x, y, z)$ also lies on the plane $2x - 3y + 4z - 6 = 0$, these coordinates must satisfy the plane's equation.

Substitute the values of x, y, and z from the parametric form into the plane equation:

$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) - 6 = 0$

$4\lambda + 9\lambda + 16\lambda - 6 = 0$

$29\lambda - 6 = 0$

$29\lambda = 6$

$\lambda = \frac{6}{29}$

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Now, substitute the value of $\lambda = \frac{6}{29}$ back into the parametric coordinates of the point on the line to find the coordinates of the foot of the perpendicular:

$x = 2\left(\frac{6}{29}\right) = \frac{12}{29}$

$y = -3\left(\frac{6}{29}\right) = -\frac{18}{29}$

$z = 4\left(\frac{6}{29}\right) = \frac{24}{29}$

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The coordinates of the foot of the perpendicular drawn from the origin to the plane are $\left(\frac{12}{29}, -\frac{18}{29}, \frac{24}{29}\right)$.

Given:

The plane passes through the point P(5, 2, – 4).

The plane is perpendicular to a line with direction ratios (2, 3, – 1).

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To Find:

The vector equation and the Cartesian equation of the plane.

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Solution:

The position vector of the point P(5, 2, – 4) on the plane is $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$.

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A line perpendicular to the plane has its direction ratios proportional to the direction ratios of the normal vector to the plane.

Since the line is perpendicular to the plane and has direction ratios (2, 3, – 1), these are the direction ratios of the normal vector to the plane.

The normal vector to the plane is $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$.

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The vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a vector $\vec{n}$ is given by:

$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$

or equivalently,

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

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Substitute the values of $\vec{a}$ and $\vec{n}$ into the equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

First, calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (5\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k})$

$\vec{a} \cdot \vec{n} = (5)(2) + (2)(3) + (-4)(-1)$

$\vec{a} \cdot \vec{n} = 10 + 6 + 4 = 20$

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So, the vector equation of the plane is:

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To find the Cartesian equation, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20$

Performing the dot product, we get:

$x(2) + y(3) + z(-1) = 20$

This is the required Cartesian equation of the plane.

Given:

The plane passes through three points:

R (2, 5, – 3)

S (– 2, – 3, 5)

T (5, 3,– 3)

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To Find:

The vector equation and the Cartesian equation of the plane.

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Solution (Vector Form - Parametric):

Let the position vectors of the points R, S, and T be $\vec{a}, \vec{b}, \vec{c}$ respectively.

$\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}$

$\vec{b} = -2\hat{i} - 3\hat{j} + 5\hat{k}$

$\vec{c} = 5\hat{i} + 3\hat{j} - 3\hat{k}$

The vector equation of a plane passing through three non-collinear points with position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by:

$\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) + \mu (\vec{c} - \vec{a})$

where $\lambda$ and $\mu$ are scalar parameters.

Calculate the vectors $(\vec{b} - \vec{a})$ and $(\vec{c} - \vec{a})$:

$\vec{b} - \vec{a} = (-2\hat{i} - 3\hat{j} + 5\hat{k}) - (2\hat{i} + 5\hat{j} - 3\hat{k}) = (-2-2)\hat{i} + (-3-5)\hat{j} + (5-(-3))\hat{k} = -4\hat{i} - 8\hat{j} + 8\hat{k}$

$\vec{c} - \vec{a} = (5\hat{i} + 3\hat{j} - 3\hat{k}) - (2\hat{i} + 5\hat{j} - 3\hat{k}) = (5-2)\hat{i} + (3-5)\hat{j} + (-3-(-3))\hat{k} = 3\hat{i} - 2\hat{j} + 0\hat{k}$

Substitute $\vec{a}$, $(\vec{b} - \vec{a})$, and $(\vec{c} - \vec{a})$ into the parametric vector equation:

This is one vector equation of the plane.

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Solution (Vector Form - Normal):

The normal vector to the plane is parallel to $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})$.

$\vec{n} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = (-4\hat{i} - 8\hat{j} + 8\hat{k}) \times (3\hat{i} - 2\hat{j})$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{vmatrix}$

$\vec{n} = \hat{i}((-8)(0) - (8)(-2)) - \hat{j}((-4)(0) - (8)(3)) + \hat{k}((-4)(-2) - (-8)(3))$

$\vec{n} = \hat{i}(0 + 16) - \hat{j}(0 - 24) + \hat{k}(8 + 24)$

$\vec{n} = 16\hat{i} + 24\hat{j} + 32\hat{k}$

We can use a simpler normal vector parallel to $\vec{n}$ by dividing by 8: $\vec{n}' = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal vector $\vec{n}'$ is $\vec{r} \cdot \vec{n}' = \vec{a} \cdot \vec{n}'$.

Calculate $\vec{a} \cdot \vec{n}'$:

$\vec{a} \cdot \vec{n}' = (2\hat{i} + 5\hat{j} - 3\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k})$

$\vec{a} \cdot \vec{n}' = (2)(2) + (5)(3) + (-3)(4) = 4 + 15 - 12 = 7$

The vector equation of the plane (normal form) is:

This is another vector equation of the plane.

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Solution (Cartesian Form):

Using the normal vector $\vec{n}' = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and the point R(2, 5, -3) on the plane.

The Cartesian equation of a plane passing through $(x_1, y_1, z_1)$ with normal direction ratios $(A, B, C)$ is given by:

$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$

Here, $(x_1, y_1, z_1) = (2, 5, -3)$ and $(A, B, C) = (2, 3, 4)$.

Substitute the values:

$2(x - 2) + 3(y - 5) + 4(z - (-3)) = 0$

$2(x - 2) + 3(y - 5) + 4(z + 3) = 0$

Expand and simplify:

$2x - 4 + 3y - 15 + 4z + 12 = 0$

$2x + 3y + 4z - 4 - 15 + 12 = 0$

$2x + 3y + 4z - 19 + 12 = 0$

This is the Cartesian equation of the plane.

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Alternatively, using the normal form $\vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 7$, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 7$

$2x + 3y + 4z = 7$

or

Both methods give the same Cartesian equation.

Given:

The plane intercepts the x-axis at $a = 2$.

The plane intercepts the y-axis at $b = 3$.

The plane intercepts the z-axis at $c = 4$.

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To Find:

The equation of the plane.

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Solution:

The equation of a plane with intercepts $a, b, c$ on the x, y, and z axes respectively is given by the intercept form:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

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Substitute the given intercept values $a=2$, $b=3$, and $c=4$ into the formula:

$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$

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To get the equation in the general Cartesian form $Ax + By + Cz + D = 0$, find a common denominator for the fractions. The least common multiple (LCM) of 2, 3, and 4 is 12.

Multiply the entire equation by 12:

$12 \left(\frac{x}{2} + \frac{y}{3} + \frac{z}{4}\right) = 12(1)$

$\frac{12x}{2} + \frac{12y}{3} + \frac{12z}{4} = 12$

$6x + 4y + 3z = 12$

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Rewrite the equation in the form $Ax + By + Cz + D = 0$:

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The equation of the plane with intercepts 2, 3, and 4 on the x, y, and z-axis respectively is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ or $6x + 4y + 3z - 12 = 0$.

Given:

Equation of Plane 1: $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 6$

Equation of Plane 2: $\vec{r} \;.\; (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5$

The required plane passes through the point (1, 1, 1).

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To Find:

The vector equation of the required plane.

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Solution:

The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by the formula:

$\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$

where $\lambda$ is a scalar parameter.

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From the given equations:

Plane 1: $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 6$

Here, $\vec{n}_1 = \hat{i} + \hat{j} + \hat{k}$ and $d_1 = 6$.

Plane 2: $\vec{r} \;.\; (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5$

Here, $\vec{n}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $d_2 = -5$.

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Substituting these values into the formula for the plane passing through the intersection:

$\vec{r} \cdot ((\hat{i} + \hat{j} + \hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})) = 6 + \lambda (-5)$

$\vec{r} \cdot ((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}) = 6 - 5\lambda$

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The required plane passes through the point (1, 1, 1). The position vector of this point is $\vec{r}_0 = \hat{i} + \hat{j} + \hat{k}$.

This point must satisfy the equation of the plane. Substitute $\vec{r}_0$ for $\vec{r}$ in the equation above:

$(\hat{i} + \hat{j} + \hat{k}) \cdot ((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}) = 6 - 5\lambda$

Perform the dot product:

$(1)(1 + 2\lambda) + (1)(1 + 3\lambda) + (1)(1 + 4\lambda) = 6 - 5\lambda$

$1 + 2\lambda + 1 + 3\lambda + 1 + 4\lambda = 6 - 5\lambda$

$3 + 9\lambda = 6 - 5\lambda$

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Solve for $\lambda$:

$9\lambda + 5\lambda = 6 - 3$

$14\lambda = 3$

$\lambda = \frac{3}{14}$

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Substitute the value of $\lambda = \frac{3}{14}$ back into the plane equation $\vec{r} \cdot ((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}) = 6 - 5\lambda$:

Normal vector component coefficients:

$1 + 2\lambda = 1 + 2\left(\frac{3}{14}\right) = 1 + \frac{6}{14} = 1 + \frac{3}{7} = \frac{7+3}{7} = \frac{10}{7}$

$1 + 3\lambda = 1 + 3\left(\frac{3}{14}\right) = 1 + \frac{9}{14} = \frac{14+9}{14} = \frac{23}{14}$

$1 + 4\lambda = 1 + 4\left(\frac{3}{14}\right) = 1 + \frac{12}{14} = 1 + \frac{6}{7} = \frac{7+6}{7} = \frac{13}{7}$

Right side constant:

$6 - 5\lambda = 6 - 5\left(\frac{3}{14}\right) = 6 - \frac{15}{14} = \frac{84 - 15}{14} = \frac{69}{14}$

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The vector equation of the plane is:

$\vec{r} \cdot \left(\frac{10}{7}\hat{i} + \frac{23}{14}\hat{j} + \frac{13}{7}\hat{k}\right) = \frac{69}{14}$

To clear the denominators, multiply the entire equation by the least common multiple of the denominators (which is 14):

$14 \left[ \vec{r} \cdot \left(\frac{10}{7}\hat{i} + \frac{23}{14}\hat{j} + \frac{13}{7}\hat{k}\right) \right] = 14 \left( \frac{69}{14} \right)$

$\vec{r} \cdot \left( 14 \times \frac{10}{7}\hat{i} + 14 \times \frac{23}{14}\hat{j} + 14 \times \frac{13}{7}\hat{k} \right) = 69$

This is the vector equation of the required plane.

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Cartesian Form (Optional, but can be included):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (20\hat{i} + 23\hat{j} + 26\hat{k}) = 69$

Performing the dot product:

$20x + 23y + 26z = 69$

Or in the general form:

This is the Cartesian equation of the plane.

Given:

Line 1: $\frac{x + 3}{−3} = \frac{y − 1}{1} = \frac{z − 5}{5}$

Line 2: $\frac{x + 1}{−1} = \frac{y − 2}{2} = \frac{z − 5}{5}$

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To Show:

The two lines are coplanar.

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Solution:

The Cartesian equation of a line is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are its direction ratios.

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For Line 1: $\frac{x - (-3)}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$

A point on Line 1 is $(x_1, y_1, z_1) = (-3, 1, 5)$.

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (-3, 1, 5)$.

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For Line 2: $\frac{x - (-1)}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

A point on Line 2 is $(x_2, y_2, z_2) = (-1, 2, 5)$.

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (-1, 2, 5)$.

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Calculate the vector connecting a point on Line 1 to a point on Line 2:

$(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-1 - (-3), 2 - 1, 5 - 5)$

$= (-1 + 3, 1, 0)$

$= (2, 1, 0)$

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ respectively are coplanar if and only if:

$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$

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Substitute the values into the determinant:

$\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$

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Evaluate the determinant (expanding along the first row):

$= 2 \begin{vmatrix} 1 & 5 \\ 2 & 5 \end{vmatrix} - 1 \begin{vmatrix} -3 & 5 \\ -1 & 5 \end{vmatrix} + 0 \begin{vmatrix} -3 & 1 \\ -1 & 2 \end{vmatrix}$

$= 2((1)(5) - (5)(2)) - 1((-3)(5) - (5)(-1)) + 0(...)$

$= 2(5 - 10) - 1(-15 + 5) + 0$

$= 2(-5) - 1(-10) + 0$

$= -10 + 10$

$= 0$

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Since the value of the determinant is 0, the two lines are coplanar.

Given:

Plane 1: 2x + y – 2z = 5

Plane 2: 3x – 6y – 2z = 7

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To Find:

The angle between the two planes using the vector method.

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Solution:

The equation of a plane in Cartesian form $Ax + By + Cz = D$ has a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

The angle $\theta$ between two planes is defined as the angle between their normal vectors $\vec{n}_1$ and $\vec{n}_2$.

The formula for the angle between two vectors is:

$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$

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From the equation of Plane 1 (2x + y – 2z = 5), the normal vector is $\vec{n}_1 = 2\hat{i} + \hat{j} - 2\hat{k}$.

From the equation of Plane 2 (3x – 6y – 2z = 7), the normal vector is $\vec{n}_2 = 3\hat{i} - 6\hat{j} - 2\hat{k}$.

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Calculate the dot product $\vec{n}_1 \cdot \vec{n}_2$:

$\vec{n}_1 \cdot \vec{n}_2 = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (3\hat{i} - 6\hat{j} - 2\hat{k})$

$= (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4$

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Calculate the magnitudes $|\vec{n}_1|$ and $|\vec{n}_2|$:

$|\vec{n}_1| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

$|\vec{n}_2| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|4|}{(3)(7)} = \frac{4}{21}$

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The angle $\theta$ between the two planes is $\theta = \cos^{-1}\left(\frac{4}{21}\right)$.

Given:

Plane 1: 3x – 6y + 2z = 7

Plane 2: 2x + 2y – 2z = 5

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To Find:

The angle between the two planes.

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Solution:

The equation of a plane in Cartesian form $Ax + By + Cz = D$ has a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

The angle $\theta$ between two planes is defined as the angle between their normal vectors $\vec{n}_1$ and $\vec{n}_2$.

The formula for the angle between two vectors is:

$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$

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From the equation of Plane 1 (3x – 6y + 2z = 7), the normal vector is $\vec{n}_1 = 3\hat{i} - 6\hat{j} + 2\hat{k}$.

From the equation of Plane 2 (2x + 2y – 2z = 5), the normal vector is $\vec{n}_2 = 2\hat{i} + 2\hat{j} - 2\hat{k}$.

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Calculate the dot product $\vec{n}_1 \cdot \vec{n}_2$:

$\vec{n}_1 \cdot \vec{n}_2 = (3\hat{i} - 6\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 2\hat{j} - 2\hat{k})$

$= (3)(2) + (-6)(2) + (2)(-2) = 6 - 12 - 4 = -10$

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Calculate the magnitudes $|\vec{n}_1|$ and $|\vec{n}_2|$:

$|\vec{n}_1| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$

$|\vec{n}_2| = \sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12}$

Simplify $|\vec{n}_2|$: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.

$|\vec{n}_2| = 2\sqrt{3}$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|-10|}{(7)(2\sqrt{3})} = \frac{10}{14\sqrt{3}} = \frac{5}{7\sqrt{3}}$

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Rationalize the denominator:

$\cos \theta = \frac{5}{7\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{7 \times 3} = \frac{5\sqrt{3}}{21}$

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The angle $\theta$ between the two planes is $\theta = \cos^{-1}\left(\frac{5\sqrt{3}}{21}\right)$.

Given:

The point is P(2, 5, – 3).

The equation of the plane is $\vec{r} \;.\; (6\hat{i} − 3\hat{j} + 2\hat{k}) = 4$.

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To Find:

The distance of the point from the plane.

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Solution (Using Vector Form):

The position vector of the point P(2, 5, – 3) is $\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}$.

The equation of the plane is $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = 6\hat{i} - 3\hat{j} + 2\hat{k}$ and $d = 4$.

The distance of a point with position vector $\vec{a}$ from a plane $\vec{r} \cdot \vec{n} = d$ is given by the formula:

Distance $= \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$

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First, calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (2\hat{i} + 5\hat{j} - 3\hat{k}) \cdot (6\hat{i} - 3\hat{j} + 2\hat{k})$

$= (2)(6) + (5)(-3) + (-3)(2) = 12 - 15 - 6 = -9$

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Next, calculate the magnitude of the normal vector $|\vec{n}|$:

$|\vec{n}| = |6\hat{i} - 3\hat{j} + 2\hat{k}| = \sqrt{6^2 + (-3)^2 + 2^2}$

$= \sqrt{36 + 9 + 4} = \sqrt{49} = 7$

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Substitute the values into the distance formula:

Distance $= \frac{|-9 - 4|}{7} = \frac{|-13|}{7} = \frac{13}{7}$

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Solution (Using Cartesian Form):

Convert the vector equation of the plane to Cartesian form. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4$

$6x - 3y + 2z = 4$

Rewrite in the form $Ax + By + Cz + D = 0$: $6x - 3y + 2z - 4 = 0$.

Here, $A=6$, $B=-3$, $C=2$, $D=-4$.

The given point is $(x_1, y_1, z_1) = (2, 5, -3)$.

The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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Substitute the values into the formula:

Distance $= \frac{|(6)(2) + (-3)(5) + (2)(-3) + (-4)|}{\sqrt{6^2 + (-3)^2 + 2^2}}$

Distance $= \frac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}}$

Distance $= \frac{|-13|}{\sqrt{49}} = \frac{13}{7}$

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Both methods give the same result.

The distance of the point (2, 5, – 3) from the plane is $\frac{13}{7}$ units.

Given:

Equation of the line: $\frac{x + 1}{2} = \frac{y}{3} = \frac{z − 3}{6}$

Equation of the plane: 10x + 2y – 11z = 3

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To Find:

The angle between the line and the plane.

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Solution:

The angle $\theta$ between a line with direction ratios $(a, b, c)$ and a plane with normal direction ratios $(A, B, C)$ is given by the formula for the angle $\phi$ between the line and the normal to the plane, where $\theta = 90^\circ - \phi$.

The angle $\phi$ between the direction vector of the line and the normal vector of the plane is given by:

$\cos \phi = \frac{|aA + bB + cC|}{\sqrt{a^2 + b^2 + c^2} \sqrt{A^2 + B^2 + C^2}}$

Since $\theta = 90^\circ - \phi$, $\sin \theta = \cos \phi$.

So, the angle $\theta$ between the line and the plane is given by:

$\sin \theta = \frac{|aA + bB + cC|}{\sqrt{a^2 + b^2 + c^2} \sqrt{A^2 + B^2 + C^2}}$

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From the equation of the line $\frac{x - (-1)}{2} = \frac{y - 0}{3} = \frac{z - 3}{6}$, the direction ratios of the line are $(a, b, c) = (2, 3, 6)$.

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From the equation of the plane $10x + 2y - 11z = 3$, the normal vector has direction ratios $(A, B, C) = (10, 2, -11)$.

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Calculate $aA + bB + cC$:

$aA + bB + cC = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = 26 - 66 = -40$

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Calculate the magnitude of the direction vector of the line:

$\sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

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Calculate the magnitude of the normal vector of the plane:

$\sqrt{A^2 + B^2 + C^2} = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$

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Substitute the values into the formula for $\sin \theta$:

$\sin \theta = \frac{|-40|}{7 \times 15} = \frac{40}{105}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 5:

$\sin \theta = \frac{\cancel{40}^{8}}{\cancel{105}_{21}} = \frac{8}{21}$

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The angle between the line and the plane is $\theta = \sin^{-1}\left(\frac{8}{21}\right)$.

The general equation of a plane is $Ax + By + Cz + D = 0$. The normal vector to the plane is $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$.

To find the direction cosines of the unit normal vector from the origin and the distance from the origin, we write the equation in the form $lx + my + nz = d$, where $(l, m, n)$ are the direction cosines and $d$ is the distance (which must be non-negative). This is achieved by dividing the equation $Ax + By + Cz = -D$ by $\pm\sqrt{A^2+B^2+C^2}$, where the sign is chosen such that the constant term on the right side is positive.

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(a) z = 2

The equation is $0x + 0y + 1z = 2$.

Comparing with $Ax + By + Cz = D$, we have $A=0, B=0, C=1, D=2$.

The normal vector is $\vec{N} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$.

The magnitude is $|\vec{N}| = \sqrt{0^2 + 0^2 + 1^2} = \sqrt{1} = 1$.

Since the constant term on the right side (2) is already positive, we divide by $+1$.

The direction cosines are $\left(\frac{A}{|\vec{N}|}, \frac{B}{|\vec{N}|}, \frac{C}{|\vec{N}|}\right) = \left(\frac{0}{1}, \frac{0}{1}, \frac{1}{1}\right) = (0, 0, 1)$.

The distance from the origin is $d = \frac{|D|}{|\vec{N}|} = \frac{|2|}{1} = 2$.

Direction cosines: $(0, 0, 1)$.

Distance from the origin: 2.

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(b) x + y + z = 1

The equation is $1x + 1y + 1z = 1$.

Comparing with $Ax + By + Cz = D$, we have $A=1, B=1, C=1, D=1$.

The normal vector is $\vec{N} = \hat{i} + \hat{j} + \hat{k}$.

The magnitude is $|\vec{N}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.

Since the constant term on the right side (1) is already positive, we divide by $+\sqrt{3}$.

The direction cosines are $\left(\frac{A}{|\vec{N}|}, \frac{B}{|\vec{N}|}, \frac{C}{|\vec{N}|}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

The distance from the origin is $d = \frac{|D|}{|\vec{N}|} = \frac{|1|}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

Direction cosines: $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Distance from the origin: $\frac{1}{\sqrt{3}}$.

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(c) 2x + 3y – z = 5

The equation is $2x + 3y - 1z = 5$.

Comparing with $Ax + By + Cz = D$, we have $A=2, B=3, C=-1, D=5$.

The normal vector is $\vec{N} = 2\hat{i} + 3\hat{j} - \hat{k}$.

The magnitude is $|\vec{N}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.

Since the constant term on the right side (5) is already positive, we divide by $+\sqrt{14}$.

The direction cosines are $\left(\frac{A}{|\vec{N}|}, \frac{B}{|\vec{N}|}, \frac{C}{|\vec{N}|}\right) = \left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.

The distance from the origin is $d = \frac{|D|}{|\vec{N}|} = \frac{|5|}{\sqrt{14}} = \frac{5}{\sqrt{14}}$.

Direction cosines: $\left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.

Distance from the origin: $\frac{5}{\sqrt{14}}$.

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(d) 5y + 8 = 0

The equation is $0x + 5y + 0z + 8 = 0$, which can be written as $0x + 5y + 0z = -8$.

Comparing with $Ax + By + Cz = D$, we have $A=0, B=5, C=0, D=-8$.

The normal vector is $\vec{N} = 0\hat{i} + 5\hat{j} + 0\hat{k} = 5\hat{j}$.

The magnitude is $|\vec{N}| = \sqrt{0^2 + 5^2 + 0^2} = \sqrt{25} = 5$.

The constant term on the right side is -8, which is negative. To make it positive, we multiply the equation $0x + 5y + 0z = -8$ by -1:

$0x - 5y + 0z = 8$.

Now, the coefficients are $A' = 0, B' = -5, C' = 0$ and the distance is $D' = 8$. The magnitude of the normal vector is still $\sqrt{0^2 + (-5)^2 + 0^2} = 5$.

The direction cosines of the unit normal vector from the origin are $\left(\frac{A'}{5}, \frac{B'}{5}, \frac{C'}{5}\right) = \left(\frac{0}{5}, \frac{-5}{5}, \frac{0}{5}\right) = (0, -1, 0)$.

The distance from the origin is $d = \frac{|D'|}{5} = \frac{|8|}{5} = \frac{8}{5}$.

Direction cosines: $(0, -1, 0)$.

Distance from the origin: $\frac{8}{5}$.

Given:

Distance of the plane from the origin, $d = 7$ units.

The normal vector to the plane is $\vec{n} = 3\hat{i} + 5\hat{j} − 6\hat{k}$.

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To Find:

The vector equation of the plane.

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Solution:

The vector equation of a plane at a distance $d$ from the origin and having a normal vector $\vec{n}$ is given by the normal form:

$\vec{r} \cdot \hat{n} = d$

where $\hat{n}$ is the unit normal vector to the plane.

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First, find the unit normal vector $\hat{n}$ from the given normal vector $\vec{n}$.

$|\vec{n}| = |3\hat{i} + 5\hat{j} − 6\hat{k}|$

$|\vec{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70}$

$\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\hat{i} + 5\hat{j} − 6\hat{k}}{\sqrt{70}}$

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Now, substitute $\hat{n}$ and $d$ into the vector equation formula:

$\vec{r} \cdot \left(\frac{3\hat{i} + 5\hat{j} − 6\hat{k}}{\sqrt{70}}\right) = 7$

Multiply both sides by $\sqrt{70}$:

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This is the required vector equation of the plane.

To convert a vector equation of a plane into its Cartesian form, we let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and perform the dot product.

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(a) $\vec{r} \;.\; (\hat{i} + \hat{j} − \hat{k}) = 2$

Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} − \hat{k}) = 2$

Perform the dot product:

$(x)(1) + (y)(1) + (z)(-1) = 2$

This is the Cartesian equation of the plane.

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(b) $\vec{r} \;.\; (2\hat{i} + 3\hat{j} − 4\hat{k}) = 1$

Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} − 4\hat{k}) = 1$

Perform the dot product:

$(x)(2) + (y)(3) + (z)(-4) = 1$

This is the Cartesian equation of the plane.

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(c) $\vec{r} \;.\; [(s−2t)\hat{i} + (3−t)\hat{j} + (s+t)\hat{k}] = 15$

This equation is not in the standard normal form $\vec{r} \cdot \vec{n} = d$. It appears to involve parameters s and t within the normal vector expression, which is unusual for a single plane equation unless s and t are fixed values or represent the coefficients of $\hat{i}, \hat{j}, \hat{k}$ in the normal vector.

Assuming the vector $(s−2t)\hat{i} + (3−t)\hat{j} + (s+t)\hat{k}$ represents the normal vector $\vec{n} = (s-2t)\hat{i} + (3-t)\hat{j} + (s+t)\hat{k}$, and s and t are meant to define the direction ratios of the normal, this form suggests a dependency on parameters which is not typical for a unique plane equation unless s and t are specific given constants.

If s and t are intended to represent constants defining the components of the normal vector, we would substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and perform the dot product:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot ((s−2t)\hat{i} + (3−t)\hat{j} + (s+t)\hat{k}) = 15$

$(x)(s-2t) + (y)(3-t) + (z)(s+t) = 15$

$(s-2t)x + (3-t)y + (s+t)z = 15$

This is the Cartesian equation in terms of s and t.

Assumption: If the question intended s and t to define the normal vector, and this is the final form, then the Cartesian equation is $(s-2t)x + (3-t)y + (s+t)z = 15$. However, this is unlikely for a standard problem unless specific values of s and t are given or can be determined from context not provided.

Alternative Interpretation: Perhaps the term $[(s−2t)\hat{i} + (3−t)\hat{j} + (s+t)\hat{k}]$ is a fixed normal vector where s and t are specific, but unstated, values. In that case, the Cartesian form is derived as above with those fixed values.

Most Likely Interpretation for this Exercise context: It is possible there is a typo in the question and the form is different, or s and t define the direction ratios of the normal vector in a fixed way. Given the form, it's possible the expression is intended as a single normal vector $\vec{n} = (s_0-2t_0)\hat{i} + (3-t_0)\hat{j} + (s_0+t_0)\hat{k}$ for some fixed $s_0, t_0$. Without specific values for s and t, the Cartesian equation is as derived: $(s-2t)x + (3-t)y + (s+t)z = 15$.

Let's assume for the purpose of providing a concrete Cartesian form that s and t are constants that define the coefficients of the normal vector. Then the equation is:

If the question implies a family of planes or some other construction involving s and t, further information would be needed.

The equation of a plane is $Ax + By + Cz + D = 0$. The normal vector to the plane has direction ratios $(A, B, C)$. The line drawn from the origin $(0, 0, 0)$ perpendicular to the plane is parallel to the normal vector.

The equation of this line is $\frac{x - 0}{A} = \frac{y - 0}{B} = \frac{z - 0}{C}$.

Let the foot of the perpendicular be the point P, which lies on this line and the plane. Any point on the line can be represented as $(A\lambda, B\lambda, C\lambda)$ for some scalar $\lambda$.

Since P lies on the plane $Ax + By + Cz + D = 0$, its coordinates must satisfy the plane equation:

$A(A\lambda) + B(B\lambda) + C(C\lambda) + D = 0$

$A^2\lambda + B^2\lambda + C^2\lambda + D = 0$

$(A^2 + B^2 + C^2)\lambda = -D$

$\lambda = \frac{-D}{A^2 + B^2 + C^2}$

The coordinates of the foot of the perpendicular $(x, y, z)$ are $(A\lambda, B\lambda, C\lambda) = \left(\frac{-AD}{A^2 + B^2 + C^2}, \frac{-BD}{A^2 + B^2 + C^2}, \frac{-CD}{A^2 + B^2 + C^2}\right)$.

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(a) 2x + 3y + 4z – 12 = 0

Comparing with $Ax + By + Cz + D = 0$, we have $A=2, B=3, C=4, D=-12$.

$A^2 + B^2 + C^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$.

$\lambda = \frac{-D}{A^2 + B^2 + C^2} = \frac{-(-12)}{29} = \frac{12}{29}$.

Coordinates of the foot of the perpendicular:

$x = A\lambda = 2 \times \frac{12}{29} = \frac{24}{29}$

$y = B\lambda = 3 \times \frac{12}{29} = \frac{36}{29}$

$z = C\lambda = 4 \times \frac{12}{29} = \frac{48}{29}$

The coordinates of the foot of the perpendicular are $\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)$.

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(b) 3y + 4z – 6 = 0

This can be written as $0x + 3y + 4z - 6 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=0, B=3, C=4, D=-6$.

$A^2 + B^2 + C^2 = 0^2 + 3^2 + 4^2 = 0 + 9 + 16 = 25$.

$\lambda = \frac{-D}{A^2 + B^2 + C^2} = \frac{-(-6)}{25} = \frac{6}{25}$.

Coordinates of the foot of the perpendicular:

$x = A\lambda = 0 \times \frac{6}{25} = 0$

$y = B\lambda = 3 \times \frac{6}{25} = \frac{18}{25}$

$z = C\lambda = 4 \times \frac{6}{25} = \frac{24}{25}$

The coordinates of the foot of the perpendicular are $\left(0, \frac{18}{25}, \frac{24}{25}\right)$.

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(c) x + y + z = 1

This can be written as $1x + 1y + 1z - 1 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=1, B=1, C=1, D=-1$.

$A^2 + B^2 + C^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3$.

$\lambda = \frac{-D}{A^2 + B^2 + C^2} = \frac{-(-1)}{3} = \frac{1}{3}$.

Coordinates of the foot of the perpendicular:

$x = A\lambda = 1 \times \frac{1}{3} = \frac{1}{3}$

$y = B\lambda = 1 \times \frac{1}{3} = \frac{1}{3}$

$z = C\lambda = 1 \times \frac{1}{3} = \frac{1}{3}$

The coordinates of the foot of the perpendicular are $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.

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(d) 5y + 8 = 0

This can be written as $0x + 5y + 0z + 8 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=0, B=5, C=0, D=8$.

$A^2 + B^2 + C^2 = 0^2 + 5^2 + 0^2 = 0 + 25 + 0 = 25$.

$\lambda = \frac{-D}{A^2 + B^2 + C^2} = \frac{-8}{25}$.

Coordinates of the foot of the perpendicular:

$x = A\lambda = 0 \times \frac{-8}{25} = 0$

$y = B\lambda = 5 \times \frac{-8}{25} = \frac{-40}{25} = -\frac{8}{5}$

$z = C\lambda = 0 \times \frac{-8}{25} = 0$

The coordinates of the foot of the perpendicular are $\left(0, -\frac{8}{5}, 0\right)$.

The vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a vector $\vec{n}$ is given by:

$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

To find the Cartesian equation from the vector equation $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ and $d = \vec{a} \cdot \vec{n}$, we substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ to get $Ax + By + Cz = d$.

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(a)

Given:

The plane passes through the point P(1, 0, – 2).

The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} − \hat{k}$.

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To Find:

The vector and Cartesian equations of the plane.

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Solution (Vector Form):

The position vector of the point P(1, 0, – 2) on the plane is $\vec{a} = \hat{i} + 0\hat{j} - 2\hat{k} = \hat{i} - 2\hat{k}$.

The normal vector is $\vec{n} = \hat{i} + \hat{j} − \hat{k}$.

Using the formula $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:

Calculate $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (\hat{i} - 2\hat{k}) \cdot (\hat{i} + \hat{j} − \hat{k})$

$= (1)(1) + (0)(1) + (-2)(-1) = 1 + 0 + 2 = 3$

The vector equation of the plane is:

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Solution (Cartesian Form):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} − \hat{k}) = 3$

Perform the dot product:

$x(1) + y(1) + z(-1) = 3$

This is the Cartesian equation of the plane.

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(b)

Given:

The plane passes through the point Q(1, 4, 6).

The normal vector to the plane is $\vec{n} = \hat{i} − 2\hat{j} + \hat{k}$.

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To Find:

The vector and Cartesian equations of the plane.

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Solution (Vector Form):

The position vector of the point Q(1, 4, 6) on the plane is $\vec{a} = \hat{i} + 4\hat{j} + 6\hat{k}$.

The normal vector is $\vec{n} = \hat{i} − 2\hat{j} + \hat{k}$.

Using the formula $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:

Calculate $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (\hat{i} − 2\hat{j} + \hat{k})$

$= (1)(1) + (4)(-2) + (6)(1) = 1 - 8 + 6 = -1$

The vector equation of the plane is:

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Solution (Cartesian Form):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} − 2\hat{j} + \hat{k}) = -1$

Perform the dot product:

$x(1) + y(-2) + z(1) = -1$

This is the Cartesian equation of the plane. It can also be written as $x - 2y + z + 1 = 0$.

The equation of a plane passing through three non-collinear points $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is given by the determinant form:

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$

Alternatively, in vector form passing through points with position vectors $\vec{a}, \vec{b}, \vec{c}$ is $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) + \mu (\vec{c} - \vec{a})$, or in normal form $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, where $\vec{n} = (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})$. We will find the Cartesian form using the determinant and the vector form using the normal vector.

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(a) Points: (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

Let the points be $P_1(1, 1, -1)$, $P_2(6, 4, -5)$, and $P_3(-4, -2, 3)$.

$(x_1, y_1, z_1) = (1, 1, -1)$

$(x_2, y_2, z_2) = (6, 4, -5)$

$(x_3, y_3, z_3) = (-4, -2, 3)$

Calculate the difference vectors for the determinant:

$x_2 - x_1 = 6 - 1 = 5$

$y_2 - y_1 = 4 - 1 = 3$

$z_2 - z_1 = -5 - (-1) = -5 + 1 = -4$

$x_3 - x_1 = -4 - 1 = -5$

$y_3 - y_1 = -2 - 1 = -3$

$z_3 - z_1 = 3 - (-1) = 3 + 1 = 4$

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Cartesian Equation:

Substitute the values into the determinant formula:

$\begin{vmatrix} x - 1 & y - 1 & z - (-1) \\ 5 & 3 & -4 \\ -5 & -3 & 4 \end{vmatrix} = 0$

$\begin{vmatrix} x - 1 & y - 1 & z + 1 \\ 5 & 3 & -4 \\ -5 & -3 & 4 \end{vmatrix} = 0$

Notice that the second row is a multiple of the third row (R2 = -1 * R3). This implies that the three points are collinear, and thus they do not define a unique plane.

Let's verify if the points are collinear. Check if the direction ratios of the line segment $P_1P_2$ are proportional to the direction ratios of the line segment $P_1P_3$.

DRs of $P_1P_2$: $(5, 3, -4)$

DRs of $P_1P_3$: $(-5, -3, 4)$

The ratios are $\frac{5}{-5} = -1$, $\frac{3}{-3} = -1$, $\frac{-4}{4} = -1$. Since the ratios are equal, the points are collinear.

Conclusion for (a): The given three points are collinear, so they do not determine a unique plane. Any plane passing through the line formed by these points will contain all three points. Thus, there is no unique plane equation.

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(b) Points: (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

Let the points be $P_1(1, 1, 0)$, $P_2(1, 2, 1)$, and $P_3(-2, 2, -1)$.

$(x_1, y_1, z_1) = (1, 1, 0)$

$(x_2, y_2, z_2) = (1, 2, 1)$

$(x_3, y_3, z_3) = (-2, 2, -1)$

Calculate the difference vectors for the determinant:

$x_2 - x_1 = 1 - 1 = 0$

$y_2 - y_1 = 2 - 1 = 1$

$z_2 - z_1 = 1 - 0 = 1$

$x_3 - x_1 = -2 - 1 = -3$

$y_3 - y_1 = 2 - 1 = 1$

$z_3 - z_1 = -1 - 0 = -1$

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Cartesian Equation:

Substitute the values into the determinant formula:

$\begin{vmatrix} x - 1 & y - 1 & z - 0 \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0$

$\begin{vmatrix} x - 1 & y - 1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0$

Evaluate the determinant (expanding along the first row):

$(x - 1) \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} - (y - 1) \begin{vmatrix} 0 & 1 \\ -3 & -1 \end{vmatrix} + z \begin{vmatrix} 0 & 1 \\ -3 & 1 \end{vmatrix} = 0$

$(x - 1)((1)(-1) - (1)(1)) - (y - 1)((0)(-1) - (1)(-3)) + z((0)(1) - (1)(-3)) = 0$

$(x - 1)(-1 - 1) - (y - 1)(0 + 3) + z(0 + 3) = 0$

$(x - 1)(-2) - (y - 1)(3) + 3z = 0$

$-2x + 2 - 3y + 3 + 3z = 0$

$-2x - 3y + 3z + 5 = 0$

Multiplying by -1 to make the leading coefficient positive (optional):

This is the Cartesian equation of the plane.

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Vector Equation (Normal Form):

Let $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{c} = -2\hat{i} + 2\hat{j} - \hat{k}$.

$\vec{b} - \vec{a} = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j}) = 0\hat{i} + \hat{j} + \hat{k}$

$\vec{c} - \vec{a} = (-2\hat{i} + 2\hat{j} - \hat{k}) - (\hat{i} + \hat{j}) = -3\hat{i} + \hat{j} - \hat{k}$

The normal vector $\vec{n}$ is parallel to $(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})$.

$\vec{n} = (0\hat{i} + \hat{j} + \hat{k}) \times (-3\hat{i} + \hat{j} - \hat{k})$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix}$

$\vec{n} = \hat{i}((1)(-1) - (1)(1)) - \hat{j}((0)(-1) - (1)(-3)) + \hat{k}((0)(1) - (1)(-3))$

$\vec{n} = \hat{i}(-1 - 1) - \hat{j}(0 + 3) + \hat{k}(0 + 3)$

$\vec{n} = -2\hat{i} - 3\hat{j} + 3\hat{k}$

The vector equation is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$. Using $\vec{a} = \hat{i} + \hat{j}$ and $\vec{n} = -2\hat{i} - 3\hat{j} + 3\hat{k}$:

$\vec{a} \cdot \vec{n} = (\hat{i} + \hat{j}) \cdot (-2\hat{i} - 3\hat{j} + 3\hat{k})$

$= (1)(-2) + (1)(-3) + (0)(3) = -2 - 3 + 0 = -5$

The vector equation is:

Multiplying by -1:

This is the vector equation of the plane.

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Vector Equation (Parametric Form):

Using the formula $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}) + \mu (\vec{c} - \vec{a})$:

This is also a valid vector equation of the plane.

Given:

The equation of the plane is $2x + y - z = 5$.

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To Find:

The intercepts cut off by the plane on the x, y, and z axes.

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Solution:

To find the x-intercept, set $y=0$ and $z=0$ in the equation of the plane and solve for x.

$2x + 0 - 0 = 5$

$2x = 5$

$x = \frac{5}{2}$

The x-intercept is $\frac{5}{2}$.

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To find the y-intercept, set $x=0$ and $z=0$ in the equation of the plane and solve for y.

$2(0) + y - 0 = 5$

$y = 5$

The y-intercept is 5.

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To find the z-intercept, set $x=0$ and $y=0$ in the equation of the plane and solve for z.

$2(0) + 0 - z = 5$

$-z = 5$

$z = -5$

The z-intercept is -5.

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Alternatively, we can write the equation of the plane in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where a, b, and c are the x, y, and z intercepts respectively.

Divide the given equation $2x + y - z = 5$ by 5:

$\frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = \frac{5}{5}$

$\frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{-5} = 1$

Comparing this with the intercept form, we get:

x-intercept $a = \frac{5}{2}$

y-intercept $b = 5$

z-intercept $c = -5$

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The intercepts cut off by the plane are $\frac{5}{2}$ on the x-axis, 5 on the y-axis, and -5 on the z-axis.

Given:

The plane has an intercept of 3 on the y-axis.

The plane is parallel to the ZOX plane.

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To Find:

The equation of the plane.

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Solution:

A plane parallel to the ZOX plane has an equation of the form $y = k$, where $k$ is a constant.

This is because for any point on the ZOX plane (where $y=0$), the normal vector is parallel to the y-axis ($\hat{j}$). A plane parallel to the ZOX plane will also have a normal vector parallel to the y-axis. The equation $\vec{r} \cdot \hat{j} = d$ or $y = d$ represents such a plane.

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The plane has an intercept of 3 on the y-axis. This means the plane passes through the point $(0, 3, 0)$.

Since the point $(0, 3, 0)$ lies on the plane $y = k$, substitute the coordinates into the equation:

$3 = k$

So, the value of the constant $k$ is 3.

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Therefore, the equation of the plane is $y = 3$.

In the general form $Ax + By + Cz + D = 0$, this can be written as:

or

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The equation of the plane is $y = 3$.

Given:

Equation of Plane 1: $P_1: 3x - y + 2z - 4 = 0$

Equation of Plane 2: $P_2: x + y + z - 2 = 0$

The required plane passes through the point (2, 2, 1).

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To Find:

The equation of the required plane.

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Solution:

The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by the linear combination:

$P_1 + \lambda P_2 = 0$

$(3x - y + 2z - 4) + \lambda (x + y + z - 2) = 0$

where $\lambda$ is a scalar parameter.

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The required plane passes through the point (2, 2, 1). This point must satisfy the equation of the plane.

Substitute $x=2, y=2, z=1$ into the equation:

$(3(2) - 2 + 2(1) - 4) + \lambda (2 + 2 + 1 - 2) = 0$

$(6 - 2 + 2 - 4) + \lambda (5 - 2) = 0$

$(4 + 2 - 4) + \lambda (3) = 0$

$2 + 3\lambda = 0$

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Solve for $\lambda$:

$3\lambda = -2$

$\lambda = -\frac{2}{3}$

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Substitute the value of $\lambda = -\frac{2}{3}$ back into the equation of the plane:

$(3x - y + 2z - 4) - \frac{2}{3} (x + y + z - 2) = 0$

Multiply the entire equation by 3 to clear the denominator:

$3(3x - y + 2z - 4) - 2 (x + y + z - 2) = 0$

$9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0$

Combine like terms:

$(9x - 2x) + (-3y - 2y) + (6z - 2z) + (-12 + 4) = 0$

$7x - 5y + 4z - 8 = 0$

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The equation of the plane passing through the intersection of the given planes and the point (2, 2, 1) is $7x - 5y + 4z - 8 = 0$.

Given:

Equation of Plane 1: $\vec{r} \;.\; (2\hat{i} + 2\hat{j} − 3\hat{k}) = 7$

Equation of Plane 2: $\vec{r} \;.\; (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$

The required plane passes through the point (2, 1, 3).

---

To Find:

The vector equation of the required plane.

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Solution:

The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by:

$\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$

where $\lambda$ is a scalar parameter.

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From the given equations:

Plane 1: $\vec{r} \;.\; (2\hat{i} + 2\hat{j} − 3\hat{k}) = 7$

Here, $\vec{n}_1 = 2\hat{i} + 2\hat{j} − 3\hat{k}$ and $d_1 = 7$.

Plane 2: $\vec{r} \;.\; (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$

Here, $\vec{n}_2 = 2\hat{i} + 5\hat{j} + 3\hat{k}$ and $d_2 = 9$.

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Substituting these values into the formula for the plane passing through the intersection:

$\vec{r} \cdot ((2\hat{i} + 2\hat{j} − 3\hat{k}) + \lambda (2\hat{i} + 5\hat{j} + 3\hat{k})) = 7 + \lambda (9)$

$\vec{r} \cdot ((2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}) = 7 + 9\lambda$

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The required plane passes through the point (2, 1, 3). The position vector of this point is $\vec{r}_0 = 2\hat{i} + \hat{j} + 3\hat{k}$.

This point must satisfy the equation of the plane. Substitute $\vec{r}_0$ for $\vec{r}$ in the equation above:

$(2\hat{i} + \hat{j} + 3\hat{k}) \cdot ((2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}) = 7 + 9\lambda$

Perform the dot product:

$(2)(2 + 2\lambda) + (1)(2 + 5\lambda) + (3)(-3 + 3\lambda) = 7 + 9\lambda$

$4 + 4\lambda + 2 + 5\lambda - 9 + 9\lambda = 7 + 9\lambda$

$(-3) + 18\lambda = 7 + 9\lambda$

---

Solve for $\lambda$:

$18\lambda - 9\lambda = 7 + 3$

$9\lambda = 10$

$\lambda = \frac{10}{9}$

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Substitute the value of $\lambda = \frac{10}{9}$ back into the plane equation $\vec{r} \cdot ((2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}) = 7 + 9\lambda$:

Normal vector component coefficients:

$2 + 2\lambda = 2 + 2\left(\frac{10}{9}\right) = 2 + \frac{20}{9} = \frac{18 + 20}{9} = \frac{38}{9}$

$2 + 5\lambda = 2 + 5\left(\frac{10}{9}\right) = 2 + \frac{50}{9} = \frac{18 + 50}{9} = \frac{68}{9}$

$-3 + 3\lambda = -3 + 3\left(\frac{10}{9}\right) = -3 + \frac{30}{9} = -3 + \frac{10}{3} = \frac{-9 + 10}{3} = \frac{1}{3} = \frac{3}{9}$

Right side constant:

$7 + 9\lambda = 7 + 9\left(\frac{10}{9}\right) = 7 + 10 = 17$

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The vector equation of the plane is:

$\vec{r} \cdot \left(\frac{38}{9}\hat{i} + \frac{68}{9}\hat{j} + \frac{3}{9}\hat{k}\right) = 17$

Multiply the entire equation by 9 to clear the denominator:

$9 \left[ \vec{r} \cdot \left(\frac{38}{9}\hat{i} + \frac{68}{9}\hat{j} + \frac{3}{9}\hat{k}\right) \right] = 9 (17)$

$\vec{r} \cdot \left( 9 \times \frac{38}{9}\hat{i} + 9 \times \frac{68}{9}\hat{j} + 9 \times \frac{3}{9}\hat{k} \right) = 153$

This is the vector equation of the required plane.

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Cartesian Form (Optional):

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153$

Performing the dot product:

$38x + 68y + 3z = 153$

Or in the general form:

This is the Cartesian equation of the plane.

Given:

Equation of Plane 1: $P_1: x + y + z = 1$, or $x + y + z - 1 = 0$.

Equation of Plane 2: $P_2: 2x + 3y + 4z = 5$, or $2x + 3y + 4z - 5 = 0$.

The required plane is perpendicular to Plane 3: $P_3: x - y + z = 0$.

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To Find:

The equation of the plane passing through the intersection of $P_1$ and $P_2$ and perpendicular to $P_3$.

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Solution:

The equation of any plane passing through the line of intersection of planes $P_1 = 0$ and $P_2 = 0$ is given by the linear combination:

$P_1 + \lambda P_2 = 0$

$(x + y + z - 1) + \lambda (2x + 3y + 4z - 5) = 0$

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Rearrange this equation to find the normal vector of this new plane:

$x + y + z - 1 + 2\lambda x + 3\lambda y + 4\lambda z - 5\lambda = 0$

$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0$

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The normal vector of this plane is $\vec{n} = (1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}$. The direction ratios of the normal are $(A_1, B_1, C_1) = (1 + 2\lambda, 1 + 3\lambda, 1 + 4\lambda)$.

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The given plane $x - y + z = 0$ has a normal vector $\vec{n}_3 = \hat{i} - \hat{j} + \hat{k}$. The direction ratios of this normal are $(A_2, B_2, C_2) = (1, -1, 1)$.

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Since the required plane is perpendicular to the plane $x - y + z = 0$, their normal vectors must be perpendicular.

The condition for perpendicularity of two vectors with direction ratios $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$ is $A_1 A_2 + B_1 B_2 + C_1 C_2 = 0$.

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Substitute the direction ratios of the two normal vectors into the perpendicularity condition:

$(1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0$

$1 + 2\lambda - (1 + 3\lambda) + 1 + 4\lambda = 0$

$1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$

$(2\lambda - 3\lambda + 4\lambda) + (1 - 1 + 1) = 0$

$3\lambda + 1 = 0$

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Solve for $\lambda$:

$3\lambda = -1$

$\lambda = -\frac{1}{3}$

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Substitute the value of $\lambda = -\frac{1}{3}$ back into the equation of the plane $(x + y + z - 1) + \lambda (2x + 3y + 4z - 5) = 0$:

$(x + y + z - 1) + \left(-\frac{1}{3}\right) (2x + 3y + 4z - 5) = 0$

Multiply the entire equation by 3 to eliminate the fraction:

$3(x + y + z - 1) - 1 (2x + 3y + 4z - 5) = 0$

$3x + 3y + 3z - 3 - 2x - 3y - 4z + 5 = 0$

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Combine like terms:

$(3x - 2x) + (3y - 3y) + (3z - 4z) + (-3 + 5) = 0$

$x + 0y - z + 2 = 0$

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The equation of the required plane is $x - z + 2 = 0$.

Given:

Equation of Plane 1: $\vec{r} \;.\; (2\hat{i} + 2\hat{j} − 3\hat{k}) = 5$

Equation of Plane 2: $\vec{r} \;.\; (3\hat{i} − 3\hat{j} + 5\hat{k}) = 3$

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To Find:

The angle between the two planes.

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Solution:

The vector equation of a plane in the normal form is $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector to the plane.

The angle $\theta$ between two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ is given by the angle between their normal vectors.

The formula for the angle between two vectors is:

$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$

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From the equation of Plane 1, the normal vector is $\vec{n}_1 = 2\hat{i} + 2\hat{j} − 3\hat{k}$.

From the equation of Plane 2, the normal vector is $\vec{n}_2 = 3\hat{i} − 3\hat{j} + 5\hat{k}$.

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Calculate the dot product $\vec{n}_1 \cdot \vec{n}_2$:

$\vec{n}_1 \cdot \vec{n}_2 = (2\hat{i} + 2\hat{j} − 3\hat{k}) \cdot (3\hat{i} − 3\hat{j} + 5\hat{k})$

$= (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15$

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Calculate the magnitudes of the normal vectors $|\vec{n}_1|$ and $|\vec{n}_2|$:

$|\vec{n}_1| = |2\hat{i} + 2\hat{j} − 3\hat{k}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$

$|\vec{n}_2| = |3\hat{i} − 3\hat{j} + 5\hat{k}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43}$

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Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|-15|}{\sqrt{17} \sqrt{43}} = \frac{15}{\sqrt{17 \times 43}} = \frac{15}{\sqrt{731}}$

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The angle $\theta$ between the two planes is $\theta = \cos^{-1}\left(\frac{15}{\sqrt{731}}\right)$.

The equation of a plane is $Ax + By + Cz + D = 0$. The normal vector to the plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. The direction ratios of the normal are $(A, B, C)$.

Two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ are:

• Parallel if $\vec{n}_1$ is parallel to $\vec{n}_2$, i.e., their direction ratios are proportional: $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
• Perpendicular if $\vec{n}_1$ is perpendicular to $\vec{n}_2$, i.e., their dot product is zero: $A_1 A_2 + B_1 B_2 + C_1 C_2 = 0$.

If they are neither parallel nor perpendicular, the angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|} = \frac{|A_1 A_2 + B_1 B_2 + C_1 C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}$.

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(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Normal vector to Plane 1: $\vec{n}_1 = 7\hat{i} + 5\hat{j} + 6\hat{k}$, direction ratios $(A_1, B_1, C_1) = (7, 5, 6)$.

Normal vector to Plane 2: $\vec{n}_2 = 3\hat{i} - \hat{j} - 10\hat{k}$, direction ratios $(A_2, B_2, C_2) = (3, -1, -10)$.

Check for Parallelism: $\frac{7}{3} \neq \frac{5}{-1}$. Not parallel.

Check for Perpendicularity: $A_1 A_2 + B_1 B_2 + C_1 C_2 = (7)(3) + (5)(-1) + (6)(-10) = 21 - 5 - 60 = 16 - 60 = -44$. Not perpendicular since $-44 \neq 0$.

Find the angle:

$\vec{n}_1 \cdot \vec{n}_2 = -44$

$|\vec{n}_1| = \sqrt{7^2 + 5^2 + 6^2} = \sqrt{49 + 25 + 36} = \sqrt{110}$

$|\vec{n}_2| = \sqrt{3^2 + (-1)^2 + (-10)^2} = \sqrt{9 + 1 + 100} = \sqrt{110}$

$\cos \theta = \frac{|-44|}{\sqrt{110} \sqrt{110}} = \frac{44}{110}$

Simplify the fraction by dividing by 22: $\frac{\cancel{44}^{2}}{\cancel{110}_{5}} = \frac{2}{5}$.

The angle is $\theta = \cos^{-1}\left(\frac{2}{5}\right)$.

Conclusion for (a): Neither parallel nor perpendicular. Angle is $\cos^{-1}\left(\frac{2}{5}\right)$.

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(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Normal vector to Plane 1: $\vec{n}_1 = 2\hat{i} + \hat{j} + 3\hat{k}$, direction ratios $(A_1, B_1, C_1) = (2, 1, 3)$.

Normal vector to Plane 2: $\vec{n}_2 = \hat{i} - 2\hat{j} + 0\hat{k}$, direction ratios $(A_2, B_2, C_2) = (1, -2, 0)$.

Check for Parallelism: $\frac{2}{1} \neq \frac{1}{-2}$. Not parallel.

Check for Perpendicularity: $A_1 A_2 + B_1 B_2 + C_1 C_2 = (2)(1) + (1)(-2) + (3)(0) = 2 - 2 + 0 = 0$.

Conclusion for (b): The planes are perpendicular.

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(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Normal vector to Plane 1: $\vec{n}_1 = 2\hat{i} - 2\hat{j} + 4\hat{k}$, direction ratios $(A_1, B_1, C_1) = (2, -2, 4)$.

Normal vector to Plane 2: $\vec{n}_2 = 3\hat{i} - 3\hat{j} + 6\hat{k}$, direction ratios $(A_2, B_2, C_2) = (3, -3, 6)$.

Check for Parallelism: $\frac{A_1}{A_2} = \frac{2}{3}$, $\frac{B_1}{B_2} = \frac{-2}{-3} = \frac{2}{3}$, $\frac{C_1}{C_2} = \frac{4}{6} = \frac{2}{3}$.

Since $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = \frac{2}{3}$, the direction ratios are proportional.

Conclusion for (c): The planes are parallel.

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(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Normal vector to Plane 1: $\vec{n}_1 = 2\hat{i} - \hat{j} + 3\hat{k}$, direction ratios $(A_1, B_1, C_1) = (2, -1, 3)$.

Normal vector to Plane 2: $\vec{n}_2 = 2\hat{i} - \hat{j} + 3\hat{k}$, direction ratios $(A_2, B_2, C_2) = (2, -1, 3)$.

Check for Parallelism: $\frac{A_1}{A_2} = \frac{2}{2} = 1$, $\frac{B_1}{B_2} = \frac{-1}{-1} = 1$, $\frac{C_1}{C_2} = \frac{3}{3} = 1$.

Since $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = 1$, the direction ratios are proportional.

Conclusion for (d): The planes are parallel.

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(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Normal vector to Plane 1: $\vec{n}_1 = 4\hat{i} + 8\hat{j} + \hat{k}$, direction ratios $(A_1, B_1, C_1) = (4, 8, 1)$.

Normal vector to Plane 2: $\vec{n}_2 = 0\hat{i} + \hat{j} + \hat{k}$, direction ratios $(A_2, B_2, C_2) = (0, 1, 1)$.

Check for Parallelism: $\frac{4}{0}$ is undefined, while $\frac{8}{1} = 8$. Not parallel.

Check for Perpendicularity: $A_1 A_2 + B_1 B_2 + C_1 C_2 = (4)(0) + (8)(1) + (1)(1) = 0 + 8 + 1 = 9$. Not perpendicular since $9 \neq 0$.

Find the angle:

$\vec{n}_1 \cdot \vec{n}_2 = 9$

$|\vec{n}_1| = \sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9$

$|\vec{n}_2| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$

$\cos \theta = \frac{|9|}{9 \sqrt{2}} = \frac{9}{9 \sqrt{2}} = \frac{1}{\sqrt{2}}$

The angle is $\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ$ or $\frac{\pi}{4}$ radians.

Conclusion for (e): Neither parallel nor perpendicular. Angle is $45^\circ$ or $\frac{\pi}{4}$.

The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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(i) Distance of point $(0, 0, 0)$ from plane $3x - 4y + 12z = 3$

Given: Point $(x_1, y_1, z_1) = (0, 0, 0)$, Plane $3x - 4y + 12z = 3$ or $3x - 4y + 12z - 3 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=3, B=-4, C=12, D=-3$.

To Find: Distance of the point from the plane.

Solution:

Distance $= \frac{|(3)(0) + (-4)(0) + (12)(0) + (-3)|}{\sqrt{3^2 + (-4)^2 + 12^2}}$

Distance $= \frac{|-3|}{\sqrt{9 + 16 + 144}} = \frac{3}{\sqrt{169}} = \frac{3}{13}$

The distance is $\frac{3}{13}$ units.

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(ii) Distance of point $(3, -2, 1)$ from plane $2x - y + 2z + 3 = 0$

Given: Point $(x_1, y_1, z_1) = (3, -2, 1)$, Plane $2x - y + 2z + 3 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=2, B=-1, C=2, D=3$.

To Find: Distance of the point from the plane.

Solution:

Distance $= \frac{|(2)(3) + (-1)(-2) + (2)(1) + 3|}{\sqrt{2^2 + (-1)^2 + 2^2}}$

Distance $= \frac{|6 + 2 + 2 + 3|}{\sqrt{4 + 1 + 4}} = \frac{|13|}{\sqrt{9}} = \frac{13}{3}$

The distance is $\frac{13}{3}$ units.

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(iii) Distance of point $(2, 3, -5)$ from plane $x + 2y - 2z = 9$

Given: Point $(x_1, y_1, z_1) = (2, 3, -5)$, Plane $x + 2y - 2z = 9$ or $x + 2y - 2z - 9 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=1, B=2, C=-2, D=-9$.

To Find: Distance of the point from the plane.

Solution:

Distance $= \frac{|(1)(2) + (2)(3) + (-2)(-5) + (-9)|}{\sqrt{1^2 + 2^2 + (-2)^2}}$

Distance $= \frac{|2 + 6 + 10 - 9|}{\sqrt{1 + 4 + 4}} = \frac{|9|}{\sqrt{9}} = \frac{9}{3} = 3$

The distance is 3 units.

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(iv) Distance of point $(-6, 0, 0)$ from plane $2x - 3y + 6z - 2 = 0$

Given: Point $(x_1, y_1, z_1) = (-6, 0, 0)$, Plane $2x - 3y + 6z - 2 = 0$.

Comparing with $Ax + By + Cz + D = 0$, we have $A=2, B=-3, C=6, D=-2$.

To Find: Distance of the point from the plane.

Solution:

Distance $= \frac{|(2)(-6) + (-3)(0) + (6)(0) + (-2)|}{\sqrt{2^2 + (-3)^2 + 6^2}}$

Distance $= \frac{|-12 + 0 + 0 - 2|}{\sqrt{4 + 9 + 36}} = \frac{|-14|}{\sqrt{49}} = \frac{14}{7} = 2$

The distance is 2 units.

Given:

A line makes angles α, β, γ, and δ with the four diagonals of a cube.

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To Prove:

$\cos^2 α + \cos^2 β + \cos^2 γ + \cos^2 δ = \frac{4}{3}$

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Proof:

Consider a cube with side length $a$. Let one vertex be at the origin O(0, 0, 0). The coordinates of the vertices can be taken as (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a), (a, a, 0), (a, 0, a), (0, a, a), (a, a, a).

The four diagonals of the cube are connecting opposite vertices.

Diagonal 1 ($d_1$): From O(0, 0, 0) to A(a, a, a).

Vector along $d_1$: $\vec{v}_1 = a\hat{i} + a\hat{j} + a\hat{k}$. Direction ratios are $(a, a, a)$, or simpler, $(1, 1, 1)$.

Direction cosines of $d_1$: $\left(\frac{1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Diagonal 2 ($d_2$): From $(a, 0, 0)$ to $(0, a, a)$.

Vector along $d_2$: $\vec{v}_2 = (0-a)\hat{i} + (a-0)\hat{j} + (a-0)\hat{k} = -a\hat{i} + a\hat{j} + a\hat{k}$. Direction ratios are $(-a, a, a)$, or simpler, $(-1, 1, 1)$.

Direction cosines of $d_2$: $\left(\frac{-1}{\sqrt{(-1)^2+1^2+1^2}}, \frac{1}{\sqrt{(-1)^2+1^2+1^2}}, \frac{1}{\sqrt{(-1)^2+1^2+1^2}}\right) = \left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Diagonal 3 ($d_3$): From $(0, a, 0)$ to $(a, 0, a)$.

Vector along $d_3$: $\vec{v}_3 = (a-0)\hat{i} + (0-a)\hat{j} + (a-0)\hat{k} = a\hat{i} - a\hat{j} + a\hat{k}$. Direction ratios are $(a, -a, a)$, or simpler, $(1, -1, 1)$.

Direction cosines of $d_3$: $\left(\frac{1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{-1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1}{\sqrt{1^2+(-1)^2+1^2}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Diagonal 4 ($d_4$): From $(0, 0, a)$ to $(a, a, 0)$.

Vector along $d_4$: $\vec{v}_4 = (a-0)\hat{i} + (a-0)\hat{j} + (0-a)\hat{k} = a\hat{i} + a\hat{j} - a\hat{k}$. Direction ratios are $(a, a, -a)$, or simpler, $(1, 1, -1)$.

Direction cosines of $d_4$: $\left(\frac{1}{\sqrt{1^2+1^2+(-1)^2}}, \frac{1}{\sqrt{1^2+1^2+(-1)^2}}, \frac{-1}{\sqrt{1^2+1^2+(-1)^2}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$.

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Let the direction cosines of the given line be $(l, m, n)$. We know that $l^2 + m^2 + n^2 = 1$.

The cosine of the angle between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$.

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The angle $\alpha$ is between the line $(l, m, n)$ and $d_1 \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$:

$\cos \alpha = l\left(\frac{1}{\sqrt{3}}\right) + m\left(\frac{1}{\sqrt{3}}\right) + n\left(\frac{1}{\sqrt{3}}\right) = \frac{l+m+n}{\sqrt{3}}$

$\cos^2 \alpha = \frac{(l+m+n)^2}{3}$

The angle $\beta$ is between the line $(l, m, n)$ and $d_2 \left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$:

$\cos \beta = l\left(\frac{-1}{\sqrt{3}}\right) + m\left(\frac{1}{\sqrt{3}}\right) + n\left(\frac{1}{\sqrt{3}}\right) = \frac{-l+m+n}{\sqrt{3}}$

$\cos^2 \beta = \frac{(-l+m+n)^2}{3}$

The angle $\gamma$ is between the line $(l, m, n)$ and $d_3 \left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$:

$\cos \gamma = l\left(\frac{1}{\sqrt{3}}\right) + m\left(\frac{-1}{\sqrt{3}}\right) + n\left(\frac{1}{\sqrt{3}}\right) = \frac{l-m+n}{\sqrt{3}}$

$\cos^2 \gamma = \frac{(l-m+n)^2}{3}$

The angle $\delta$ is between the line $(l, m, n)$ and $d_4 \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$:

$\cos \delta = l\left(\frac{1}{\sqrt{3}}\right) + m\left(\frac{1}{\sqrt{3}}\right) + n\left(\frac{-1}{\sqrt{3}}\right) = \frac{l+m-n}{\sqrt{3}}$

$\cos^2 \delta = \frac{(l+m-n)^2}{3}$

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Now, sum the squared cosines:

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3}[(l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2]$

Expand the squares: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$

$(l+m+n)^2 = l^2 + m^2 + n^2 + 2lm + 2mn + 2nl$

$(-l+m+n)^2 = (-l)^2 + m^2 + n^2 + 2(-l)m + 2mn + 2n(-l) = l^2 + m^2 + n^2 - 2lm + 2mn - 2nl$

$(l-m+n)^2 = l^2 + (-m)^2 + n^2 + 2l(-m) + 2(-m)n + 2nl = l^2 + m^2 + n^2 - 2lm - 2mn + 2nl$

$(l+m-n)^2 = l^2 + m^2 + (-n)^2 + 2lm + 2m(-n) + 2(-n)l = l^2 + m^2 + n^2 + 2lm - 2mn - 2nl$

Sum the expanded terms:

$(l^2 + m^2 + n^2) + (l^2 + m^2 + n^2) + (l^2 + m^2 + n^2) + (l^2 + m^2 + n^2) = 4(l^2 + m^2 + n^2)$

$(2lm - 2lm - 2lm + 2lm) = 0lm$

$(2mn + 2mn - 2mn - 2mn) = 0mn$

$(2nl - 2nl + 2nl - 2nl) = 0nl$

Sum of the squares = $4(l^2 + m^2 + n^2)$.

Since $(l, m, n)$ are direction cosines, $l^2 + m^2 + n^2 = 1$.

Sum of the squares = $4(1) = 4$.

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Therefore, $\cos^2 α + \cos^2 β + \cos^2 γ + \cos^2 δ = \frac{1}{3}[4(l^2 + m^2 + n^2)] = \frac{1}{3}[4(1)] = \frac{4}{3}$.

Hence Proved.

Given:

The required plane passes through the point P(1, – 1, 2).

The required plane is perpendicular to Plane 1: $2x + 3y - 2z = 5$.

The required plane is perpendicular to Plane 2: $x + 2y - 3z = 8$.

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To Find:

The equation of the required plane.

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Solution:

Let the equation of the required plane be $Ax + By + Cz + D = 0$.

The normal vector to this plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

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The normal vector to Plane 1 ($2x + 3y - 2z = 5$) is $\vec{n}_1 = 2\hat{i} + 3\hat{j} - 2\hat{k}$.

The normal vector to Plane 2 ($x + 2y - 3z = 8$) is $\vec{n}_2 = \hat{i} + 2\hat{j} - 3\hat{k}$.

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Since the required plane is perpendicular to Plane 1, their normal vectors are perpendicular. The dot product of their normal vectors is zero:

$\vec{n} \cdot \vec{n}_1 = 0$

$(A\hat{i} + B\hat{j} + C\hat{k}) \cdot (2\hat{i} + 3\hat{j} - 2\hat{k}) = 0$

Since the required plane is perpendicular to Plane 2, their normal vectors are perpendicular. The dot product of their normal vectors is zero:

$\vec{n} \cdot \vec{n}_2 = 0$

$(A\hat{i} + B\hat{j} + C\hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k}) = 0$

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We have a system of two linear equations in A, B, and C:

$2A + 3B - 2C = 0$

$A + 2B - 3C = 0$

We can solve for the ratios of A, B, and C using cross-multiplication:

$\frac{A}{(3)(-3) - (-2)(2)} = \frac{-B}{(2)(-3) - (-2)(1)} = \frac{C}{(2)(2) - (3)(1)}$

$\frac{A}{-9 + 4} = \frac{-B}{-6 + 2} = \frac{C}{4 - 3}$

$\frac{A}{-5} = \frac{-B}{-4} = \frac{C}{1}$

$\frac{A}{-5} = \frac{B}{4} = \frac{C}{1}$

The direction ratios of the normal vector are proportional to $(-5, 4, 1)$. We can take $A = -5k, B = 4k, C = 1k$ for some constant k. We can choose $k=1$ (or any non-zero value) for simplicity, so $A = -5, B = 4, C = 1$.

The normal vector to the required plane is $\vec{n} = -5\hat{i} + 4\hat{j} + \hat{k}$.

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The equation of the required plane is of the form $-5x + 4y + z + D = 0$.

The plane contains the point P(1, – 1, 2). Substitute the coordinates of this point into the plane equation to find D:

$-5(1) + 4(-1) + (2) + D = 0$

$-5 - 4 + 2 + D = 0$

$-9 + 2 + D = 0$

$-7 + D = 0$

$D = 7$

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The equation of the required plane is $-5x + 4y + z + 7 = 0$.

Multiplying by -1 (optional, for a positive leading coefficient):

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Vector Equation (Optional):

The position vector of the point P is $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$.

The normal vector is $\vec{n} = -5\hat{i} + 4\hat{j} + \hat{k}$.

The vector equation is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

$\vec{a} \cdot \vec{n} = (\hat{i} - \hat{j} + 2\hat{k}) \cdot (-5\hat{i} + 4\hat{j} + \hat{k})$

$= (1)(-5) + (-1)(4) + (2)(1) = -5 - 4 + 2 = -7$

The vector equation is $\vec{r} \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = -7$.

Multiplying by -1:

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The equation of the plane is $5x - 4y - z - 7 = 0$.

Given:

Point P(6, 5, 9).

The plane is determined by points A(3, – 1, 2), B(5, 2, 4), and C(– 1, – 1, 6).

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To Find:

The distance between point P and the plane determined by A, B, and C.

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Solution:

First, we find the equation of the plane passing through the points A, B, and C.

Let $(x_1, y_1, z_1) = (3, -1, 2)$, $(x_2, y_2, z_2) = (5, 2, 4)$, and $(x_3, y_3, z_3) = (-1, -1, 6)$.

The Cartesian equation of a plane passing through three non-collinear points is given by:

$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$

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Calculate the difference vectors:

$x_2 - x_1 = 5 - 3 = 2$

$y_2 - y_1 = 2 - (-1) = 3$

$z_2 - z_1 = 4 - 2 = 2$

$x_3 - x_1 = -1 - 3 = -4$

$y_3 - y_1 = -1 - (-1) = 0$

$z_3 - z_1 = 6 - 2 = 4$

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Substitute into the determinant:

$\begin{vmatrix} x - 3 & y - (-1) & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0$

$\begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0$

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Evaluate the determinant (expanding along the second row or third column for simplicity):

Using the first row expansion:

$(x - 3) \begin{vmatrix} 3 & 2 \\ 0 & 4 \end{vmatrix} - (y + 1) \begin{vmatrix} 2 & 2 \\ -4 & 4 \end{vmatrix} + (z - 2) \begin{vmatrix} 2 & 3 \\ -4 & 0 \end{vmatrix} = 0$

$(x - 3)((3)(4) - (2)(0)) - (y + 1)((2)(4) - (2)(-4)) + (z - 2)((2)(0) - (3)(-4)) = 0$

$(x - 3)(12 - 0) - (y + 1)(8 + 8) + (z - 2)(0 + 12) = 0$

$12(x - 3) - 16(y + 1) + 12(z - 2) = 0$

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Divide the entire equation by the greatest common divisor, 4:

$3(x - 3) - 4(y + 1) + 3(z - 2) = 0$

Expand and simplify:

$3x - 9 - 4y - 4 + 3z - 6 = 0$

$3x - 4y + 3z - 9 - 4 - 6 = 0$

$3x - 4y + 3z - 19 = 0$

This is the equation of the plane in the form $Ax + By + Cz + D = 0$, with $A=3, B=-4, C=3, D=-19$.

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Now, find the distance of the point P(6, 5, 9) from the plane $3x - 4y + 3z - 19 = 0$.

The formula for the distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Here, $(x_1, y_1, z_1) = (6, 5, 9)$ and $A=3, B=-4, C=3, D=-19$.

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Calculate the numerator and the denominator:

$|Ax_1 + By_1 + Cz_1 + D| = |(3)(6) + (-4)(5) + (3)(9) + (-19)|$

$= |18 - 20 + 27 - 19| = |-2 + 27 - 19| = |25 - 19| = |6| = 6$

$\sqrt{A^2 + B^2 + C^2} = \sqrt{3^2 + (-4)^2 + 3^2} = \sqrt{9 + 16 + 9} = \sqrt{34}$

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Distance $= \frac{6}{\sqrt{34}}$

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To rationalize the denominator:

Distance $= \frac{6}{\sqrt{34}} \times \frac{\sqrt{34}}{\sqrt{34}} = \frac{6\sqrt{34}}{34} = \frac{3\sqrt{34}}{17}$

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The distance between the point P(6, 5, 9) and the plane is $\frac{6}{\sqrt{34}}$ units or $\frac{3\sqrt{34}}{17}$ units.

Given:

Line 1: $\frac{x − a + d}{α − δ} = \frac{y − a}{α} = \frac{z − a − d}{α + δ}$

Line 2: $\frac{x − b + c}{β − γ} = \frac{y − b}{β} = \frac{z − b − c}{β + γ}$

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To Show:

The two lines are coplanar.

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Solution:

The Cartesian equation of a line is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are its direction ratios.

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For Line 1: $\frac{x − (a - d)}{α − δ} = \frac{y − a}{α} = \frac{z − (a + d)}{α + δ}$

A point on Line 1 is $(x_1, y_1, z_1) = (a - d, a, a + d)$.

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (α - δ, α, α + δ)$.

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For Line 2: $\frac{x − (b - c)}{β − γ} = \frac{y − b}{β} = \frac{z − (b + c)}{β + γ}$

A point on Line 2 is $(x_2, y_2, z_2) = (b - c, b, b + c)$.

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (β − γ, β, β + γ)$.

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Calculate the vector connecting a point on Line 1 to a point on Line 2:

$x_2 - x_1 = (b - c) - (a - d) = b - c - a + d$

$y_2 - y_1 = b - a$

$z_2 - z_1 = (b + c) - (a + d) = b + c - a - d$

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ respectively are coplanar if and only if:

$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$

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Substitute the values into the determinant:

$\begin{vmatrix} b - c - a + d & b - a & b + c - a - d \\ α - δ & α & α + δ \\ β - γ & β & β + γ \end{vmatrix}$

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Perform the column operation $C_1 \to C_1 + C_3$ on the determinant:

The first element of the new first column is $(b - c - a + d) + (b + c - a - d) = 2b - 2a = 2(b - a)$.

The second element of the new first column is $(α - δ) + (α + δ) = 2α$.

The third element of the new first column is $(β - γ) + (β + γ) = 2β$.

The determinant becomes:

$\begin{vmatrix} 2(b - a) & b - a & b + c - a - d \\ 2α & α & α + δ \\ 2β & β & β + γ \end{vmatrix}$

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Factor out the common factor of 2 from the first column:

$2 \begin{vmatrix} b - a & b - a & b + c - a - d \\ α & α & α + δ \\ β & β & β + γ \end{vmatrix}$

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Observe that the first column $\begin{pmatrix} b - a \\ α \\ β \end{pmatrix}$ and the second column $\begin{pmatrix} b - a \\ α \\ β \end{pmatrix}$ of this determinant are identical.

A determinant with two identical columns is equal to zero.

Therefore, the value of the determinant is $2 \times 0 = 0$.

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Since the determinant is 0, the two lines are coplanar.

Hence Showed.

Given:

Line passes through points A(3, 4, 1) and B(5, 1, 6).

The plane is the XY-plane.

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To Find:

The coordinates of the point where the line crosses the XY-plane.

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Solution:

First, find the Cartesian equation of the line passing through points A(3, 4, 1) and B(5, 1, 6).

The formula for the equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of points A(3, 4, 1) and B(5, 1, 6):

$\frac{x - 3}{5 - 3} = \frac{y - 4}{1 - 4} = \frac{z - 1}{6 - 1}$

This is the equation of the line AB.

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The equation of the XY-plane is $z = 0$.

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To find the point where the line crosses the XY-plane, substitute $z = 0$ into the equation of the line:

$\frac{x - 3}{2} = \frac{y - 4}{-3} = \frac{0 - 1}{5}$

$\frac{x - 3}{2} = \frac{y - 4}{-3} = \frac{-1}{5}$

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Equate the first part to $\frac{-1}{5}$ and solve for x:

$\frac{x - 3}{2} = \frac{-1}{5}$

$5(x - 3) = 2(-1)$

$5x - 15 = -2$

$5x = 15 - 2$

$5x = 13$

$x = \frac{13}{5}$

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Equate the second part to $\frac{-1}{5}$ and solve for y:

$\frac{y - 4}{-3} = \frac{-1}{5}$

$5(y - 4) = (-3)(-1)$

$5y - 20 = 3$

$5y = 20 + 3$

$5y = 23$

$y = \frac{23}{5}$

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The z-coordinate of the point is given by the equation of the XY-plane, which is $z = 0$.

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The coordinates of the point where the line crosses the XY-plane are $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$.

Given:

Line 1 passes through points O(0, 0, 0) and A(2, 1, 1).

Line 2 passes through points B(3, 5, – 1) and C(4, 3, – 1).

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To Show:

Line 1 is perpendicular to Line 2.

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Solution:

The direction ratios (DRs) of a line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

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Direction ratios of Line 1 (through O(0, 0, 0) and A(2, 1, 1)):

$a_1 = 2 - 0 = 2$

$b_1 = 1 - 0 = 1$

$c_1 = 1 - 0 = 1$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (2, 1, 1)$.

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Direction ratios of Line 2 (through B(3, 5, – 1) and C(4, 3, – 1)):

$a_2 = 4 - 3 = 1$

$b_2 = 3 - 5 = -2$

$c_2 = -1 - (-1) = -1 + 1 = 0$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (1, -2, 0)$.

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Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Let's calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$ for the given lines:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(1) + (1)(-2) + (1)(0)$

$= 2 - 2 + 0$

$= 0$

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Since the sum of the products of their corresponding direction ratios is zero, i.e., $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, the two lines are perpendicular.

Hence, the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

Given:

Two mutually perpendicular lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.

Since the lines are perpendicular, $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

Also, as they are direction cosines, $l_1^2 + m_1^2 + n_1^2 = 1$ and $l_2^2 + m_2^2 + n_2^2 = 1$.

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To Show:

The direction cosines of the line perpendicular to both are $(m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1)$.

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Proof:

Let the direction cosines of the line perpendicular to both given lines be $(l, m, n)$.

Since this line is perpendicular to the first line (with d.c.s $(l_1, m_1, n_1)$), the dot product of their direction cosines must be zero:

Since this line is perpendicular to the second line (with d.c.s $(l_2, m_2, n_2)$), the dot product of their direction cosines must be zero:

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We can solve equations (1) and (2) for the ratios $l:m:n$ using cross-multiplication:

From the coefficients of $l, m, n$ in equations (1) and (2):

$\frac{l}{m_1 n_2 - m_2 n_1} = \frac{m}{n_1 l_2 - n_2 l_1} = \frac{n}{l_1 m_2 - l_2 m_1}$

Let the common ratio be $k$. Then the direction ratios of the line perpendicular to both are proportional to $(m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1)$.

The direction cosines $(l, m, n)$ are obtained by dividing the direction ratios by the magnitude of the vector formed by these direction ratios.

Let $a = m_1 n_2 - m_2 n_1$, $b = n_1 l_2 - n_2 l_1$, $c = l_1 m_2 - l_2 m_1$.

The direction cosines are $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$.

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We need to show that $\sqrt{a^2+b^2+c^2} = 1$. This is the magnitude of the cross product of the vectors $\vec{d}_1 = l_1\hat{i} + m_1\hat{j} + n_1\hat{k}$ and $\vec{d}_2 = l_2\hat{i} + m_2\hat{j} + n_2\hat{k}$.

$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = (m_1 n_2 - m_2 n_1)\hat{i} - (l_1 n_2 - n_1 l_2)\hat{j} + (l_1 m_2 - m_1 l_2)\hat{k}$

$\vec{d}_1 \times \vec{d}_2 = (m_1 n_2 - m_2 n_1)\hat{i} + (n_1 l_2 - n_2 l_1)\hat{j} + (l_1 m_2 - l_2 m_1)\hat{k}$

So, $a = m_1 n_2 - m_2 n_1$, $b = n_1 l_2 - n_2 l_1$, $c = l_1 m_2 - l_2 m_1$.

The magnitude is $|\vec{d}_1 \times \vec{d}_2| = \sqrt{(m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 m_2 - l_2 m_1)^2}$.

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We also know that $|\vec{d}_1 \times \vec{d}_2| = |\vec{d}_1| |\vec{d}_2| \sin \phi$, where $\phi$ is the angle between $\vec{d}_1$ and $\vec{d}_2$.

Since $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are direction cosines, $|\vec{d}_1| = \sqrt{l_1^2+m_1^2+n_1^2} = 1$ and $|\vec{d}_2| = \sqrt{l_2^2+m_2^2+n_2^2} = 1$.

Since the two lines are mutually perpendicular, the angle between them is $\phi = 90^\circ$. So, $\sin \phi = \sin 90^\circ = 1$.

$|\vec{d}_1 \times \vec{d}_2| = (1)(1)(1) = 1$.

---

Thus, $\sqrt{(m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 m_2 - l_2 m_1)^2} = 1$.

The direction cosines $(l, m, n)$ are $\left(\frac{a}{1}, \frac{b}{1}, \frac{c}{1}\right) = (a, b, c)$.

So, the direction cosines are $(m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1)$.

---

Hence Proved.

Given:

Direction ratios of Line 1: $(a_1, b_1, c_1) = (a, b, c)$.

Direction ratios of Line 2: $(a_2, b_2, c_2) = (b - c, c - a, a - b)$.

---

To Find:

The angle between the two lines.

---

Solution:

The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:

$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$

---

Calculate the numerator $a_1 a_2 + b_1 b_2 + c_1 c_2$:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (a)(b - c) + (b)(c - a) + (c)(a - b)$

$= ab - ac + bc - ba + ca - cb$

$= (ab - ba) + (bc - cb) + (-ac + ca)$

$= 0 + 0 + 0 = 0$

---

Since $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$, the dot product of the direction vectors is zero. This means the direction vectors are perpendicular.

The angle $\theta$ such that $\cos \theta = 0$ is $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians.

---

The angle between the lines is $90^\circ$ or $\frac{\pi}{2}$ radians.

Given:

The line is parallel to the x-axis.

The line passes through the origin (0, 0, 0).

---

To Find:

The equation of the line (in vector and Cartesian form).

---

Solution:

A line parallel to the x-axis has a direction vector parallel to the positive or negative x-axis. The direction vector of the x-axis is $\hat{i}$.

So, the direction vector of the required line can be taken as $\vec{b} = \hat{i}$. The direction ratios are (1, 0, 0).

---

The line passes through the origin (0, 0, 0). The position vector of this point is $\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$.

---

Vector Equation:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

Substitute $\vec{a} = \vec{0}$ and $\vec{b} = \hat{i}$:

$\vec{r} = \vec{0} + \lambda (\hat{i})$

where $\lambda$ is a scalar parameter.

---

Cartesian Equation:

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

From the vector equation: $x\hat{i} + y\hat{j} + z\hat{k} = \lambda \hat{i} = \lambda \hat{i} + 0\hat{j} + 0\hat{k}$.

Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$:

$x = \lambda$

$y = 0$

$z = 0$

The Cartesian equations of the line are $y=0$ and $z=0$. The x-coordinate can be any real number, represented by $\lambda$. This defines the x-axis itself.

Using the symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ with point (0, 0, 0) and direction ratios (1, 0, 0):

$\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}$

This notation with zeros in the denominator means that the numerator must be zero for the corresponding term, while the other term represents the varying coordinate.

So, $y - 0 = 0 \implies y = 0$.

And $z - 0 = 0 \implies z = 0$.

The term $\frac{x-0}{1}$ indicates that $x$ is the varying parameter (which is $\lambda$).

The Cartesian equations are $y = 0$ and $z = 0$.

---

The equation of the line parallel to the x-axis and passing through the origin is $\vec{r} = \lambda \hat{i}$ (vector form) or $y = 0, z = 0$ (Cartesian form).

Given:

Point A(1, 2, 3)

Point B(4, 5, 7)

Point C(– 4, 3, – 6)

Point D(2, 9, 2)

---

To Find:

The angle between the line AB and the line CD.

---

Solution:

The direction ratios (DRs) of a line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

---

Direction ratios of Line AB (through A(1, 2, 3) and B(4, 5, 7)):

$a_1 = 4 - 1 = 3$

$b_1 = 5 - 2 = 3$

$c_1 = 7 - 3 = 4$

The direction ratios of Line AB are $(a_1, b_1, c_1) = (3, 3, 4)$.

---

Direction ratios of Line CD (through C(– 4, 3, – 6) and D(2, 9, 2)):

$a_2 = 2 - (-4) = 2 + 4 = 6$

$b_2 = 9 - 3 = 6$

$c_2 = 2 - (-6) = 2 + 6 = 8$

The direction ratios of Line CD are $(a_2, b_2, c_2) = (6, 6, 8)$.

---

Observe that the direction ratios of CD $(6, 6, 8)$ are proportional to the direction ratios of AB $(3, 3, 4)$ since $6 = 2 \times 3$, $6 = 2 \times 3$, and $8 = 2 \times 4$. This indicates that the lines AB and CD are parallel.

The angle between two parallel lines is $0^\circ$ or $\pi$ radians (depending on the direction sense, but typically the smallest angle is considered). Let's confirm this using the formula for the angle between lines.

---

The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:

$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$

---

Calculate $a_1 a_2 + b_1 b_2 + c_1 c_2$:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (3)(6) + (3)(6) + (4)(8)$

$= 18 + 18 + 32 = 68$

---

Calculate the magnitudes of the direction vectors:

$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{3^2 + 3^2 + 4^2} = \sqrt{9 + 9 + 16} = \sqrt{34}$

$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{6^2 + 6^2 + 8^2} = \sqrt{36 + 36 + 64} = \sqrt{136}$

Simplify $\sqrt{136}$: $\sqrt{136} = \sqrt{4 \times 34} = \sqrt{4} \times \sqrt{34} = 2\sqrt{34}$.

---

Substitute the values into the formula for $\cos \theta$:

$\cos \theta = \frac{|68|}{\sqrt{34} \times 2\sqrt{34}}$

$\cos \theta = \frac{68}{2 \times (\sqrt{34})^2} = \frac{68}{2 \times 34} = \frac{68}{68} = 1$

---

Since $\cos \theta = 1$, the angle $\theta = \cos^{-1}(1) = 0^\circ$.

---

The angle between the lines AB and CD is $0^\circ$. This confirms that the lines are parallel.

Given:

Line 1: $\frac{x − 1}{−3} = \frac{y − 2}{2k} = \frac{z − 3}{2}$

Line 2: $\frac{x − 1}{3k} = \frac{y − 1}{1} = \frac{z − 6}{−5}$

The lines are perpendicular.

---

To Find:

The value of k.

---

Solution:

The Cartesian equation of a line in the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ has direction ratios $(a, b, c)$.

---

For Line 1: $\frac{x − 1}{−3} = \frac{y − 2}{2k} = \frac{z − 3}{2}$

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (-3, 2k, 2)$.

---

For Line 2: $\frac{x − 1}{3k} = \frac{y − 1}{1} = \frac{z − 6}{−5}$

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (3k, 1, -5)$.

---

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if the sum of the products of their corresponding direction ratios is zero:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

---

Substitute the direction ratios into the perpendicularity condition:

$(-3)(3k) + (2k)(1) + (2)(-5) = 0$

$-9k + 2k - 10 = 0$

$-7k - 10 = 0$

---

Solve for k:

$-7k = 10$

$k = -\frac{10}{7}$

---

The value of k for which the lines are perpendicular is $-\frac{10}{7}$.

Given:

The line passes through the point P(1, 2, 3).

The line is perpendicular to the plane $\vec{r} \;.\; (\hat{i} + 2\hat{j} − 5\hat{k}) + 9 = 0$.

---

To Find:

The vector equation of the line.

---

Solution:

The position vector of the point P(1, 2, 3) on the line is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

---

The equation of the plane is $\vec{r} \;.\; (\hat{i} + 2\hat{j} − 5\hat{k}) + 9 = 0$. This can be written as $\vec{r} \;.\; (\hat{i} + 2\hat{j} − 5\hat{k}) = -9$.

The normal vector to this plane is $\vec{n} = \hat{i} + 2\hat{j} − 5\hat{k}$.

---

A line perpendicular to a plane is parallel to the normal vector of the plane.

Therefore, the direction vector of the required line is parallel to the normal vector $\vec{n}$. We can take the direction vector $\vec{b}$ of the line to be equal to the normal vector $\vec{n}$.

So, $\vec{b} = \hat{i} + 2\hat{j} − 5\hat{k}$.

---

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

$\vec{r} = \vec{a} + \lambda \vec{b}$

where $\lambda$ is a scalar parameter.

---

Substitute the values of $\vec{a}$ and $\vec{b}$ into the formula:

---

This is the required vector equation of the line.

Given:

The plane passes through the point P(a, b, c).

The plane is parallel to the plane $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 2$.

---

To Find:

The equation of the required plane (in vector and Cartesian form).

---

Solution:

The position vector of the point P(a, b, c) on the required plane is $\vec{a}_0 = a\hat{i} + b\hat{j} + c\hat{k}$.

---

The equation of the given plane is $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 2$.

The normal vector to this plane is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.

---

Since the required plane is parallel to the given plane, their normal vectors are parallel. We can use the same normal vector $\vec{n} = \hat{i} + \hat{j} + \hat{k}$ for the required plane.

---

Vector Equation:

The vector equation of a plane passing through a point with position vector $\vec{a}_0$ and having a normal vector $\vec{n}$ is given by:

$\vec{r} \cdot \vec{n} = \vec{a}_0 \cdot \vec{n}$

Calculate the dot product $\vec{a}_0 \cdot \vec{n}$:

$\vec{a}_0 \cdot \vec{n} = (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$

$= (a)(1) + (b)(1) + (c)(1) = a + b + c$

Substitute the values into the vector equation formula:

This is the required vector equation of the plane.

---

Cartesian Equation:

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = a + b + c$

Perform the dot product:

$x(1) + y(1) + z(1) = a + b + c$

This is the required Cartesian equation of the plane.

Given:

Equation of Line 1:

$\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + λ (\hat{i} − 2\hat{j} + 2\hat{k})$

Equation of Line 2:

$\vec{r} = −4\hat{i} − \hat{k} + µ (3\hat{i} − 2\hat{j} − 2\hat{k})$

---

To Find:

The shortest distance between the two lines.

---

Solution:

The given lines are in the form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$.

From the equation of Line 1, we have:

$\vec{a}_1 = 6\hat{i} + 2\hat{j} + 2\hat{k}$

$\vec{b}_1 = \hat{i} − 2\hat{j} + 2\hat{k}$

From the equation of Line 2, we have:

$\vec{a}_2 = -4\hat{i} - \hat{k}$

$\vec{b}_2 = 3\hat{i} − 2\hat{j} − 2\hat{k}$

---

Check if the lines are parallel. The direction ratios of $\vec{b}_1$ are $(1, -2, 2)$ and the direction ratios of $\vec{b}_2$ are $(3, -2, -2)$.

The ratios of the corresponding direction ratios are $\frac{1}{3}$, $\frac{-2}{-2} = 1$, $\frac{2}{-2} = -1$. Since these ratios are not equal, the lines are skew lines.

The formula for the shortest distance ($d$) between two skew lines is:

$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$

---

First, calculate the vector $(\vec{a}_2 - \vec{a}_1)$:

$\vec{a}_2 - \vec{a}_1 = (-4\hat{i} - \hat{k}) - (6\hat{i} + 2\hat{j} + 2\hat{k})$

$\vec{a}_2 - \vec{a}_1 = (-4-6)\hat{i} + (0-2)\hat{j} + (-1-2)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = -10\hat{i} - 2\hat{j} - 3\hat{k}$

---

Next, calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$:

$\vec{b}_1 \times \vec{b}_2 = (\hat{i} − 2\hat{j} + 2\hat{k}) \times (3\hat{i} − 2\hat{j} − 2\hat{k})$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}((-2)(-2) - (2)(-2)) - \hat{j}((1)(-2) - (2)(3)) + \hat{k}((1)(-2) - (-2)(3))$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}(4 + 4) - \hat{j}(-2 - 6) + \hat{k}(-2 + 6)$

$\vec{b}_1 \times \vec{b}_2 = 8\hat{i} + 8\hat{j} + 4\hat{k}$

---

Now, calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-10\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (8\hat{i} + 8\hat{j} + 4\hat{k})$

$= (-10)(8) + (-2)(8) + (-3)(4)$

$= -80 - 16 - 12 = -108$

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Calculate the magnitude of the cross product $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = |8\hat{i} + 8\hat{j} + 4\hat{k}|$

$= \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$

---

Substitute the values into the shortest distance formula:

$d = \left| \frac{-108}{12} \right| = \frac{|-108|}{|12|} = \frac{108}{12}$

$d = 9$

---

The shortest distance between the lines is 9 units.

Given:

Line passes through points A(5, 1, 6) and B(3, 4, 1).

The plane is the YZ-plane.

---

To Find:

The coordinates of the point where the line crosses the YZ-plane.

---

Solution:

First, find the Cartesian equation of the line passing through points A(5, 1, 6) and B(3, 4, 1).

The formula for the equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of points A(5, 1, 6) and B(3, 4, 1):

$\frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6}$

This is the equation of the line AB.

---

The equation of the YZ-plane is $x = 0$.

---

To find the point where the line crosses the YZ-plane, substitute $x = 0$ into the equation of the line:

$\frac{0 - 5}{-2} = \frac{y - 1}{3} = \frac{z - 6}{-5}$

$\frac{-5}{-2} = \frac{y - 1}{3} = \frac{z - 6}{-5}$

$\frac{5}{2} = \frac{y - 1}{3} = \frac{z - 6}{-5}$

---

Equate the second part to $\frac{5}{2}$ and solve for y:

$\frac{y - 1}{3} = \frac{5}{2}$

$2(y - 1) = 3(5)$

$2y - 2 = 15$

$2y = 15 + 2$

$2y = 17$

$y = \frac{17}{2}$

---

Equate the third part to $\frac{5}{2}$ and solve for z:

$\frac{z - 6}{-5} = \frac{5}{2}$

$2(z - 6) = (-5)(5)$

$2z - 12 = -25$

$2z = -25 + 12$

$2z = -13$

$z = -\frac{13}{2}$

---

The x-coordinate of the point is given by the equation of the YZ-plane, which is $x = 0$.

---

The coordinates of the point where the line crosses the YZ-plane are $\left(0, \frac{17}{2}, -\frac{13}{2}\right)$.

Given:

Line passes through points A(5, 1, 6) and B(3, 4, 1).

The plane is the ZX-plane.

---

To Find:

The coordinates of the point where the line crosses the ZX-plane.

---

Solution:

First, find the Cartesian equation of the line passing through points A(5, 1, 6) and B(3, 4, 1).

The formula for the equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of points A(5, 1, 6) and B(3, 4, 1):

$\frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6}$

This is the equation of the line AB.

---

The equation of the ZX-plane is $y = 0$.

---

To find the point where the line crosses the ZX-plane, substitute $y = 0$ into the equation of the line:

$\frac{x - 5}{-2} = \frac{0 - 1}{3} = \frac{z - 6}{-5}$

$\frac{x - 5}{-2} = \frac{-1}{3} = \frac{z - 6}{-5}$

---

Equate the first part to $\frac{-1}{3}$ and solve for x:

$\frac{x - 5}{-2} = \frac{-1}{3}$

$3(x - 5) = (-2)(-1)$

$3x - 15 = 2$

$3x = 15 + 2$

$3x = 17$

$x = \frac{17}{3}$

---

Equate the third part to $\frac{-1}{3}$ and solve for z:

$\frac{z - 6}{-5} = \frac{-1}{3}$

$3(z - 6) = (-5)(-1)$

$3z - 18 = 5$

$3z = 18 + 5$

$3z = 23$

$z = \frac{23}{3}$

---

The y-coordinate of the point is given by the equation of the ZX-plane, which is $y = 0$.

---

The coordinates of the point where the line crosses the ZX-plane are $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

Given:

Line passes through points A(3, – 4, – 5) and B(2, – 3, 1).

The plane is $2x + y + z = 7$.

---

To Find:

The coordinates of the point where the line crosses the plane.

---

Solution:

First, find the Cartesian equation of the line passing through points A(3, – 4, – 5) and B(2, – 3, 1).

The formula for the equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

Substitute the coordinates of points A(3, – 4, – 5) and B(2, – 3, 1):

$\frac{x - 3}{2 - 3} = \frac{y - (-4)}{-3 - (-4)} = \frac{z - (-5)}{1 - (-5)}$

$\frac{x - 3}{-1} = \frac{y + 4}{-3 + 4} = \frac{z + 5}{1 + 5}$

This is the equation of the line.

---

Let the coordinates of the point of intersection be $(x, y, z)$. This point lies on the line and the plane.

From the equation of the line, any point on the line can be represented parametrically. Let:

$\frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda$

Then, the coordinates of a point on the line are:

$x - 3 = -1\lambda \implies x = 3 - \lambda$

$y + 4 = 1\lambda \implies y = -4 + \lambda$

$z + 5 = 6\lambda \implies z = -5 + 6\lambda$

---

Since the point $(x, y, z)$ lies on the plane $2x + y + z = 7$, substitute the parametric expressions for x, y, and z into the plane equation:

$2(3 - \lambda) + (-4 + \lambda) + (-5 + 6\lambda) = 7$

$6 - 2\lambda - 4 + \lambda - 5 + 6\lambda = 7$

Combine like terms:

$(-2\lambda + \lambda + 6\lambda) + (6 - 4 - 5) = 7$

$5\lambda + (-3) = 7$

$5\lambda - 3 = 7$

---

Solve for $\lambda$:

$5\lambda = 7 + 3$

$5\lambda = 10$

$\lambda = \frac{10}{5} = 2$

---

Substitute the value of $\lambda = 2$ back into the parametric coordinates to find the point of intersection:

$x = 3 - \lambda = 3 - 2 = 1$

$y = -4 + \lambda = -4 + 2 = -2$

$z = -5 + 6\lambda = -5 + 6(2) = -5 + 12 = 7$

---

The coordinates of the point where the line crosses the plane are $(1, -2, 7)$.

Given:

The required plane passes through the point P(– 1, 3, 2).

The required plane is perpendicular to Plane 1: $x + 2y + 3z = 5$.

The required plane is perpendicular to Plane 2: $3x + 3y + z = 0$.

---

To Find:

The equation of the required plane (in Cartesian and vector form).

---

Solution:

Let the normal vector to the required plane be $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

The normal vector to Plane 1 ($x + 2y + 3z = 5$) is $\vec{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$.

The normal vector to Plane 2 ($3x + 3y + z = 0$) is $\vec{n}_2 = 3\hat{i} + 3\hat{j} + \hat{k}$.

---

Since the required plane is perpendicular to both Plane 1 and Plane 2, its normal vector $\vec{n}$ must be perpendicular to both $\vec{n}_1$ and $\vec{n}_2$.

Therefore, the normal vector $\vec{n}$ is parallel to the cross product $\vec{n}_1 \times \vec{n}_2$.

Calculate the cross product $\vec{n}_1 \times \vec{n}_2$:

$\vec{n}_1 \times \vec{n}_2 = (\hat{i} + 2\hat{j} + 3\hat{k}) \times (3\hat{i} + 3\hat{j} + \hat{k})$

$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{vmatrix}$

$\vec{n}_1 \times \vec{n}_2 = \hat{i}((2)(1) - (3)(3)) - \hat{j}((1)(1) - (3)(3)) + \hat{k}((1)(3) - (2)(3))$

$\vec{n}_1 \times \vec{n}_2 = \hat{i}(2 - 9) - \hat{j}(1 - 9) + \hat{k}(3 - 6)$

$\vec{n}_1 \times \vec{n}_2 = -7\hat{i} - (-8)\hat{j} - 3\hat{k}$

$\vec{n}_1 \times \vec{n}_2 = -7\hat{i} + 8\hat{j} - 3\hat{k}$

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We can take the normal vector of the required plane as $\vec{n} = -7\hat{i} + 8\hat{j} - 3\hat{k}$. The direction ratios of the normal are $(-7, 8, -3)$.

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The required plane passes through the point P(– 1, 3, 2). The position vector of this point is $\vec{a} = -\hat{i} + 3\hat{j} + 2\hat{k}$.

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Vector Equation:

The vector equation of a plane passing through a point with position vector $\vec{a}$ and having a normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

Calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (-\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (-7\hat{i} + 8\hat{j} - 3\hat{k})$

$= (-1)(-7) + (3)(8) + (2)(-3)$

$= 7 + 24 - 6 = 25$

Substitute the values into the vector equation formula:

Multiplying by -1 (optional):

This is the vector equation of the required plane.

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Cartesian Equation:

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation $\vec{r} \cdot (-7\hat{i} + 8\hat{j} - 3\hat{k}) = 25$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-7\hat{i} + 8\hat{j} - 3\hat{k}) = 25$

Perform the dot product:

$x(-7) + y(8) + z(-3) = 25$

$-7x + 8y - 3z = 25$

Rewrite in the general form $Ax + By + Cz + D = 0$:

Multiplying by -1 (optional):

This is the Cartesian equation of the required plane.

Given:

Point $P_1$(1, 1, p).

Point $P_2$(– 3, 0, 1).

Plane: $\vec{r} \;.\; (3\hat{i} + 4\hat{j} − 12\hat{k}) + 13 = 0$.

Points $P_1$ and $P_2$ are equidistant from the plane.

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To Find:

The value of p.

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Solution:

Convert the vector equation of the plane to Cartesian form. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} + 4\hat{j} − 12\hat{k}) + 13 = 0$

$3x + 4y - 12z + 13 = 0$

This is the equation of the plane in the form $Ax + By + Cz + D = 0$, with $A=3, B=4, C=-12, D=13$.

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The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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Calculate the distance of point $P_1$(1, 1, p) from the plane:

Distance of $P_1 = \frac{|(3)(1) + (4)(1) + (-12)(p) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}}$

Distance of $P_1 = \frac{|3 + 4 - 12p + 13|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12p|}{\sqrt{169}} = \frac{|20 - 12p|}{13}$

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Calculate the distance of point $P_2$(– 3, 0, 1) from the plane:

Distance of $P_2 = \frac{|(3)(-3) + (4)(0) + (-12)(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}}$

Distance of $P_2 = \frac{|-9 + 0 - 12 + 13|}{\sqrt{9 + 16 + 144}} = \frac{|-21 + 13|}{\sqrt{169}} = \frac{|-8|}{13} = \frac{8}{13}$

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Since the two points are equidistant from the plane, their distances are equal:

$\frac{|20 - 12p|}{13} = \frac{8}{13}$

$|20 - 12p| = 8$

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This gives two possibilities:

Case 1: $20 - 12p = 8$

$-12p = 8 - 20$

$-12p = -12$

$p = \frac{-12}{-12} = 1$

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Case 2: $20 - 12p = -8$

$-12p = -8 - 20$

$-12p = -28$

$p = \frac{-28}{-12} = \frac{28}{12}$

Simplify the fraction by dividing by 4: $p = \frac{\cancel{28}^{7}}{\cancel{12}_{3}} = \frac{7}{3}$

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The possible values of p are 1 and $\frac{7}{3}$.

Given:

Equation of Plane 1: $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 1$

Equation of Plane 2: $\vec{r} \;.\; (2\hat{i} + 3\hat{j} − \hat{k}) + 4 = 0$

The required plane is parallel to the x-axis.

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To Find:

The equation of the required plane (in vector and Cartesian form).

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Solution:

The equation of a plane passing through the line of intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by:

$\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$

where $\lambda$ is a scalar parameter.

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From the given equations:

Plane 1: $\vec{r} \;.\; (\hat{i} + \hat{j} + \hat{k}) = 1$

Here, $\vec{n}_1 = \hat{i} + \hat{j} + \hat{k}$ and $d_1 = 1$.

Plane 2: $\vec{r} \;.\; (2\hat{i} + 3\hat{j} − \hat{k}) + 4 = 0$, which is $\vec{r} \;.\; (2\hat{i} + 3\hat{j} − \hat{k}) = -4$.

Here, $\vec{n}_2 = 2\hat{i} + 3\hat{j} − \hat{k}$ and $d_2 = -4$.

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Substituting these values into the formula for the plane passing through the intersection:

$\vec{r} \cdot ((\hat{i} + \hat{j} + \hat{k}) + \lambda (2\hat{i} + 3\hat{j} − \hat{k})) = 1 + \lambda (-4)$

$\vec{r} \cdot ((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 - \lambda)\hat{k}) = 1 - 4\lambda$

This is the vector equation of the family of planes passing through the line of intersection.

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The required plane is parallel to the x-axis. The direction vector of the x-axis is $\vec{b} = \hat{i}$.

A plane is parallel to a line if the normal vector of the plane is perpendicular to the direction vector of the line.

The normal vector of the required plane is $\vec{n} = (1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 - \lambda)\hat{k}$.

The condition for perpendicularity is $\vec{n} \cdot \hat{i} = 0$.

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Perform the dot product:

$((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 - \lambda)\hat{k}) \cdot (\hat{i}) = 0$

$(1 + 2\lambda)(1) + (1 + 3\lambda)(0) + (1 - \lambda)(0) = 0$

$1 + 2\lambda = 0$

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Solve for $\lambda$:

$2\lambda = -1$

$\lambda = -\frac{1}{2}$

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Substitute the value of $\lambda = -\frac{1}{2}$ back into the vector equation of the family of planes:

$\vec{r} \cdot ((1 + 2(-\frac{1}{2}))\hat{i} + (1 + 3(-\frac{1}{2}))\hat{j} + (1 - (-\frac{1}{2}))\hat{k}) = 1 - 4(-\frac{1}{2})$

$\vec{r} \cdot ((1 - 1)\hat{i} + (1 - \frac{3}{2})\hat{j} + (1 + \frac{1}{2})\hat{k}) = 1 + 2$

$\vec{r} \cdot (0\hat{i} - \frac{1}{2}\hat{j} + \frac{3}{2}\hat{k}) = 3$

Multiply by 2 to simplify the coefficients:

This is the required vector equation of the plane.

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Cartesian Equation:

To find the Cartesian equation, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation $\vec{r} \cdot (0\hat{i} - \hat{j} + 3\hat{k}) = 6$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (0\hat{i} - \hat{j} + 3\hat{k}) = 6$

Perform the dot product:

$x(0) + y(-1) + z(3) = 6$

$-y + 3z = 6$

Rewrite in the general form $Ax + By + Cz + D = 0$:

This is the required Cartesian equation of the plane.

Given:

Origin O(0, 0, 0).

Point P(1, 2, – 3).

The plane passes through point P.

The plane is perpendicular to the line segment OP.

---

To Find:

The equation of the required plane (in vector and Cartesian form).

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Solution:

The position vector of the point P(1, 2, – 3) is $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$.

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The line segment OP connects the origin O(0, 0, 0) to the point P(1, 2, – 3). The direction vector of the line OP is given by the vector $\vec{OP}$.

$\vec{OP} = (1 - 0)\hat{i} + (2 - 0)\hat{j} + (-3 - 0)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k}$.

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Since the plane is perpendicular to the line OP, the direction vector of OP serves as the normal vector to the plane.

So, the normal vector to the required plane is $\vec{n} = \hat{i} + 2\hat{j} - 3\hat{k}$.

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The required plane passes through the point P, whose position vector is $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$.

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Vector Equation:

The vector equation of a plane passing through a point with position vector $\vec{a}$ and having a normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

Calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k})$

$= (1)(1) + (2)(2) + (-3)(-3) = 1 + 4 + 9 = 14$

Substitute the values into the vector equation formula:

This is the required vector equation of the plane.

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Cartesian Equation:

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Substitute this into the vector equation $\vec{r} \cdot (\hat{i} + 2\hat{j} - 3\hat{k}) = 14$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k}) = 14$

Perform the dot product:

$x(1) + y(2) + z(-3) = 14$

Rewrite in the general form $Ax + By + Cz + D = 0$:

This is the required Cartesian equation of the plane.

Given:

Equation of the first plane: $\vec{r} \;.\; (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 = 0$. In Cartesian form, this is $x + 2y + 3z - 4 = 0$. Let this be $P_1 = 0$.

Equation of the second plane: $\vec{r} \;.\; (2\hat{i} + \hat{j} − \hat{k}) + 5 = 0$. In Cartesian form, this is $2x + y - z + 5 = 0$. Let this be $P_2 = 0$.

Equation of the third plane (to which the required plane is perpendicular): $\vec{r} \;.\; (5\hat{i} + 3\hat{j} − 6\hat{k}) + 8 = 0$. The normal vector to this plane is $\vec{n_3} = 5\hat{i} + 3\hat{j} - 6\hat{k}$.

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To Find:

The equation of the plane which contains the line of intersection of the first two planes and is perpendicular to the third plane.

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Solution:

The equation of a plane containing the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a constant.

Substituting the Cartesian forms of $P_1$ and $P_2$, we get:

$(x + 2y + 3z - 4) + \lambda (2x + y - z + 5) = 0$

Grouping terms by $x, y, z$ and the constant term:

$(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$

This is the equation of the required plane in Cartesian form. The normal vector to this plane is $\vec{n} = (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k}$.

The required plane is perpendicular to the plane $\vec{r} \;.\; (5\hat{i} + 3\hat{j} − 6\hat{k}) + 8 = 0$, whose normal vector is $\vec{n_3} = 5\hat{i} + 3\hat{j} - 6\hat{k}$.

For two planes to be perpendicular, their normal vectors must be perpendicular. This means the dot product of their normal vectors is zero:

$\vec{n} \cdot \vec{n_3} = 0$

$((1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k}) \cdot (5\hat{i} + 3\hat{j} - 6\hat{k}) = 0$

$5(1 + 2\lambda) + 3(2 + \lambda) - 6(3 - \lambda) = 0$

$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$

Combining like terms:

$(10\lambda + 3\lambda + 6\lambda) + (5 + 6 - 18) = 0$

$19\lambda - 7 = 0$

$19\lambda = 7$

$\lambda = \frac{7}{19}$

Now, substitute the value of $\lambda$ back into the equation of the required plane $(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$:

$(1 + 2(\frac{7}{19}))x + (2 + \frac{7}{19})y + (3 - \frac{7}{19})z + (-4 + 5(\frac{7}{19})) = 0$

$(\frac{19}{19} + \frac{14}{19})x + (\frac{38}{19} + \frac{7}{19})y + (\frac{57}{19} - \frac{7}{19})z + (\frac{-76}{19} + \frac{35}{19}) = 0$

$\frac{33}{19}x + \frac{45}{19}y + \frac{50}{19}z - \frac{41}{19} = 0$

Multiplying the entire equation by 19 to clear the denominator:

$33x + 45y + 50z - 41 = 0$

This is the equation of the required plane in Cartesian form.

In vector form, the equation is $\vec{r} \cdot (33\hat{i} + 45\hat{j} + 50\hat{k}) - 41 = 0$.

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Final Answer:

The equation of the required plane is $33x + 45y + 50z - 41 = 0$.

In vector form, the equation is $\vec{r} \cdot (33\hat{i} + 45\hat{j} + 50\hat{k}) - 41 = 0$.

Given:

The given point is $A(-1, -5, -10)$.

The equation of the line is $\vec{r} = 2\hat{i} − \hat{j} + 2\hat{k} + \lambda (3\hat{i} + 4\hat{j} + 2\hat{k})$.

The equation of the plane is $\vec{r} \;.\; (\hat{i} − \hat{j} + \hat{k}) = 5$.

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To Find:

The distance of the point $A(-1, -5, -10)$ from the point of intersection of the given line and plane.

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Solution:

First, we need to find the coordinates of the point of intersection of the line and the plane.

The position vector of any point on the line is given by $\vec{r} = (2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}$.

Let the coordinates of a point on the line be $(x, y, z)$. Then, $x = 2 + 3\lambda$, $y = -1 + 4\lambda$, and $z = 2 + 2\lambda$.

For the point of intersection, this point must also lie on the plane $\vec{r} \;.\; (\hat{i} − \hat{j} + \hat{k}) = 5$.

Substitute the position vector of the point on the line into the plane equation:

$((2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}) \;.\; (\hat{i} − \hat{j} + \hat{k}) = 5$

Taking the dot product:

$(2 + 3\lambda)(1) + (-1 + 4\lambda)(-1) + (2 + 2\lambda)(1) = 5$

$2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda = 5$

Combine like terms:

$(3\lambda - 4\lambda + 2\lambda) + (2 + 1 + 2) = 5$

$\lambda + 5 = 5$

$\lambda = 0$

Now substitute $\lambda = 0$ back into the parametric equations of the line to find the coordinates of the intersection point (let's call it $B$).

$x = 2 + 3(0) = 2$

$y = -1 + 4(0) = -1$

$z = 2 + 2(0) = 2$

So, the point of intersection is $B(2, -1, 2)$.

Now, we need to find the distance between the given point $A(-1, -5, -10)$ and the intersection point $B(2, -1, 2)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Distance $AB = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2}$

Distance $AB = \sqrt{(2 + 1)^2 + (-1 + 5)^2 + (2 + 10)^2}$

Distance $AB = \sqrt{(3)^2 + (4)^2 + (12)^2}$

Distance $AB = \sqrt{9 + 16 + 144}$

Distance $AB = \sqrt{25 + 144}$

Distance $AB = \sqrt{169}$

Distance $AB = 13$ units.

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Final Answer:

The distance of the point $(-1, -5, -10)$ from the point of intersection of the line and the plane is $13$ units.

Given:

The line passes through the point with coordinates $(1, 2, 3)$. The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The line is parallel to the plane $\vec{r} \;.\; (\hat{i} − \hat{j} + 2\hat{k}) = 5$. The normal vector to this plane is $\vec{n_1} = \hat{i} − \hat{j} + 2\hat{k}$.

The line is parallel to the plane $\vec{r} \;.\; (3\hat{i} + \hat{j} + \hat{k}) = 6$. The normal vector to this plane is $\vec{n_2} = 3\hat{i} + \hat{j} + \hat{k}$.

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To Find:

The vector equation of the line.

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Solution:

The equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.

We are given the point $(1, 2, 3)$, so $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The line is parallel to both given planes. This means the direction vector of the line, $\vec{b}$, must be perpendicular to the normal vectors of both planes, $\vec{n_1}$ and $\vec{n_2}$.

Therefore, the direction vector $\vec{b}$ is parallel to the cross product of the normal vectors $\vec{n_1}$ and $\vec{n_2}$.

$\vec{b} || (\vec{n_1} \times \vec{n_2})$

Let's compute the cross product $\vec{n_1} \times \vec{n_2}$:

$\vec{n_1} \times \vec{n_2} = (\hat{i} − \hat{j} + 2\hat{k}) \times (3\hat{i} + \hat{j} + \hat{k})$

$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix}$

$\vec{n_1} \times \vec{n_2} = \hat{i}((-1)(1) - (2)(1)) - \hat{j}((1)(1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3))$

$\vec{n_1} \times \vec{n_2} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3)$

$\vec{n_1} \times \vec{n_2} = -3\hat{i} - (-5\hat{j}) + 4\hat{k}$

$\vec{n_1} \times \vec{n_2} = -3\hat{i} + 5\hat{j} + 4\hat{k}$

We can take the direction vector $\vec{b}$ to be any vector parallel to $-3\hat{i} + 5\hat{j} + 4\hat{k}$. The simplest choice is $\vec{b} = -3\hat{i} + 5\hat{j} + 4\hat{k}$.

Now, substitute the values of $\vec{a}$ and $\vec{b}$ into the line equation $\vec{r} = \vec{a} + \lambda \vec{b}$:

$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (-3\hat{i} + 5\hat{j} + 4\hat{k})$

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Final Answer:

The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (-3\hat{i} + 5\hat{j} + 4\hat{k})$.

Given:

The required line passes through the point $(1, 2, -4)$. The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.

The required line is perpendicular to the line $\frac{x − 8}{3} = \frac{y + 19}{−16} = \frac{z − 10}{7}$. The direction vector of this line is $\vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$.

The required line is perpendicular to the line $\frac{x − 15}{3} = \frac{y − 29}{8} = \frac{z − 5}{−5}$. The direction vector of this line is $\vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.

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To Find:

The vector equation of the line passing through $(1, 2, -4)$ and perpendicular to the two given lines.

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Solution:

The equation of a line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.

We are given $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.

The required line is perpendicular to both lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$. Therefore, the direction vector $\vec{b}$ of the required line must be perpendicular to both $\vec{b_1}$ and $\vec{b_2}$.

A vector that is perpendicular to two vectors is their cross product. Thus, the direction vector $\vec{b}$ is parallel to $\vec{b_1} \times \vec{b_2}$.

Let's calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = (3\hat{i} - 16\hat{j} + 7\hat{k}) \times (3\hat{i} + 8\hat{j} - 5\hat{k})$

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

$\vec{b_1} \times \vec{b_2} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$

$\vec{b_1} \times \vec{b_2} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$

$\vec{b_1} \times \vec{b_2} = 24\hat{i} + 36\hat{j} + 72\hat{k}$

The direction vector $\vec{b}$ can be any vector parallel to $24\hat{i} + 36\hat{j} + 72\hat{k}$. We can simplify this vector by dividing by the common factor 12:

$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

Now, substitute $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$ into the line equation $\vec{r} = \vec{a} + \lambda \vec{b}$:

$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})$

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Final Answer:

The vector equation of the required line is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})$.

Given:

A plane has intercepts $a, b, c$ on the x, y, and z axes respectively.

The distance of the plane from the origin is $p$ units.

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To Prove:

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$

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Proof:

The equation of a plane having intercepts $a, b, c$ on the x, y, and z axes respectively is given by the intercept form:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

We can rewrite this equation in the general form $Ax + By + Cz + D = 0$:

$\frac{1}{a}x + \frac{1}{b}y + \frac{1}{c}z - 1 = 0$

Comparing this to the standard form $Ax + By + Cz + D = 0$, we have:

$A = \frac{1}{a}$, $B = \frac{1}{b}$, $C = \frac{1}{c}$, and $D = -1$.

The perpendicular distance $p$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by the formula:

$p = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

For the origin $(x_0, y_0, z_0) = (0, 0, 0)$, the formula simplifies to:

$p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$

Substitute the values of $A, B, C,$ and $D$ into this formula:

$p = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}}$

$p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$

To eliminate the square root, square both sides of the equation:

$p^2 = \left(\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}\right)^2$

$p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$

Now, take the reciprocal of both sides of the equation:

$\frac{1}{p^2} = \frac{1}{\frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$

$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$

This is the required result.

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Hence, proved.

Given:

Equation of the first plane: $2x + 3y + 4z = 4$, which can be written as $2x + 3y + 4z - 4 = 0$.

Equation of the second plane: $4x + 6y + 8z = 12$, which can be written as $4x + 6y + 8z - 12 = 0$.

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To Find:

The distance between the two given planes.

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Solution:

We observe that the normal vectors of the two planes are $\vec{n_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{n_2} = 4\hat{i} + 6\hat{j} + 8\hat{k}$.

Since $\vec{n_2} = 2(2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\vec{n_1}$, the normal vectors are parallel, which means the planes are parallel.

To find the distance between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$, we first ensure the coefficients of $x, y,$ and $z$ are the same.

Divide the second equation by 2:

$\frac{4x}{2} + \frac{6y}{2} + \frac{8z}{2} = \frac{12}{2}$

$2x + 3y + 4z = 6$

Rewrite this as $2x + 3y + 4z - 6 = 0$.

Now the two planes are $2x + 3y + 4z - 4 = 0$ and $2x + 3y + 4z - 6 = 0$.

Here, $A=2$, $B=3$, $C=4$, $D_1=-4$, and $D_2=-6$.

The distance between two parallel planes is given by the formula:

Distance $= \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

Substitute the values:

Distance $= \frac{|-4 - (-6)|}{\sqrt{2^2 + 3^2 + 4^2}}$

Distance $= \frac{|-4 + 6|}{\sqrt{4 + 9 + 16}}$

Distance $= \frac{|2|}{\sqrt{29}}$

Distance $= \frac{2}{\sqrt{29}}$ units.

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Final Answer:

The distance between the two planes is $\frac{2}{\sqrt{29}}$ units.

The correct option is (D) $\frac{2}{\sqrt{29}}$ units.

Given:

Equation of the first plane: $2x – y + 4z = 5$.

Equation of the second plane: $5x – 2.5y + 10z = 6$.

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To Determine:

The relationship between the two planes.

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Solution:

The general equation of a plane is $Ax + By + Cz + D = 0$. The normal vector to this plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

For the first plane, $2x - y + 4z - 5 = 0$. The normal vector is $\vec{n_1} = 2\hat{i} - \hat{j} + 4\hat{k}$.

For the second plane, $5x - 2.5y + 10z - 6 = 0$. The normal vector is $\vec{n_2} = 5\hat{i} - 2.5\hat{j} + 10\hat{k}$.

Two planes are parallel if their normal vectors are parallel, i.e., $\vec{n_2} = k\vec{n_1}$ for some scalar $k$.

Let's check if $\vec{n_2}$ is a scalar multiple of $\vec{n_1}$:

$\vec{n_2} = 5\hat{i} - 2.5\hat{j} + 10\hat{k}$

We can factor out 2.5 from $\vec{n_2}$:

$\vec{n_2} = 2.5 ( \frac{5}{2.5}\hat{i} - \frac{2.5}{2.5}\hat{j} + \frac{10}{2.5}\hat{k} )$

$\vec{n_2} = 2.5 ( 2\hat{i} - 1\hat{j} + 4\hat{k} )$

$\vec{n_2} = 2.5 (2\hat{i} - \hat{j} + 4\hat{k})$

We see that $\vec{n_2} = 2.5 \vec{n_1}$. Since the normal vectors are parallel, the planes are parallel.

To check if they are coincident, we check if the constant terms are also related by the same scalar factor. The equations are $2x - y + 4z - 5 = 0$ ($D_1 = -5$) and $5x - 2.5y + 10z - 6 = 0$ ($D_2 = -6$).

If they were coincident, $D_2$ should be $2.5 \times D_1$.

$2.5 \times (-5) = -12.5$

Since $-6 \neq -12.5$, the planes are parallel but not coincident.

Let's check the other options:

(A) Perpendicular: Planes are perpendicular if $\vec{n_1} \cdot \vec{n_2} = 0$.

$\vec{n_1} \cdot \vec{n_2} = (2)(5) + (-1)(-2.5) + (4)(10) = 10 + 2.5 + 40 = 52.5$. Since $52.5 \neq 0$, they are not perpendicular.

(C) intersect y-axis: Both planes intersect the y-axis (when $x=0, z=0$), but this describes a property of each plane, not their relationship to each other.

(D) passes through $(0, 0, \frac{5}{4})$: For the first plane, $2(0) - 0 + 4(\frac{5}{4}) = 5$. So, the first plane passes through this point. For the second plane, $5(0) - 2.5(0) + 10(\frac{5}{4}) = \frac{50}{4} = 12.5 \neq 6$. The second plane does not pass through this point.

The correct description of the relationship between the two planes is that they are parallel.

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Final Answer:

The planes are parallel.

The correct option is (B) Parallel.