Chapter 11 Three Dimensional Geometry

Given:

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

To Find:

The direction cosines of the line.

Solution:

Let the direction ratios of the line be $a, b, c$. The direction cosines $l, m, n$ 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, we calculate the magnitude of the direction ratios, which is $\sqrt{a^2 + b^2 + c^2}$.

Substituting the given values $a=1, b=1, c=2$:

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

$\sqrt{a^2 + b^2 + c^2} = \sqrt{1 + 1 + 4}$

$\sqrt{a^2 + b^2 + c^2} = \sqrt{6}$

Now, we can find the direction cosines using the formulas:

$l = \frac{1}{\sqrt{6}}$

$m = \frac{1}{\sqrt{6}}$

$n = \frac{2}{\sqrt{6}}$

Answer:

The direction cosines of the line are $\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}$.

Given:

The points are $P(2, 3, 5)$ and $Q(-1, 2, 4)$.

To Find:

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

Solution:

Let the coordinates of point $P$ be $(x_1, y_1, z_1) = (2, 3, 5)$.

Let the coordinates of point $Q$ be $(x_2, y_2, z_2) = (-1, 2, 4)$.

The direction ratios of the line passing through points $P$ and $Q$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

So, the direction ratios $(a, b, c)$ are:

$a = x_2 - x_1 = -1 - 2 = -3$

$b = y_2 - y_1 = 2 - 3 = -1$

$c = z_2 - z_1 = 4 - 5 = -1$

The direction ratios of the line $PQ$ are $(-3, -1, -1)$.

The magnitude of the direction ratios is $\sqrt{a^2 + b^2 + c^2}$.

Magnitude $= \sqrt{(-3)^2 + (-1)^2 + (-1)^2}$

Magnitude $= \sqrt{9 + 1 + 1}$

Magnitude $= \sqrt{11}$

The direction cosines $(l, m, n)$ are given by:

$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = \frac{-3}{\sqrt{11}}$

$m = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{-1}{\sqrt{11}}$

$n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} = \frac{-1}{\sqrt{11}}$

Answer:

The direction cosines of the line passing through $P(2, 3, 5)$ and $Q(-1, 2, 4)$ are $\left(\frac{-3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}\right)$.

Alternate Solution:

Alternatively, we could consider the direction ratios from $Q$ to $P$, which would be $(x_1 - x_2, y_1 - y_2, z_1 - z_2)$.

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

$b' = 3 - 2 = 1$

$c' = 5 - 4 = 1$

The direction ratios of the line $QP$ are $(3, 1, 1)$.

The magnitude is the same: $\sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$.

The direction cosines $(l', m', n')$ are:

$l' = \frac{3}{\sqrt{11}}$

$m' = \frac{1}{\sqrt{11}}$

$n' = \frac{1}{\sqrt{11}}$

Both sets of direction cosines $\left(\frac{-3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}\right)$ and $\left(\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right)$ are valid as they represent opposite directions along the same line.

Given:

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

To Find:

The direction cosines of the line.

Solution:

The direction cosines of a line are the cosines of the angles that the line makes with the positive x, y, and z-axes.

Let the direction cosines be $l, m, n$. Then,

$l = \cos \alpha$

$m = \cos \beta$

$n = \cos \gamma$

Substitute the given angle values:

$l = \cos 30^\circ$

$m = \cos 60^\circ$

$n = \cos 90^\circ$

Using standard trigonometric values:

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

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

$n = 0$

Thus, the direction cosines of the line are $\frac{\sqrt{3}}{2}, \frac{1}{2}, 0$.

We can verify that the sum of the squares of the direction cosines is 1:

$l^2 + m^2 + n^2 = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0^2$

$l^2 + m^2 + n^2 = \frac{3}{4} + \frac{1}{4} + 0$

$l^2 + m^2 + n^2 = \frac{3+1}{4} = \frac{4}{4} = 1$

This confirms our calculation is correct.

Answer:

The direction cosines of the line are $\frac{\sqrt{3}}{2}, \frac{1}{2}, 0$.

Given:

Points $Q(2, 2, 1)$ and $R(5, 1, -2)$.

A point $P$ lies on the line joining $Q$ and $R$.

The x-coordinate of point $P$ is 4.

To Find:

The z-coordinate of point $P$.

Solution:

Let the coordinates of point $Q$ be $(x_1, y_1, z_1) = (2, 2, 1)$.

Let the coordinates of point $R$ be $(x_2, y_2, z_2) = (5, 1, -2)$.

Let the point $P$ with coordinates $(x, y, z)$ lie on the line passing through $Q$ and $R$.

We can use the section formula. Assume $P$ divides the line segment $QR$ in the ratio $k:1$.

The coordinates of $P$ are given by:

$x = \frac{k x_2 + 1 x_1}{k+1}$

$y = \frac{k y_2 + 1 y_1}{k+1}$

$z = \frac{k z_2 + 1 z_1}{k+1}$

We are given that the x-coordinate of point $P$ is 4. Substitute $x=4$ and the coordinates of $Q$ and $R$ into the formula for the x-coordinate:

$4 = \frac{k(5) + 1(2)}{k+1}$

Now, we solve for $k$:

$4(k+1) = 5k + 2$

$4k + 4 = 5k + 2$

$4 - 2 = 5k - 4k$

$k = 2$

So, the point $P$ divides the line segment $QR$ in the ratio $2:1$.

Now, we can find the z-coordinate of $P$ using the value of $k=2$ and the formula for the z-coordinate:

$z = \frac{k z_2 + 1 z_1}{k+1}$

$z = \frac{2(-2) + 1(1)}{2+1}$

$z = \frac{-4 + 1}{3}$

$z = \frac{-3}{3}$

$z = -1$

The z-coordinate of the point $P$ is $-1$.

Answer:

The z-coordinate of the point is $-1$.

Given:

Position vector of the point is $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$.

The equation of the plane is $\vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9$.

To Find:

The distance of the point from the plane.

Solution:

The equation of the plane is given in the form $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = \hat{i} - 2\hat{j} + 4\hat{k}$ is the normal vector to the plane and $d=9$.

The position vector of the given point is $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$.

The distance $D$ of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula:

$D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$

First, calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k})$

$\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4)$

$\vec{a} \cdot \vec{n} = 2 - 2 - 4$

$\vec{a} \cdot \vec{n} = -4$

Next, calculate the magnitude of the normal vector $|\vec{n}|$.

$|\vec{n}| = |\hat{i} - 2\hat{j} + 4\hat{k}|$

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

$|\vec{n}| = \sqrt{1 + 4 + 16}$

$|\vec{n}| = \sqrt{21}$

Now, substitute the values into the distance formula:

$D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$

$D = \frac{|-4 - 9|}{\sqrt{21}}$

$D = \frac{|-13|}{\sqrt{21}}$

$D = \frac{13}{\sqrt{21}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{21}$:

$D = \frac{13}{\sqrt{21}} \times \frac{\sqrt{21}}{\sqrt{21}}$

$D = \frac{13\sqrt{21}}{21}$

Answer:

The distance of the point from the plane is $\frac{13}{\sqrt{21}}$ or $\frac{13\sqrt{21}}{21}$ units.

Given:

The point is $P(-2, 4, -5)$.

The equation of the line is $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$.

To Find:

The distance of the point $P$ from the given line.

Solution:

The equation of the line in symmetric form is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

Comparing the given equation $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$ with the general form, we can identify:

A point $A$ on the line with coordinates $(x_0, y_0, z_0) = (-3, 4, -8)$.

The direction ratios of the line $(a, b, c) = (3, 5, 6)$.

The direction vector of the line is $\vec{b} = 3\hat{i} + 5\hat{j} + 6\hat{k}$.

The given point is $P(-2, 4, -5)$. Its position vector is $\vec{p} = -2\hat{i} + 4\hat{j} - 5\hat{k}$.

The position vector of point $A$ on the line is $\vec{a} = -3\hat{i} + 4\hat{j} - 8\hat{k}$.

The vector joining point $A$ on the line to the point $P$ is $\vec{AP} = \vec{p} - \vec{a}$.

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

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

$\vec{AP} = (-2 + 3)\hat{i} + 0\hat{j} + (-5 + 8)\hat{k}$

$\vec{AP} = 1\hat{i} + 0\hat{j} + 3\hat{k}$

The distance $d$ of the point $P$ from the line is given by the formula:

$d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$

First, calculate the cross product $\vec{AP} \times \vec{b}$:

$\vec{AP} \times \vec{b} = (1\hat{i} + 0\hat{j} + 3\hat{k}) \times (3\hat{i} + 5\hat{j} + 6\hat{k})$

$\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 3 \\ 3 & 5 & 6 \end{vmatrix}$

$\vec{AP} \times \vec{b} = \hat{i}((0)(6) - (3)(5)) - \hat{j}((1)(6) - (3)(3)) + \hat{k}((1)(5) - (0)(3))$

$\vec{AP} \times \vec{b} = \hat{i}(0 - 15) - \hat{j}(6 - 9) + \hat{k}(5 - 0)$

$\vec{AP} \times \vec{b} = -15\hat{i} + 3\hat{j} + 5\hat{k}$

Next, calculate the magnitude of the cross product $|\vec{AP} \times \vec{b}|$.

$|\vec{AP} \times \vec{b}| = \sqrt{(-15)^2 + 3^2 + 5^2}$

$|\vec{AP} \times \vec{b}| = \sqrt{225 + 9 + 25}$

$|\vec{AP} \times \vec{b}| = \sqrt{259}$

Now, calculate the magnitude of the direction vector $|\vec{b}|$.

$|\vec{b}| = \sqrt{3^2 + 5^2 + 6^2}$

$|\vec{b}| = \sqrt{9 + 25 + 36}$

$|\vec{b}| = \sqrt{70}$

Finally, substitute these magnitudes into the distance formula:

$d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{259}}{\sqrt{70}}$

The distance can also be written by rationalizing the denominator:

$d = \frac{\sqrt{259}}{\sqrt{70}} = \sqrt{\frac{259}{70}}$

Or, $d = \frac{\sqrt{259}}{\sqrt{70}} \times \frac{\sqrt{70}}{\sqrt{70}} = \frac{\sqrt{259 \times 70}}{70} = \frac{\sqrt{18130}}{70}$.

Answer:

The distance of the point $(-2, 4, -5)$ from the given line is $\frac{\sqrt{259}}{\sqrt{70}}$ units.

Given:

Points on the line: $A(3, -4, -5)$ and $B(2, -3, 1)$.

Points on the plane: $P(2, 2, 1)$, $Q(3, 0, 1)$, and $R(4, -1, 0)$.

To Find:

The coordinates of the point where the line $AB$ crosses the plane $PQR$.

Solution:

1. Find the equation of the line passing through A(3, -4, -5) and B(2, -3, 1).

The direction ratios of the line passing through $(x_1, y_1, z_1)$ 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 are $(2 - 3, -3 - (-4), 1 - (-5)) = (-1, 1, 6)$.

The equation of the line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.

Using point $A(3, -4, -5)$ and direction ratios $(-1, 1, 6)$, the equation of the line is:

$\frac{x - 3}{-1} = \frac{y - (-4)}{1} = \frac{z - (-5)}{6}$

$\frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6}$

Let this ratio be equal to a parameter $\lambda$. Any point on the line can be represented as:

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

$\frac{y + 4}{1} = \lambda \implies y + 4 = \lambda \implies y = -4 + \lambda$

$\frac{z + 5}{6} = \lambda \implies z + 5 = 6\lambda \implies z = -5 + 6\lambda$

So, any point on the line is of the form $(3 - \lambda, -4 + \lambda, -5 + 6\lambda)$.

2. Find the equation of the plane passing through P(2, 2, 1), Q(3, 0, 1), and R(4, -1, 0).

Let the equation of the plane be $Ax + By + Cz + D = 0$. Since the points P, Q, and R lie on the plane, they must satisfy the equation.

For P(2, 2, 1): $2A + 2B + C + D = 0$ ... (1)

For Q(3, 0, 1): $3A + 0B + C + D = 0 \implies 3A + C + D = 0$ ... (2)

For R(4, -1, 0): $4A - B + 0C + D = 0 \implies 4A - B + D = 0$ ... (3)

Subtract equation (2) from equation (1):

$(2A + 2B + C + D) - (3A + C + D) = 0 - 0$

$2A + 2B + C + D - 3A - C - D = 0$

$-A + 2B = 0 \implies A = 2B$ ... (4)

From equation (3), $D = B - 4A$. Substitute $A = 2B$ into this equation:

$D = B - 4(2B) = B - 8B = -7B$ ... (5)

Substitute $A = 2B$ and $D = -7B$ into equation (2):

$3(2B) + C + (-7B) = 0$

$6B + C - 7B = 0$

$-B + C = 0 \implies C = B$ ... (6)

Substitute $A=2B, C=B, D=-7B$ into the plane equation $Ax + By + Cz + D = 0$:

$(2B)x + By + (B)z + (-7B) = 0$

Divide by $B$ (assuming $B \neq 0$, if $B=0$, then $A=0, C=0, D=0$ which means the points are collinear or same, which is not the case here).

$2x + y + z - 7 = 0$

So, the equation of the plane is $2x + y + z = 7$.

Alternatively, find the equation of the plane using vectors:

Vector $\vec{PQ} = Q - P = (3-2)\hat{i} + (0-2)\hat{j} + (1-1)\hat{k} = \hat{i} - 2\hat{j} + 0\hat{k}$.

Vector $\vec{PR} = R - P = (4-2)\hat{i} + (-1-2)\hat{j} + (0-1)\hat{k} = 2\hat{i} - 3\hat{j} - \hat{k}$.

The normal vector to the plane is $\vec{n} = \vec{PQ} \times \vec{PR}$.

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix} = \hat{i}((-2)(-1) - (0)(-3)) - \hat{j}((1)(-1) - (0)(2)) + \hat{k}((1)(-3) - (-2)(2))$

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

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

The equation of the plane is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, where $\vec{a}$ is the position vector of a point on the plane (e.g., P, Q, or R). Using P(2, 2, 1), $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$.

$\vec{a} \cdot \vec{n} = (2\hat{i} + 2\hat{j} + \hat{k}) \cdot (2\hat{i} + \hat{j} + \hat{k}) = (2)(2) + (2)(1) + (1)(1) = 4 + 2 + 1 = 7$.

The vector equation of the plane is $\vec{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 7$.

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. The Cartesian equation is $2x + y + z = 7$.

3. Find the intersection point.

The coordinates of any point on the line are $(3 - \lambda, -4 + \lambda, -5 + 6\lambda)$. If this point lies on the plane $2x + y + z = 7$, it must satisfy the plane equation.

Substitute the coordinates of the point on the line into the plane equation:

$2(3 - \lambda) + (-4 + \lambda) + (-5 + 6\lambda) = 7$

$6 - 2\lambda - 4 + \lambda - 5 + 6\lambda = 7$

Combine constant terms and terms with $\lambda$:

$(6 - 4 - 5) + (-2\lambda + \lambda + 6\lambda) = 7$

$-3 + 5\lambda = 7$

$5\lambda = 7 + 3$

$5\lambda = 10$

$\lambda = 2$

Now substitute the value of $\lambda = 2$ back into the parametric equation of the line to find the coordinates of the intersection point:

$x = 3 - \lambda = 3 - 2 = 1$

$y = -4 + \lambda = -4 + 2 = -2$

$z = -5 + 6\lambda = -5 + 6(2) = -5 + 12 = 7$

The coordinates of the intersection point are $(1, -2, 7)$.

Answer:

The coordinates of the point where the line crosses the plane are $(1, -2, 7)$.

Given:

The given point is $P(-1, -5, -10)$.

The equation of the line is $\vec{r}= 2\hat{i}− \hat{j}+ 2\hat{k}+ λ \left( 3\hat{i}+ 4\hat{j}+ 2\hat{k} \right)$.

The equation of the plane is $\vec{r}. \left( \hat{i}− \hat{j}+ \hat{k} \right) = 5$.

To Find:

The distance of the point $P(-1, -5, -10)$ from the point of intersection of the given line and the given plane.

Solution:

First, we find the coordinates of the point of intersection of the line and the plane.

Any point on the line $\vec{r}= 2\hat{i}− \hat{j}+ 2\hat{k}+ λ \left( 3\hat{i}+ 4\hat{j}+ 2\hat{k} \right)$ has the position vector:

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

where $\lambda$ is a scalar parameter.

The equation of the plane is $\vec{r}. \left( \hat{i}− \hat{j}+ \hat{k} \right) = 5$.

To find the point of intersection, substitute the expression for $\vec{r}$ from the line equation into the plane equation:

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

Take 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 the constant terms and the $\lambda$ terms:

$(2 + 1 + 2) + (3\lambda - 4\lambda + 2\lambda) = 5$

$5 + \lambda = 5$

Solving for $\lambda$:

$\lambda = 5 - 5$

$\lambda = 0$

Now substitute the value of $\lambda = 0$ back into the line equation to find the position vector of the intersection point, say $Q$:

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

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

The coordinates of the point of intersection $Q$ are $(2, -1, 2)$.

Now, we need to find the distance between the given point $P(-1, -5, -10)$ and the intersection point $Q(2, -1, 2)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Let $(x_1, y_1, z_1) = (-1, -5, -10)$ and $(x_2, y_2, z_2) = (2, -1, 2)$.

Distance $PQ = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2}$

$PQ = \sqrt{(2 + 1)^2 + (-1 + 5)^2 + (2 + 10)^2}$

$PQ = \sqrt{(3)^2 + (4)^2 + (12)^2}$

$PQ = \sqrt{9 + 16 + 144}$

$PQ = \sqrt{25 + 144}$

$PQ = \sqrt{169}$

$PQ = 13$

Answer:

The distance of the point $(-1, -5, -10)$ from the point of intersection of the line and the plane is $13$ units.

Given:

A plane meets the x, y, and z-axes at points A, B, and C respectively.

The centroid of the triangle $\triangle ABC$ is the point $(\alpha, \beta, \gamma)$.

To Prove:

The equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

Proof:

Let the equation of the plane in intercept form be:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

where $a, b, c$ are the intercepts on the x, y, and z-axes respectively.

The points where the plane meets the coordinate axes are:

Point A on the x-axis: $(a, 0, 0)$

Point B on the y-axis: $(0, b, 0)$

Point C on the z-axis: $(0, 0, c)$

The coordinates of the vertices of $\triangle ABC$ are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.

The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the formula:

$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

The centroid of $\triangle ABC$ is given as $(\alpha, \beta, \gamma)$. Using the centroid formula with the coordinates of A, B, and C:

$\alpha = \frac{a + 0 + 0}{3} = \frac{a}{3}$

$\beta = \frac{0 + b + 0}{3} = \frac{b}{3}$

$\gamma = \frac{0 + 0 + c}{3} = \frac{c}{3}$

From these relationships, we can express the intercepts $a, b, c$ in terms of $\alpha, \beta, \gamma$:

Now, substitute these values of $a, b, c$ into the intercept form of the plane equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:

$\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$

Multiply the entire equation by 3:

$3 \left( \frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} \right) = 3(1)$

$\frac{3x}{3\alpha} + \frac{3y}{3\beta} + \frac{3z}{3\gamma} = 3$

$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$

Thus, the equation of the plane is $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

This proves the required result.

Hence proved.

Given:

The direction cosines $(l, m, n)$ of the two lines satisfy the following equations:

1) $3l + m + 5n = 0$

2) $6mn – 2nl + 5lm = 0$

We also know that the sum of the squares of direction cosines is 1:

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

To Find:

The angle between the two lines.

Solution:

From equation (1), we can express $m$ in terms of $l$ and $n$:

$m = -3l - 5n$

Substitute this expression for $m$ into equation (2):

$6n(-3l - 5n) - 2nl + 5l(-3l - 5n) = 0$

$-18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0$

Rearrange and combine like terms:

$-15l^2 + (-18ln - 2nl - 25ln) - 30n^2 = 0$

$-15l^2 - 45ln - 30n^2 = 0$

Divide the entire equation by $-15$:

$l^2 + 3ln + 2n^2 = 0$

This is a quadratic equation in terms of $l$ and $n$. We can solve for the ratio $\frac{l}{n}$ by dividing by $n^2$ (assuming $n \neq 0$):

$\left(\frac{l}{n}\right)^2 + 3\left(\frac{l}{n}\right) + 2 = 0$

Let $x = \frac{l}{n}$. The equation becomes $x^2 + 3x + 2 = 0$.

Factoring the quadratic equation:

$(x+1)(x+2) = 0$

This gives two possible values for the ratio $\frac{l}{n}$:

$x+1 = 0 \implies x = -1 \implies \frac{l}{n} = -1 \implies l = -n$

$x+2 = 0 \implies x = -2 \implies \frac{l}{n} = -2 \implies l = -2n$

These two relationships between $l$ and $n$ correspond to the two lines.

Case 1: $l = -n$

Substitute $l = -n$ into the expression for $m$: $m = -3l - 5n = -3(-n) - 5n = 3n - 5n = -2n$.

So, the direction cosines for the first line are proportional to $(-n, -2n, n)$, or $(1, 2, -1)$ by taking $n=-1$ as a proportionality constant for direction ratios. However, we need direction cosines, so we use $l^2 + m^2 + n^2 = 1$.

$(-n)^2 + (-2n)^2 + n^2 = 1$

$n^2 + 4n^2 + n^2 = 1$

$6n^2 = 1 \implies n^2 = \frac{1}{6} \implies n = \pm \frac{1}{\sqrt{6}}$

Let's choose the positive value for $n$ for the first line's direction cosines $(l_1, m_1, n_1)$:

$n_1 = \frac{1}{\sqrt{6}}$

$l_1 = -n_1 = -\frac{1}{\sqrt{6}}$

$m_1 = -2n_1 = -\frac{2}{\sqrt{6}}$

So, the direction cosines of the first line are $\left(-\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$.

Case 2: $l = -2n$

Substitute $l = -2n$ into the expression for $m$: $m = -3l - 5n = -3(-2n) - 5n = 6n - 5n = n$.

So, the direction cosines for the second line are proportional to $(-2n, n, n)$, or $(-2, 1, 1)$ by taking $n=1$ as a proportionality constant for direction ratios. Using $l^2 + m^2 + n^2 = 1$ for direction cosines $(l_2, m_2, n_2)$:

$(-2n)^2 + n^2 + n^2 = 1$

$4n^2 + n^2 + n^2 = 1$

$6n^2 = 1 \implies n^2 = \frac{1}{6} \implies n = \pm \frac{1}{\sqrt{6}}$

Let's choose the positive value for $n$ for the second line's direction cosines $(l_2, m_2, n_2)$:

$n_2 = \frac{1}{\sqrt{6}}$

$l_2 = -2n_2 = -\frac{2}{\sqrt{6}}$

$m_2 = n_2 = \frac{1}{\sqrt{6}}$

So, the direction cosines of the second line are $\left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$.

Let $\theta$ be the angle between the two lines. 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:

$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$

Substitute the direction cosines we found:

$\cos \theta = \left| \left(-\frac{1}{\sqrt{6}}\right) \left(-\frac{2}{\sqrt{6}}\right) + \left(-\frac{2}{\sqrt{6}}\right) \left(\frac{1}{\sqrt{6}}\right) + \left(\frac{1}{\sqrt{6}}\right) \left(\frac{1}{\sqrt{6}}\right) \right|$

$\cos \theta = \left| \frac{2}{6} - \frac{2}{6} + \frac{1}{6} \right|$

$\cos \theta = \left| 0 + \frac{1}{6} \right|$

$\cos \theta = \frac{1}{6}$

The angle $\theta$ is the arc cosine of $\frac{1}{6}$.

$\theta = \cos^{-1}\left(\frac{1}{6}\right)$

Answer:

The angle between the two lines is $\cos^{-1}\left(\frac{1}{6}\right)$.

Given:

Point A is $(1, 8, 4)$.

The line passes through points B $(0, -1, 3)$ and C $(2, -3, -1)$.

To Find:

The coordinates of the foot of the perpendicular drawn from point A to the line BC.

Solution:

First, find the equation of the line passing through points B $(0, -1, 3)$ and C $(2, -3, -1)$.

The direction ratios of the line BC are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

Direction ratios $=(2 - 0, -3 - (-1), -1 - 3) = (2, -2, -4)$.

We can use simplified direction ratios by dividing by 2: $(1, -1, -2)$.

The equation of the line passing through point B $(0, -1, 3)$ with direction ratios $(1, -1, -2)$ is:

$\frac{x - 0}{1} = \frac{y - (-1)}{-1} = \frac{z - 3}{-2}$

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

Let this ratio be equal to a parameter $\lambda$. Any point M on the line BC can be represented as:

$\frac{x}{1} = \lambda \implies x = \lambda$

$\frac{y + 1}{-1} = \lambda \implies y + 1 = -\lambda \implies y = -1 - \lambda$

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

So, the coordinates of any point M on the line are $(\lambda, -1 - \lambda, 3 - 2\lambda)$.

Let M be the foot of the perpendicular from point A $(1, 8, 4)$ to the line BC.

The vector $\vec{AM}$ connects point A to point M.

$\vec{AM} = M - A = (\lambda - 1, (-1 - \lambda) - 8, (3 - 2\lambda) - 4)$

$\vec{AM} = (\lambda - 1, -9 - \lambda, -1 - 2\lambda)$

The line AM is perpendicular to the line BC. This means the vector $\vec{AM}$ is orthogonal to the direction vector of the line BC.

The direction vector of line BC is $\vec{v} = \hat{i} - \hat{j} - 2\hat{k}$ (from the direction ratios $(1, -1, -2)$).

The dot product of $\vec{AM}$ and $\vec{v}$ must be zero for perpendicularity:

$\vec{AM} \cdot \vec{v} = 0$

$(\lambda - 1)(1) + (-9 - \lambda)(-1) + (-1 - 2\lambda)(-2) = 0$

$\lambda - 1 + 9 + \lambda + 2 + 4\lambda = 0$

Combine terms with $\lambda$ and constant terms:

$(\lambda + \lambda + 4\lambda) + (-1 + 9 + 2) = 0$

$6\lambda + 10 = 0$

Solve for $\lambda$:

$6\lambda = -10$

$\lambda = -\frac{10}{6} = -\frac{5}{3}$

Now substitute the value of $\lambda = -\frac{5}{3}$ back into the coordinates of point M to find the foot of the perpendicular:

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

$y = -1 - \lambda = -1 - \left(-\frac{5}{3}\right) = -1 + \frac{5}{3} = \frac{-3 + 5}{3} = \frac{2}{3}$

$z = 3 - 2\lambda = 3 - 2\left(-\frac{5}{3}\right) = 3 + \frac{10}{3} = \frac{9 + 10}{3} = \frac{19}{3}$

The coordinates of the foot of the perpendicular M are $\left(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3}\right)$.

Answer:

The coordinates of the foot of the perpendicular from point A to the line BC are $\left(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3}\right)$.

Given:

The given point is $P(1, 6, 3)$.

The equation of the line is $\frac{x}{1} = \frac{y − 1}{2} = \frac{z − 2}{3}$.

To Find:

The coordinates of the image of point $P$ in the given line.

Solution:

The equation of the line is $\frac{x}{1} = \frac{y − 1}{2} = \frac{z − 2}{3}$.

This line passes through the point $A(0, 1, 2)$ and has direction ratios $(1, 2, 3)$.

Let the direction vector of the line be $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Any point M on the line can be represented parametrically by setting the ratios equal to $\lambda$:

$\frac{x}{1} = \lambda \implies x = \lambda$

$\frac{y - 1}{2} = \lambda \implies y = 1 + 2\lambda$

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

So, the coordinates of any point M on the line are $(\lambda, 1 + 2\lambda, 2 + 3\lambda)$.

Let M be the foot of the perpendicular from point $P(1, 6, 3)$ to the line.

The vector $\vec{PM}$ connects point P to point M.

$\vec{PM} = M - P = (\lambda - 1, (1 + 2\lambda) - 6, (2 + 3\lambda) - 3)$

$\vec{PM} = (\lambda - 1, 2\lambda - 5, 3\lambda - 1)$

The line segment PM is perpendicular to the given line. This means the vector $\vec{PM}$ is orthogonal to the direction vector of the line, $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The dot product of $\vec{PM}$ and $\vec{v}$ must be zero for perpendicularity:

$\vec{PM} \cdot \vec{v} = 0$

$(\lambda - 1)(1) + (2\lambda - 5)(2) + (3\lambda - 1)(3) = 0$

$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$

Combine the terms:

$(\lambda + 4\lambda + 9\lambda) + (-1 - 10 - 3) = 0$

$14\lambda - 14 = 0$

$14\lambda = 14$

$\lambda = 1$

Now, substitute the value of $\lambda = 1$ into the coordinates of point M to find the foot of the perpendicular:

$x_M = \lambda = 1$

$y_M = 1 + 2\lambda = 1 + 2(1) = 3$

$z_M = 2 + 3\lambda = 2 + 3(1) = 5$

The coordinates of the foot of the perpendicular M are $(1, 3, 5)$.

Let $P'(x', y', z')$ be the image of point $P(1, 6, 3)$ in the line. The foot of the perpendicular M is the midpoint of the line segment PP'.

Using the midpoint formula:

$\frac{x_P + x_{P'}}{2} = x_M \implies \frac{1 + x'}{2} = 1$

$1 + x' = 2 \implies x' = 1$

$\frac{y_P + y_{P'}}{2} = y_M \implies \frac{6 + y'}{2} = 3$

$6 + y' = 6 \implies y' = 0$

$\frac{z_P + z_{P'}}{2} = z_M \implies \frac{3 + z'}{2} = 5$

$3 + z' = 10 \implies z' = 7$

The coordinates of the image point $P'$ are $(1, 0, 7)$.

Answer:

The coordinates of the image of the point $(1, 6, 3)$ in the given line are $(1, 0, 7)$.

Given:

Position vector of the point is $\vec{p} = \hat{i} + 3\hat{j} + 4\hat{k}$. Let the point be $P(1, 3, 4)$.

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

To Find:

The image of the point $P$ in the given plane.

Solution:

The equation of the plane is $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = -3$.

Comparing this with the standard form $\vec{r} \cdot \vec{n} = d$, the normal vector to the plane is $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$, and $d = -3$.

Let $P'(x', y', z')$ be the image of the point $P(1, 3, 4)$ in the plane. Let $M$ be the foot of the perpendicular from $P$ to the plane. $M$ is the midpoint of the line segment $PP'$.

The line $PP'$ is perpendicular to the plane, so it is parallel to the normal vector $\vec{n}$. The equation of the line passing through $P(\vec{p})$ and parallel to $\vec{n}$ is given by $\vec{r} = \vec{p} + \lambda \vec{n}$, where $\lambda$ is a scalar.

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

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

Any point on this line has position vector given by $\vec{r}$. The foot of the perpendicular $M$ lies on this line, so its position vector $\vec{m}$ is of this form.

Since $M$ also lies on the plane, its position vector $\vec{m}$ must satisfy the plane equation $\vec{r} \cdot \vec{n} = -3$. Substitute the expression for $\vec{r}$ into the plane equation:

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

Take the dot product:

$(1 + 2\lambda)(2) + (3 - \lambda)(-1) + (4 + \lambda)(1) = -3$

$2 + 4\lambda - 3 + \lambda + 4 + \lambda = -3$

Combine constant terms and $\lambda$ terms:

$(2 - 3 + 4) + (4\lambda + \lambda + \lambda) = -3$

$3 + 6\lambda = -3$

Solve for $\lambda$:

$6\lambda = -3 - 3$

$6\lambda = -6$

$\lambda = -1$

Now substitute the value of $\lambda = -1$ back into the expression for $\vec{r}$ to find the position vector of the foot of the perpendicular $M$:

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

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

$\vec{m} = -\hat{i} + 4\hat{j} + 3\hat{k}$

The coordinates of $M$ are $(-1, 4, 3)$.

Since M is the midpoint of $P(1, 3, 4)$ and its image $P'(x', y', z')$, we use the midpoint formula:

$\vec{m} = \frac{\vec{p} + \vec{p'}}{2}$

$\vec{p'} = 2\vec{m} - \vec{p}$

$\vec{p'} = 2(-\hat{i} + 4\hat{j} + 3\hat{k}) - (\hat{i} + 3\hat{j} + 4\hat{k})$

$\vec{p'} = (-2\hat{i} + 8\hat{j} + 6\hat{k}) - (\hat{i} + 3\hat{j} + 4\hat{k})$

$\vec{p'} = (-2 - 1)\hat{i} + (8 - 3)\hat{j} + (6 - 4)\hat{k}$

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

The coordinates of the image point $P'$ are $(-3, 5, 2)$.

Answer:

The image of the point having position vector $\hat{i} + 3\hat{j} + 4\hat{k}$ in the given plane has the position vector $-3\hat{i} + 5\hat{j} + 2\hat{k}$, and its coordinates are $(-3, 5, 2)$.

Given:

The given point is $(2, 5, 7)$.

To Find:

The coordinates of the foot of the perpendicular from the point $(2, 5, 7)$ to the x-axis.

Solution:

The x-axis is the line where the y and z coordinates are always zero.

Any point on the x-axis can be represented in the form $(x, 0, 0)$.

To find the foot of the perpendicular from a point $(x_0, y_0, z_0)$ to the x-axis, we project the point onto the x-axis.

The projection of the point $(x_0, y_0, z_0)$ onto the x-axis has the coordinates $(x_0, 0, 0)$.

In this case, the given point is $(2, 5, 7)$, so $(x_0, y_0, z_0) = (2, 5, 7)$.

The coordinates of the foot of the perpendicular on the x-axis are $(2, 0, 0)$.

Answer:

The coordinates of the foot of the perpendicular are $(2, 0, 0)$.

This corresponds to option (A).

The correct option is (A).

Given:

Point P lies on the line segment joining $A(3, 2, -1)$ and $B(6, 2, -2)$.

The x-coordinate of point P is 5.

To Find:

The y-coordinate of point P.

Solution:

Let point P divide the line segment joining $A(3, 2, -1)$ and $B(6, 2, -2)$ in the ratio $k:1$.

The coordinates of point P $(x_P, y_P, z_P)$ are given by the section formula:

$x_P = \frac{k x_B + 1 x_A}{k+1}$

$y_P = \frac{k y_B + 1 y_A}{k+1}$

$z_P = \frac{k z_B + 1 z_A}{k+1}$

We are given that the x-coordinate of P is 5. Substitute the x-coordinates of A and B:

$5 = \frac{k(6) + 1(3)}{k+1}$

$5(k+1) = 6k + 3$

$5k + 5 = 6k + 3$

$5 - 3 = 6k - 5k$

$k = 2$

Since $k=2$ is positive, the point P divides the line segment AB internally in the ratio $2:1$.

Now, we find the y-coordinate of P using the section formula with $k=2$ and the y-coordinates of A and B:

$y_P = \frac{k y_B + 1 y_A}{k+1}$

$y_P = \frac{2(2) + 1(2)}{2+1}$

$y_P = \frac{4 + 2}{3}$

$y_P = \frac{6}{3}$

$y_P = 2$

The y-coordinate of point P is 2.

Answer:

The y-coordinate of P is 2.

This corresponds to option (A).

The correct option is (A).

Given:

The angles that a line makes with the positive x, y, and z-axes are $\alpha, \beta, \gamma$ respectively.

To Find:

The direction cosines of the line.

Solution:

By definition, the direction cosines of a line that makes angles $\alpha, \beta, \gamma$ with the positive direction of the x, y, and z-axes are the cosines of these angles.

Let the direction cosines be $l, m, n$. Then,

$l = \cos \alpha$

$m = \cos \beta$

$n = \cos \gamma$

Therefore, the direction cosines of the line are $\cos \alpha, \cos \beta, \cos \gamma$.

Answer:

The direction cosines of the line are $\cos \alpha, \cos \beta, \cos \gamma$.

This matches option (B).

The correct option is (B).

Given:

The given point is $P(a, b, c)$.

To Find:

The distance of the point $P$ from the x-axis.

Solution:

To find the distance of the point $P(a, b, c)$ from the x-axis, we find the coordinates of the foot of the perpendicular from $P$ to the x-axis. Let this foot of the perpendicular be $M$.

As shown in Example 14, the foot of the perpendicular from $(x_0, y_0, z_0)$ to the x-axis is $(x_0, 0, 0)$.

So, the coordinates of the foot of the perpendicular $M$ from $P(a, b, c)$ to the x-axis are $(a, 0, 0)$.

Now, we find the distance between the point $P(a, b, c)$ and the foot of the perpendicular $M(a, 0, 0)$ using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Distance $PM = \sqrt{(a - a)^2 + (0 - b)^2 + (0 - c)^2}$

$PM = \sqrt{0^2 + (-b)^2 + (-c)^2}$

$PM = \sqrt{0 + b^2 + c^2}$

$PM = \sqrt{b^2 + c^2}$

The distance of the point $(a, b, c)$ from the x-axis is $\sqrt{b^2 + c^2}$.

Answer:

The distance of the point $(a, b, c)$ from the x-axis is $\sqrt{b^2 + c^2}$.

This matches option (C).

The correct option is (C).

Given:

The question asks for the equations that define the x-axis in three-dimensional space.

To Find:

The equations of the x-axis in space.

Solution:

In a three-dimensional Cartesian coordinate system $(x, y, z)$, the x-axis is the line where points have a specific form.

The x-axis is the set of all points whose y-coordinate is 0 and whose z-coordinate is 0, while the x-coordinate can be any real number.

For any point on the x-axis, the distance from the yz-plane is determined by the x-coordinate, but the distance from the xy-plane (where z=0) and the xz-plane (where y=0) must be zero.

Therefore, the conditions that define the x-axis are that the y and z coordinates are zero.

The equations of the x-axis in space are $y = 0$ and $z = 0$.

Answer:

The equations of the x-axis in space are $y = 0, z = 0$.

This corresponds to option (D).

The correct option is (D).

Given:

A line makes equal angles with the coordinate axes.

To Find:

The direction cosines of the line.

Solution:

Let the angles that the line makes with the positive direction of the x, y, and z-axes be $\alpha$, $\beta$, and $\gamma$, respectively.

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

We are given that the line makes equal angles with the coordinate axes, which means:

Therefore, the direction cosines must be equal:

$l = m = n = \cos \alpha$

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 $l=m=n$ into this equation:

$l^2 + l^2 + l^2 = 1$

$3l^2 = 1$

$l^2 = \frac{1}{3}$

$l = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$

Since $l = m = n$, the direction cosines are:

$l = \pm \frac{1}{\sqrt{3}}$

$m = \pm \frac{1}{\sqrt{3}}$

$n = \pm \frac{1}{\sqrt{3}}$

For a single direction of the line, the signs must be the same. Therefore, the possible sets of direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ and $\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$.

These two sets can be represented together as $± \left( \frac{1}{\sqrt{3}},\; \frac{1}{\sqrt{3}},\; \frac{1}{\sqrt{3}} \right)$.

Answer:

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

This matches option (B).

The correct option is (B).

Given:

A line makes angles $\alpha = \frac{\pi}{2}$, $\beta = \frac{3\pi}{4}$, and $\gamma = \frac{\pi}{4}$ with the x, y, and z-axes, respectively.

To Find:

The direction cosines of the line.

Solution:

The direction cosines of a line are the cosines of the angles it makes with the positive x, y, and z-axes.

Let the direction cosines be $l, m, n$. Then:

$l = \cos \alpha = \cos \left(\frac{\pi}{2}\right)$

$m = \cos \beta = \cos \left(\frac{3\pi}{4}\right)$

$n = \cos \gamma = \cos \left(\frac{\pi}{4}\right)$

Calculate the cosine values:

$\cos \left(\frac{\pi}{2}\right) = 0$

$\cos \left(\frac{3\pi}{4}\right) = \cos \left(\pi - \frac{\pi}{4}\right) = -\cos \left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

$\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

So, the direction cosines are $l=0$, $m=-\frac{1}{\sqrt{2}}$, and $n=\frac{1}{\sqrt{2}}$.

We can verify that $l^2 + m^2 + n^2 = 1$:

$0^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$. This confirms the direction cosines are correct.

Answer:

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

Fill in the blank: $0,\; -\frac{1}{\sqrt{2}},\; \frac{1}{\sqrt{2}}$

Given:

A line makes angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes.

To Find:

The value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.

Solution:

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

We know the fundamental property of direction cosines:

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

Substituting the cosine values:

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

We also know the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, which can be rewritten as $\cos^2 \theta = 1 - \sin^2 \theta$.

Apply this identity to each term in the equation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:

$(1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1$

Remove the parentheses:

$1 - \sin^2 \alpha + 1 - \sin^2 \beta + 1 - \sin^2 \gamma = 1$

Combine the constant terms:

$3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1$

Rearrange the equation to solve for $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$:

$3 - 1 = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$

$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2$

Answer:

The value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ is 2.

Fill in the blank: $2$

Given:

A line makes an angle of $\frac{\pi}{4}$ with the y-axis and an angle of $\frac{\pi}{4}$ with the z-axis.

Let the angles made by the line with the positive direction of the x, y, and z-axes be $\alpha, \beta, \gamma$ respectively.

So, we are given $\beta = \frac{\pi}{4}$ and $\gamma = \frac{\pi}{4}$.

To Find:

The angle $\alpha$ which the line makes with the x-axis.

Solution:

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

We know that for any line, the sum of the squares of its direction cosines is equal to 1.

Substitute the given values of $\beta = \frac{\pi}{4}$ and $\gamma = \frac{\pi}{4}$ into equation (i):

$\cos^2 \alpha + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \left(\frac{\pi}{4}\right) = 1$

We know that the value of $\cos \left(\frac{\pi}{4}\right)$ is $\frac{1}{\sqrt{2}}$.

So, $\cos^2 \left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

Substitute this value into the equation:

$\cos^2 \alpha + \frac{1}{2} + \frac{1}{2} = 1$

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

Subtracting 1 from both sides, we get:

$\cos^2 \alpha = 1 - 1$

$\cos^2 \alpha = 0$

Taking the square root of both sides gives:

$\cos \alpha = 0$

For an angle $\alpha$ between $0$ and $\pi$, the value of $\alpha$ for which $\cos \alpha = 0$ is $\frac{\pi}{2}$.

Thus, the angle which the line makes with the x-axis is $\frac{\pi}{2}$.

Answer:

The angle which it makes with x-axis is $\frac{π}{2}$.

Fill in the blank: $\frac{π}{2}$

Given:

Three points: $A(1, 2, 3)$, $B(-2, 3, 4)$, and $C(7, 0, 1)$.

To Determine:

Whether the given statement "The points are collinear" is True or False.

Solution:

Three points A, B, and C are collinear if the direction ratios of the line segment AB are proportional to the direction ratios of the line segment BC (or AC, or AB and AC, etc.).

Find the direction ratios of the line segment AB:

Direction ratios of AB are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$, where $(x_1, y_1, z_1) = (1, 2, 3)$ and $(x_2, y_2, z_2) = (-2, 3, 4)$.

Direction ratios of AB $= (-2 - 1, 3 - 2, 4 - 3) = (-3, 1, 1)$.

Find the direction ratios of the line segment BC:

Direction ratios of BC are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$, where $(x_1, y_1, z_1) = (-2, 3, 4)$ and $(x_2, y_2, z_2) = (7, 0, 1)$.

Direction ratios of BC $= (7 - (-2), 0 - 3, 1 - 4) = (7 + 2, -3, -3) = (9, -3, -3)$.

For the points to be collinear, the direction ratios of AB must be proportional to the direction ratios of BC. This means there exists a constant $k$ such that:

$-3 = k \times 9$

$1 = k \times (-3)$

$1 = k \times (-3)$

From the first equation: $k = -\frac{3}{9} = -\frac{1}{3}$.

From the second equation: $k = -\frac{1}{3}$.

From the third equation: $k = -\frac{1}{3}$.

Since we found a consistent non-zero constant of proportionality $k = -\frac{1}{3}$ for all three components of the direction ratios, the direction ratios are proportional.

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

Alternatively, using the determinant of the vectors:

Three points A, B, and C are collinear if the vectors $\vec{AB}$ and $\vec{AC}$ are parallel, which means one is a scalar multiple of the other. This is equivalent to the condition that the scalar triple product of vectors formed by the points is zero if they were in a plane (which is always true for collinear points). A more direct method is to check if the area of the triangle formed by the points is zero.

$\vec{AB} = B - A = (-2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = -3\hat{i} + \hat{j} + \hat{k}$.

$\vec{AC} = C - A = (7-1)\hat{i} + (0-2)\hat{j} + (1-3)\hat{k} = 6\hat{i} - 2\hat{j} - 2\hat{k}$.

Check if $\vec{AC}$ is a scalar multiple of $\vec{AB}$:

$6\hat{i} - 2\hat{j} - 2\hat{k} = c(-3\hat{i} + \hat{j} + \hat{k})$

Comparing coefficients:

$6 = -3c \implies c = -2$

$-2 = c$

$-2 = c$

Since $c = -2$ satisfies all three component equations, $\vec{AC} = -2 \vec{AB}$. This means the vectors $\vec{AB}$ and $\vec{AC}$ are parallel, and since they share a common point A, the points A, B, and C are collinear.

Conclusion:

The points $(1, 2, 3)$, $(-2, 3, 4)$, and $(7, 0, 1)$ are collinear.

Answer:

The statement is True.

Given:

Point $A(3, 5, 4)$ and point $B(5, 8, 11)$.

Given vector equation of the line: $\vec{r}= 3\hat{i}+ 4\hat{j}+ 4\hat{k} + λ \left( 2\hat{i}+ 3\hat{j}+ 7\hat{k} \right)$.

To Determine:

Whether the given statement "The vector equation represents the line passing through the points (3, 5, 4) and (5, 8, 11)" is True or False.

Solution:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{v}$ is given by $\vec{r} = \vec{a} + \lambda \vec{v}$.

The given equation is $\vec{r}= (3\hat{i}+ 4\hat{j}+ 4\hat{k}) + λ (2\hat{i}+ 3\hat{j}+ 7\hat{k})$.

This equation represents a line passing through the point with position vector $\vec{a_0} = 3\hat{i}+ 4\hat{j}+ 4\hat{k}$, which corresponds to the point $(3, 4, 4)$.

The direction vector of this line is $\vec{v} = 2\hat{i}+ 3\hat{j}+ 7\hat{k}$.

Now, let's find the direction vector of the line passing through the given points $A(3, 5, 4)$ and $B(5, 8, 11)$.

The position vector of point A is $\vec{a} = 3\hat{i} + 5\hat{j} + 4\hat{k}$.

The position vector of point B is $\vec{b} = 5\hat{i} + 8\hat{j} + 11\hat{k}$.

The direction vector of the line AB is $\vec{b} - \vec{a}$:

$\vec{AB} = (5\hat{i} + 8\hat{j} + 11\hat{k}) - (3\hat{i} + 5\hat{j} + 4\hat{k})$

$\vec{AB} = (5-3)\hat{i} + (8-5)\hat{j} + (11-4)\hat{k}$

$\vec{AB} = 2\hat{i} + 3\hat{j} + 7\hat{k}$

The direction vector of the given line equation $(2\hat{i}+ 3\hat{j}+ 7\hat{k})$ is the same as the direction vector of the line passing through the points A and B.

However, for the given equation to represent the line passing through A and B, the point $(3, 4, 4)$ (with position vector $3\hat{i}+ 4\hat{j}+ 4\hat{k}$) must lie on the line passing through A and B.

A correct vector equation for the line passing through A and B would be $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})$ or $\vec{r} = \vec{b} + \lambda (\vec{b} - \vec{a})$.

Using point A $(3, 5, 4)$ and direction vector $(2, 3, 7)$, a correct equation is:

$\vec{r} = (3\hat{i} + 5\hat{j} + 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 7\hat{k})$

The given equation is $\vec{r}= (3\hat{i}+ 4\hat{j}+ 4\hat{k}) + λ \left( 2\hat{i}+ 3\hat{j}+ 7\hat{k} \right)$. The starting point is $(3, 4, 4)$ which is different from $(3, 5, 4)$ and $(5, 8, 11)$.

To check if the point $(3, 4, 4)$ lies on the line through A and B, we can substitute its coordinates into the parametric form of the line through A and B:

$\frac{x-3}{2} = \frac{y-5}{3} = \frac{z-4}{7} = \mu$

For point $(3, 4, 4)$:

$\frac{3-3}{2} = \frac{0}{2} = 0$

$\frac{4-5}{3} = \frac{-1}{3}$

Since $0 \neq -\frac{1}{3}$, the point $(3, 4, 4)$ does not lie on the line passing through A and B.

Therefore, the given vector equation does not represent the line passing through the points (3, 5, 4) and (5, 8, 11).

Conclusion:

The statement is False.

Answer:

The statement is False.

Solution:

Given:

Point A is in space.

Position vector of A is $\overrightarrow{OA}$.

The vector $\overrightarrow{OA}$ is inclined at $60^\circ$ to the positive x-axis (OX).

The vector $\overrightarrow{OA}$ is inclined at $45^\circ$ to the positive y-axis (OY).

The magnitude of the vector is $|\overrightarrow{OA}| = 10$ units.

To Find:

The position vector of point A.

Calculation:

Let the position vector of point A be $\overrightarrow{OA} = x\hat{i} + y\hat{j} + z\hat{k}$.

The magnitude of the vector is given by $|\overrightarrow{OA}| = \sqrt{x^2 + y^2 + z^2}$.

We are given $|\overrightarrow{OA}| = 10$, so $\sqrt{x^2 + y^2 + z^2} = 10$.

Let $\alpha$, $\beta$, and $\gamma$ be the angles the vector $\overrightarrow{OA}$ makes with the positive x, y, and z axes, respectively. These are the direction angles.

We are given the direction angles $\alpha = 60^\circ$ and $\beta = 45^\circ$.

The components of the vector are related to its magnitude and direction cosines ($\cos \alpha$, $\cos \beta$, $\cos \gamma$) by:

$x = |\overrightarrow{OA}| \cos \alpha$

$y = |\overrightarrow{OA}| \cos \beta$

$z = |\overrightarrow{OA}| \cos \gamma$

We know that the sum of the squares of the direction cosines of any vector in three dimensions is equal to 1.

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$

Substitute the given angles:

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

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

Substituting these values into the direction cosine relation:

$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1$

$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$

Combine the fractions on the left side:

$\frac{1}{4} + \frac{2}{4} + \cos^2 \gamma = 1$

$\frac{3}{4} + \cos^2 \gamma = 1$

Isolate $\cos^2 \gamma$:

$\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$

Taking the square root of both sides to find $\cos \gamma$:

$\cos \gamma = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$

This indicates that the angle $\gamma$ the vector makes with the positive z-axis can be $60^\circ$ (if $\cos \gamma = 1/2$) or $120^\circ$ (if $\cos \gamma = -1/2$).

Now, we can find the components $x$, $y$, and $z$ using the magnitude $|\overrightarrow{OA}| = 10$ and the calculated direction cosines:

$x = |\overrightarrow{OA}| \cos \alpha = 10 \times \frac{1}{2} = 5$

$y = |\overrightarrow{OA}| \cos \beta = 10 \times \frac{1}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

For the z-component, we have two possibilities for $\cos \gamma$:

Case 1: $\cos \gamma = \frac{1}{2}$

$z = |\overrightarrow{OA}| \cos \gamma = 10 \times \frac{1}{2} = 5$

In this case, the position vector $\overrightarrow{OA}$ is $5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}$.

Case 2: $\cos \gamma = -\frac{1}{2}$

$z = |\overrightarrow{OA}| \cos \gamma = 10 \times \left(-\frac{1}{2}\right) = -5$

In this case, the position vector $\overrightarrow{OA}$ is $5\hat{i} + 5\sqrt{2}\hat{j} - 5\hat{k}$.

Since the question does not provide any information to restrict the location of point A relative to the xy-plane (such as being in the upper or lower half-space, or having a positive or negative z-coordinate), both possible values for $\gamma$ are valid, leading to two possible position vectors for point A.

Conclusion:

The possible position vectors of point A are $5\hat{i} + 5\sqrt{2}\hat{j} + 5\hat{k}$ or $5\hat{i} + 5\sqrt{2}\hat{j} - 5\hat{k}$.

Solution:

Given:

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

The line is parallel to the vector $\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.

To Find:

The vector equation of the line.

Calculation:

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.

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

The position vector $\vec{a}$ of this point is $1\hat{i} - 2\hat{j} + 3\hat{k}$.

So, $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$.

The line is parallel to the vector $\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.

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

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

This is the required vector equation of the line.

Conclusion:

The vector equation of the line is $\vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda (3\hat{i} - 2\hat{j} + 6\hat{k})$, where $\lambda$ is a scalar.

Solution:

Given:

Equation of the first line ($L_1$): $\frac{x − 1}{2} = \frac{y − 2}{3} = \frac{z − 3}{4}$

Equation of the second line ($L_2$): $\frac{x − 4}{5} = \frac{y − 1}{2} = z$

To Prove:

The lines $L_1$ and $L_2$ intersect.

To Find:

The point of intersection of the lines.

Calculation:

We can represent the coordinates of any point on the line $L_1$ in terms of a parameter, say $\lambda$.

Let $\frac{x − 1}{2} = \frac{y − 2}{3} = \frac{z − 3}{4} = \lambda$.

From this, we get the parametric equations for line $L_1$:

$x = 2\lambda + 1$

$y = 3\lambda + 2$

$z = 4\lambda + 3$

Thus, any point on line $L_1$ can be represented as $(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$.

Similarly, we can represent the coordinates of any point on the line $L_2$ in terms of another parameter, say $\mu$.

Let $\frac{x − 4}{5} = \frac{y − 1}{2} = \frac{z - 0}{1} = \mu$. Note that $z$ is equivalent to $\frac{z-0}{1}$.

From this, we get the parametric equations for line $L_2$:

$x = 5\mu + 4$

$y = 2\mu + 1$

$z = 1\mu + 0 = \mu$

Thus, any point on line $L_2$ can be represented as $(5\mu + 4, 2\mu + 1, \mu)$.

For the lines to intersect, there must exist values of $\lambda$ and $\mu$ such that the coordinates $(x, y, z)$ of a point on $L_1$ are the same as the coordinates of a point on $L_2$.

Equating the corresponding coordinates, we get a system of three linear equations in $\lambda$ and $\mu$:

We can solve the first two equations for $\lambda$ and $\mu$. Let's rearrange them:

Let's solve this system using elimination. Multiply equation (4) by 2 and equation (5) by 5:

Subtract equation (6) from equation (7):

$(15\lambda - 10\mu) - (4\lambda - 10\mu) = -5 - 6$

$11\lambda = -11$

$\lambda = -1$

Now substitute $\lambda = -1$ into equation (4):

$2(-1) - 5\mu = 3$

$-2 - 5\mu = 3$

$-5\mu = 3 + 2$

$-5\mu = 5$

$\mu = -1$

Now, we must check if these values of $\lambda = -1$ and $\mu = -1$ satisfy the third equation (3).

Substitute $\lambda = -1$ and $\mu = -1$ into equation (3):

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

$-4 + 3 = -1$

$-1 = -1$

Since the values $\lambda = -1$ and $\mu = -1$ satisfy all three equations, the lines intersect.

Point of Intersection:

To find the point of intersection, substitute the value of $\lambda = -1$ into the parametric equations of line $L_1$:

$x = 2\lambda + 1 = 2(-1) + 1 = -2 + 1 = -1$

$y = 3\lambda + 2 = 3(-1) + 2 = -3 + 2 = -1$

$z = 4\lambda + 3 = 4(-1) + 3 = -4 + 3 = -1$

The point of intersection is $(-1, -1, -1)$.

Alternatively, substitute the value of $\mu = -1$ into the parametric equations of line $L_2$:

$x = 5\mu + 4 = 5(-1) + 4 = -5 + 4 = -1$

$y = 2\mu + 1 = 2(-1) + 1 = -2 + 1 = -1$

$z = \mu = -1$

This gives the same point of intersection $(-1, -1, -1)$, which confirms our result.

Conclusion:

Since there exist unique values of $\lambda$ and $\mu$ that satisfy the equations for equal coordinates, the lines intersect.

The point of intersection of the two lines is $(-1, -1, -1)$.

Solution:

Given:

Equation of the first line ($L_1$): $\vec{r}= 3\hat{i}− 2\hat{j}+ 6\hat{k}+ λ \left( 2\hat{i}+ \hat{j}+ 2\hat{k} \right)$

Equation of the second line ($L_2$): $\vec{r}= (2\hat{j}− 5\hat{k}) + μ \left( 6\hat{i}+ 3\hat{j}+ 2\hat{k} \right)$

To Find:

The angle between the two lines.

Calculation:

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 $\vec{r} = \vec{a} + \lambda \vec{b}$. The vector $\vec{b}$ is the direction vector of the line.

From the equation of the first line ($L_1$), the direction vector is $\vec{b}_1 = 2\hat{i}+ \hat{j}+ 2\hat{k}$.

From the equation of the second line ($L_2$), the direction vector is $\vec{b}_2 = 6\hat{i}+ 3\hat{j}+ 2\hat{k}$.

The angle $\theta$ between two lines with direction vectors $\vec{b}_1$ and $\vec{b}_2$ is given by the formula:

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

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

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

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

$\vec{b}_1 \cdot \vec{b}_2 = 12 + 3 + 4 = 19$

Next, calculate the magnitudes of $\vec{b}_1$ and $\vec{b}_2$:

$|\vec{b}_1| = |2\hat{i}+ \hat{j}+ 2\hat{k}| = \sqrt{2^2 + 1^2 + 2^2}$

$|\vec{b}_1| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

$|\vec{b}_2| = |6\hat{i}+ 3\hat{j}+ 2\hat{k}| = \sqrt{6^2 + 3^2 + 2^2}$

$|\vec{b}_2| = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$

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:

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

Conclusion:

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

Solution:

Given:

Line $L_1$ passes through points A (0, –1, –1) and B (4, 5, 1).

Line $L_2$ passes through points C (3, 9, 4) and D (– 4, 4, 4).

To Prove:

The line through A and B intersects the line through C and D.

Proof:

First, we find the vector equation of the line $L_1$ passing through A (0, –1, –1) and B (4, 5, 1).

The position vector of point A is $\vec{a} = 0\hat{i} - 1\hat{j} - 1\hat{k} = -\hat{j} - \hat{k}$.

The direction vector of the line AB is $\vec{b}_1 = \vec{B} - \vec{A}$.

$\vec{b}_1 = (4\hat{i} + 5\hat{j} + \hat{k}) - (0\hat{i} - \hat{j} - \hat{k})$

$\vec{b}_1 = (4-0)\hat{i} + (5-(-1))\hat{j} + (1-(-1))\hat{k}$

$\vec{b}_1 = 4\hat{i} + 6\hat{j} + 2\hat{k}$.

The vector equation of line $L_1$ is $\vec{r} = \vec{a} + \lambda \vec{b}_1$, where $\lambda$ is a scalar parameter.

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

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

The parametric equations for any point $(x, y, z)$ on $L_1$ are:

$x = 4\lambda$

$y = 6\lambda - 1$

$z = 2\lambda - 1$

Next, we find the vector equation of the line $L_2$ passing through C (3, 9, 4) and D (– 4, 4, 4).

The position vector of point C is $\vec{c} = 3\hat{i} + 9\hat{j} + 4\hat{k}$.

The direction vector of the line CD is $\vec{b}_2 = \vec{D} - \vec{C}$.

$\vec{b}_2 = (-4\hat{i} + 4\hat{j} + 4\hat{k}) - (3\hat{i} + 9\hat{j} + 4\hat{k})$

$\vec{b}_2 = (-4-3)\hat{i} + (4-9)\hat{j} + (4-4)\hat{k}$

$\vec{b}_2 = -7\hat{i} - 5\hat{j} + 0\hat{k} = -7\hat{i} - 5\hat{j}$.

The vector equation of line $L_2$ is $\vec{r} = \vec{c} + \mu \vec{b}_2$, where $\mu$ is a scalar parameter.

$\vec{r} = (3\hat{i} + 9\hat{j} + 4\hat{k}) + \mu (-7\hat{i} - 5\hat{j})$

$\vec{r} = (3 - 7\mu)\hat{i} + (9 - 5\mu)\hat{j} + 4\hat{k}$.

The parametric equations for any point $(x, y, z)$ on $L_2$ are:

$x = 3 - 7\mu$

$y = 9 - 5\mu$

$z = 4$

For the lines to intersect, there must be a common point, meaning there exist values of $\lambda$ and $\mu$ such that the coordinates $(x, y, z)$ from both sets of parametric equations are equal.

Equating the coordinates:

We solve the system of these three linear equations for $\lambda$ and $\mu$.

From equation (3):

$2\lambda - 1 = 4$

$2\lambda = 4 + 1$

$2\lambda = 5$

$\lambda = \frac{5}{2}$

Substitute the value of $\lambda = \frac{5}{2}$ into equation (1):

$4\left(\frac{5}{2}\right) = 3 - 7\mu$

$10 = 3 - 7\mu$

$7\mu = 3 - 10$

$7\mu = -7$

$\mu = -1$

Now, we must check if these values $\lambda = \frac{5}{2}$ and $\mu = -1$ satisfy the remaining equation, equation (2).

Substitute $\lambda = \frac{5}{2}$ and $\mu = -1$ into equation (2):

$6\left(\frac{5}{2}\right) - 1 = 9 - 5(-1)$

$3 \times 5 - 1 = 9 + 5$

$15 - 1 = 14$

$14 = 14$

Since the values $\lambda = \frac{5}{2}$ and $\mu = -1$ satisfy all three equations, there exists a common point that lies on both lines. Therefore, the lines intersect.

Conclusion:

Since the system of equations derived from equating the coordinates of points on the two lines has a consistent solution for $\lambda$ and $\mu$, the lines intersect.

Solution:

Given:

Equation of the first line ($L_1$):

$x = py + q$

$z = ry + s$

Equation of the second line ($L_2$):

$x = p'y + q'$

$z = r'y + s'$

To Prove:

The lines $L_1$ and $L_2$ are perpendicular if $pp' + rr' + 1 = 0$.

Proof:

To find the direction vector of the first line, we can express the coordinates in terms of a parameter. Let's use $y$ as the parameter.

For line $L_1$, if we let $y = t$, then the coordinates of any point on the line are given by:

$x = pt + q$

$y = t$

$z = rt + s$

A point on the line for $t=0$ is $(q, 0, s)$.

A point on the line for $t=1$ is $(p+q, 1, r+s)$.

The direction vector $\vec{d}_1$ of line $L_1$ is the vector connecting these two points:

$\vec{d}_1 = (p+q - q)\hat{i} + (1 - 0)\hat{j} + (r+s - s)\hat{k}$

$\vec{d}_1 = p\hat{i} + 1\hat{j} + r\hat{k}$

Similarly, for the second line $L_2$, let's use $y$ as the parameter. If we let $y = u$, the coordinates of any point on the line are given by:

$x = p'u + q'$

$y = u$

$z = r'u + s'$

A point on the line for $u=0$ is $(q', 0, s')$.

A point on the line for $u=1$ is $(p'+q', 1, r'+s')$.

The direction vector $\vec{d}_2$ of line $L_2$ is the vector connecting these two points:

$\vec{d}_2 = (p'+q' - q')\hat{i} + (1 - 0)\hat{j} + (r'+s' - s')\hat{k}$

$\vec{d}_2 = p'\hat{i} + 1\hat{j} + r'\hat{k}$

Two lines are perpendicular if and only if their direction vectors are orthogonal. This means the dot product of their direction vectors must be zero.

The dot product of $\vec{d}_1$ and $\vec{d}_2$ is:

$\vec{d}_1 \cdot \vec{d}_2 = (p\hat{i} + \hat{j} + r\hat{k}) \cdot (p'\hat{i} + \hat{j} + r'\hat{k})$

$\vec{d}_1 \cdot \vec{d}_2 = (p)(p') + (1)(1) + (r)(r')$

$\vec{d}_1 \cdot \vec{d}_2 = pp' + 1 + rr'$

For the lines to be perpendicular, $\vec{d}_1 \cdot \vec{d}_2 = 0$.

Therefore, $pp' + 1 + rr' = 0$, which can be rearranged as $pp' + rr' + 1 = 0$.

This is the required condition for the lines to be perpendicular.

Conclusion:

We have shown that the direction vectors of the given lines are $\vec{d}_1 = p\hat{i} + \hat{j} + r\hat{k}$ and $\vec{d}_2 = p'\hat{i} + \hat{j} + r'\hat{k}$. The lines are perpendicular if their direction vectors are orthogonal, which means their dot product is zero. The dot product is $pp' + rr' + 1$. Setting the dot product to zero gives the condition $pp' + rr' + 1 = 0$. Thus, the lines are perpendicular if $pp' + rr' + 1 = 0$.

Solution:

Given:

Points A (2, 3, 4) and B (4, 5, 8).

The plane bisects the line segment AB perpendicularly.

To Find:

The equation of the plane.

Calculation:

A plane that bisects a line segment perpendicularly passes through the midpoint of the segment and has the segment's direction vector as its normal vector.

Step 1: Find the midpoint of the line segment AB.

Let M be the midpoint of AB. The coordinates of the midpoint of a segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

For points A (2, 3, 4) and B (4, 5, 8):

Midpoint M = $\left(\frac{2+4}{2}, \frac{3+5}{2}, \frac{4+8}{2}\right)$

Midpoint M = $\left(\frac{6}{2}, \frac{8}{2}, \frac{12}{2}\right)$

Midpoint M = (3, 4, 6)

The plane passes through the point M (3, 4, 6).

The position vector of this point is $\vec{r}_0 = 3\hat{i} + 4\hat{j} + 6\hat{k}$.

Step 2: Find the direction vector of the line segment AB.

The direction vector of the line segment AB is given by $\vec{AB} = \vec{B} - \vec{A}$.

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

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

$\vec{AB} = 2\hat{i} + 2\hat{j} + 4\hat{k}$

Since the plane is perpendicular to the line AB, the direction vector $\vec{AB}$ is a normal vector to the plane. Let the normal vector be $\vec{n} = 2\hat{i} + 2\hat{j} + 4\hat{k}$.

We can use any vector parallel to this as the normal vector, for simplicity, we can use $\vec{n} = \hat{i} + \hat{j} + 2\hat{k}$ (by dividing the components by 2). However, using the original vector is also correct and might avoid fractional coefficients initially.

Step 3: Write the equation of the plane.

The equation of a plane passing through a point with position vector $\vec{r}_0$ and having a normal vector $\vec{n}$ is given by the vector equation $(\vec{r} - \vec{r}_0) \cdot \vec{n} = 0$, or $\vec{r} \cdot \vec{n} = \vec{r}_0 \cdot \vec{n}$.

Here, $\vec{r}_0 = 3\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{n} = 2\hat{i} + 2\hat{j} + 4\hat{k}$.

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ be the position vector of any point on the plane.

The dot product $\vec{r} \cdot \vec{n}$ is:

$\vec{r} \cdot \vec{n} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 4\hat{k}) = 2x + 2y + 4z$

The dot product $\vec{r}_0 \cdot \vec{n}$ is:

$\vec{r}_0 \cdot \vec{n} = (3\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 4\hat{k}) = (3)(2) + (4)(2) + (6)(4)$

$\vec{r}_0 \cdot \vec{n} = 6 + 8 + 24 = 38$

So, the equation of the plane is:

$2x + 2y + 4z = 38$

We can simplify this equation by dividing the entire equation by 2:

$x + y + 2z = 19$

Conclusion:

The equation of the plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) is $x + y + 2z = 19$.

Solution:

Given:

The plane is at a distance of $3\sqrt{3}$ units from the origin.

The normal to the plane is equally inclined to the coordinate axes.

To Find:

The equation of the plane.

Calculation:

The equation of a plane in the normal form is given by $lx + my + nz = d$, where $(l, m, n)$ are the direction cosines of the normal to the plane and $d$ is the perpendicular distance of the plane from the origin. The distance $d$ is always taken as non-negative.

We are given that the distance from the origin is $d = 3\sqrt{3}$.

We are also given that the normal to the plane is equally inclined to the coordinate axes. Let the angles made by the normal with the positive x, y, and z axes be $\alpha$, $\beta$, and $\gamma$, respectively. Since the normal is equally inclined, we have $\alpha = \beta = \gamma$.

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

Since $\alpha = \beta = \gamma$, it follows that $\cos \alpha = \cos \beta = \cos \gamma$.

Thus, the direction cosines are equal: $l = m = n$.

The sum of the squares of the direction cosines of any vector is equal to 1. So, we have:

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

Substitute $l=m=n$ into this equation:

$l^2 + l^2 + l^2 = 1$

$3l^2 = 1$

$l^2 = \frac{1}{3}$

Taking the square root, we get:

$l = \pm \frac{1}{\sqrt{3}}$

Since $l = m = n$, the possible sets of direction cosines for the normal are:

Case 1: $(l, m, n) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Case 2: $(l, m, n) = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$

Now, we can write the equation of the plane $lx + my + nz = d$ for each case, using the given distance $d = 3\sqrt{3}$.

Case 1: $(l, m, n) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ and $d = 3\sqrt{3}$

The equation of the plane is:

$\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = 3\sqrt{3}$

Multiply the entire equation by $\sqrt{3}$ to simplify:

$\sqrt{3} \left(\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z\right) = \sqrt{3} \left(3\sqrt{3}\right)$

$x + y + z = 3 \times 3 = 9$

So, one possible equation of the plane is $x + y + z = 9$.

Case 2: $(l, m, n) = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$ and $d = 3\sqrt{3}$

The equation of the plane is:

$-\frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}}y - \frac{1}{\sqrt{3}}z = 3\sqrt{3}$

Multiply the entire equation by $-\sqrt{3}$ to simplify:

$-\sqrt{3} \left(-\frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}}y - \frac{1}{\sqrt{3}}z\right) = -\sqrt{3} \left(3\sqrt{3}\right)$

$x + y + z = -(3 \times 3) = -9$

So, the other possible equation of the plane is $x + y + z = -9$.

Both equations represent planes that are at a distance of $3\sqrt{3}$ from the origin and whose normals are equally inclined to the coordinate axes. The two planes are parallel and on opposite sides of the origin.

Conclusion:

The equations of the planes satisfying the given conditions are $x + y + z = 9$ and $x + y + z = -9$.

Solution:

Given:

A line passes through point P (–2, –1, –3).

The line meets a plane at point Q (1, –3, 3).

The line is perpendicular to the plane at point Q.

To Find:

The equation of the plane.

Calculation:

Let the point from which the line is drawn be P(–2, –1, –3) and the point where the line meets the plane at a right angle be Q(1, –3, 3).

Since the line PQ is perpendicular to the plane at Q, the vector $\overrightarrow{PQ}$ is a normal vector to the plane.

The components of the vector $\overrightarrow{PQ}$ are found by subtracting the coordinates of P from the coordinates of Q:

$\overrightarrow{PQ} = (x_Q - x_P)\hat{i} + (y_Q - y_P)\hat{j} + (z_Q - z_P)\hat{k}$

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

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

$\overrightarrow{PQ} = 3\hat{i} - 2\hat{j} + 6\hat{k}$

So, the normal vector to the plane is $\vec{n} = 3\hat{i} - 2\hat{j} + 6\hat{k}$. The direction ratios of the normal are (3, -2, 6).

The plane passes through the point Q (1, –3, 3).

The equation of a plane with normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ passing through a point $(x_0, y_0, z_0)$ is given by:

$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

Here, $(x_0, y_0, z_0) = (1, -3, 3)$ and $(a, b, c) = (3, -2, 6)$.

Substitute these values into the equation:

$3(x - 1) + (-2)(y - (-3)) + 6(z - 3) = 0$

$3(x - 1) - 2(y + 3) + 6(z - 3) = 0$

Expand and simplify the equation:

$3x - 3 - 2y - 6 + 6z - 18 = 0$

Group the terms:

$3x - 2y + 6z - 3 - 6 - 18 = 0$

$3x - 2y + 6z - 27 = 0$

Rearrange the equation:

$3x - 2y + 6z = 27$

Conclusion:

The equation of the plane is $3x - 2y + 6z = 27$.

Solution:

Given:

The plane passes through the points A (2, 1, 0), B (3, –2, –2), and C (3, 1, 7).

To Find:

The equation of the plane.

Calculation:

To find the equation of a plane passing through three non-collinear points, we can find two vectors lying in the plane and use their cross product to determine a normal vector to the plane. Then, using one of the points and the normal vector, we can write the equation of the plane.

Let the points be A(2, 1, 0), B(3, –2, –2), and C(3, 1, 7).

Two vectors in the plane can be $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Calculate the vector $\overrightarrow{AB}$:

$\overrightarrow{AB} = \vec{B} - \vec{A} = (3-2)\hat{i} + (-2-1)\hat{j} + (-2-0)\hat{k}$

$\overrightarrow{AB} = 1\hat{i} - 3\hat{j} - 2\hat{k}$

Calculate the vector $\overrightarrow{AC}$:

$\overrightarrow{AC} = \vec{C} - \vec{A} = (3-2)\hat{i} + (1-1)\hat{j} + (7-0)\hat{k}$

$\overrightarrow{AC} = 1\hat{i} + 0\hat{j} + 7\hat{k}$

The normal vector $\vec{n}$ to the plane is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{AC}$. We can find the normal vector by taking the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & -2 \\ 1 & 0 & 7 \end{vmatrix}$

Expand the determinant:

$\vec{n} = \hat{i}[(-3)(7) - (-2)(0)] - \hat{j}[(1)(7) - (-2)(1)] + \hat{k}[(1)(0) - (-3)(1)]$

$\vec{n} = \hat{i}[-21 - 0] - \hat{j}[7 + 2] + \hat{k}[0 + 3]$

$\vec{n} = -21\hat{i} - 9\hat{j} + 3\hat{k}$

The normal vector to the plane is $-21\hat{i} - 9\hat{j} + 3\hat{k}$. We can use a parallel vector as the normal vector for simplicity. Divide the components by -3:

$\vec{n}' = \frac{1}{-3}(-21\hat{i} - 9\hat{j} + 3\hat{k}) = 7\hat{i} + 3\hat{j} - \hat{k}$

The direction ratios of the normal vector are (7, 3, -1).

The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.

We can use point A (2, 1, 0) as $(x_0, y_0, z_0)$ and the normal vector $\vec{n}'$ with $(a, b, c) = (7, 3, -1)$.

$7(x - 2) + 3(y - 1) + (-1)(z - 0) = 0$

Expand and simplify the equation:

$7x - 14 + 3y - 3 - z = 0$

$7x + 3y - z - 17 = 0$

$7x + 3y - z = 17$

We can verify this equation by checking if points B and C also satisfy it.

For B(3, –2, –2): $7(3) + 3(-2) - (-2) = 21 - 6 + 2 = 17$. (Satisfied)

For C(3, 1, 7): $7(3) + 3(1) - (7) = 21 + 3 - 7 = 17$. (Satisfied)

The equation is consistent with all three points.

Conclusion:

The equation of the plane passing through the given points is $7x + 3y - z = 17$.

Solution:

Given:

The line $L_1$: $\frac{x − 3}{2} = \frac{y − 3}{1} = \frac{z}{1}$.

Two lines pass through the origin O (0, 0, 0).

These lines intersect $L_1$ at an angle of $\frac{\pi}{3}$ each.

To Find:

The equations of the two lines through the origin.

Calculation:

Let the given line be $L_1$. Its equation in symmetric form is $\frac{x − 3}{2} = \frac{y − 3}{1} = \frac{z - 0}{1}$.

This line passes through the point A(3, 3, 0) and has a direction vector $\vec{b}_1 = 2\hat{i} + \hat{j} + \hat{k}$.

Let the required line be $L_2$. Since $L_2$ passes through the origin O(0, 0, 0), its equation can be written as $\vec{r} = k \vec{d}$, where $\vec{d}$ is the direction vector of $L_2$ and $k$ is a scalar parameter.

The required line $L_2$ intersects the given line $L_1$. Let the point of intersection be Q. Since $L_2$ passes through the origin and Q, the vector $\overrightarrow{OQ}$ is parallel to the direction vector $\vec{d}$ of $L_2$.

Any point Q on the line $L_1$ can be represented parametrically. Let $\frac{x − 3}{2} = \frac{y − 3}{1} = \frac{z}{1} = t$.

The coordinates of Q are $(x, y, z) = (2t + 3, t + 3, t)$.

The position vector of Q is $\overrightarrow{OQ} = (2t + 3)\hat{i} + (t + 3)\hat{j} + t\hat{k}$.

This vector $\overrightarrow{OQ}$ serves as the direction vector for the line $L_2$ passing through the origin and Q.

So, we can take $\vec{d} = \overrightarrow{OQ} = (2t + 3)\hat{i} + (t + 3)\hat{j} + t\hat{k}$.

The angle between the lines $L_1$ and $L_2$ is the angle between their direction vectors $\vec{b}_1 = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{d} = (2t + 3)\hat{i} + (t + 3)\hat{j} + t\hat{k}$.

We are given that this angle is $\theta = \frac{\pi}{3}$. The cosine of the angle between two vectors is given by:

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

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

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

$\vec{b}_1 \cdot \vec{d} = 2(2t + 3) + 1(t + 3) + 1(t)$

$\vec{b}_1 \cdot \vec{d} = 4t + 6 + t + 3 + t = 6t + 9$

Calculate the magnitude of $\vec{b}_1$:

$|\vec{b}_1| = |2\hat{i} + \hat{j} + \hat{k}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

Calculate the magnitude of $\vec{d}$:

$|\vec{d}| = |(2t + 3)\hat{i} + (t + 3)\hat{j} + t\hat{k}| = \sqrt{(2t + 3)^2 + (t + 3)^2 + t^2}$

$|\vec{d}| = \sqrt{(4t^2 + 12t + 9) + (t^2 + 6t + 9) + t^2}$

$|\vec{d}| = \sqrt{6t^2 + 18t + 18}$

Substitute these values into the angle formula, with $\theta = \frac{\pi}{3}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$:

$\frac{1}{2} = \frac{|6t + 9|}{\sqrt{6t^2 + 18t + 18} \sqrt{6}}$

Since $6t+9$ can be positive or negative, we consider the absolute value. Squaring both sides will handle this.

$\left(\frac{1}{2}\right)^2 = \frac{(6t + 9)^2}{(6t^2 + 18t + 18)(6)}$

$\frac{1}{4} = \frac{36t^2 + 108t + 81}{36t^2 + 108t + 108}$

Cross-multiply:

$36t^2 + 108t + 108 = 4(36t^2 + 108t + 81)$

$36t^2 + 108t + 108 = 144t^2 + 432t + 324$

Rearrange the terms to form a quadratic equation:

$144t^2 - 36t^2 + 432t - 108t + 324 - 108 = 0$

$108t^2 + 324t + 216 = 0$

Divide the equation by 108:

Solve the quadratic equation (1):

Factor the quadratic expression:

$(t + 1)(t + 2) = 0$

This gives two possible values for $t$:

$t = -1$ or $t = -2$

Each value of $t$ corresponds to a point of intersection Q on line $L_1$, and the vector $\overrightarrow{OQ}$ gives the direction vector of one of the required lines passing through the origin.

Case 1: $t = -1$

The point of intersection Q$_1$ is $(2(-1) + 3, -1 + 3, -1) = (-2 + 3, 2, -1) = (1, 2, -1)$.

The direction vector of the first line is $\vec{d}_1 = \overrightarrow{OQ_1} = 1\hat{i} + 2\hat{j} - 1\hat{k}$.

The vector equation of the first line is $\vec{r} = \lambda \vec{d}_1$, where $\lambda$ is a scalar parameter.

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

The Cartesian equation of the first line (passing through origin (0,0,0) with direction ratios (1, 2, -1)) is:

$\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{-1}$

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

Case 2: $t = -2$

The point of intersection Q$_2$ is $(2(-2) + 3, -2 + 3, -2) = (-4 + 3, 1, -2) = (-1, 1, -2)$.

The direction vector of the second line is $\vec{d}_2 = \overrightarrow{OQ_2} = -1\hat{i} + 1\hat{j} - 2\hat{k}$.

The vector equation of the second line is $\vec{r} = \mu \vec{d}_2$, where $\mu$ is a scalar parameter.

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

The Cartesian equation of the second line (passing through origin (0,0,0) with direction ratios (-1, 1, -2)) is:

$\frac{x - 0}{-1} = \frac{y - 0}{1} = \frac{z - 0}{-2}$

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

Conclusion:

There are two such lines. Their equations are:

Line 1:

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

Cartesian form: $\frac{x}{1} = \frac{y}{2} = \frac{z}{-1}$

Line 2:

Vector form: $\vec{r} = \mu (-\hat{i} + \hat{j} - 2\hat{k})$

Cartesian form: $\frac{x}{-1} = \frac{y}{1} = \frac{z}{-2}$

Solution:

Given:

The direction cosines $(l, m, n)$ of the two lines satisfy the following equations:

To Find:

The angle between the two lines.

Calculation:

From equation (1), we can express $n$ in terms of $l$ and $m$:

$n = -(l + m)$

Substitute this expression for $n$ into equation (2):

$l^2 + m^2 - (-(l + m))^2 = 0$

$l^2 + m^2 - (l + m)^2 = 0$

Expand the term $(l + m)^2$:

$l^2 + m^2 - (l^2 + 2lm + m^2) = 0$

$l^2 + m^2 - l^2 - 2lm - m^2 = 0$

Simplify the equation:

$-2lm = 0$

This equation implies that either $l = 0$ or $m = 0$ (or both, but we will see this is covered by the two cases).

Case 1: $l = 0$

Substitute $l = 0$ into equation (1):

$0 + m + n = 0$

$m + n = 0 \implies n = -m$

The direction cosines are proportional to $(0, m, -m)$.

Since $(l, m, n)$ are direction cosines, they must satisfy the property $l^2 + m^2 + n^2 = 1$.

Substitute $l=0$ and $n=-m$:

$0^2 + m^2 + (-m)^2 = 1$

$m^2 + m^2 = 1$

$2m^2 = 1$

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

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

If $m = \frac{1}{\sqrt{2}}$, then $n = -m = -\frac{1}{\sqrt{2}}$. The direction cosines for one line are $\left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

If $m = -\frac{1}{\sqrt{2}}$, then $n = -m = \frac{1}{\sqrt{2}}$. The direction cosines for the same line (opposite direction) are $\left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

Let the direction cosines of the first line be $l_1 = 0, m_1 = \frac{1}{\sqrt{2}}, n_1 = -\frac{1}{\sqrt{2}}$.

Case 2: $m = 0$

Substitute $m = 0$ into equation (1):

$l + 0 + n = 0$

$l + n = 0 \implies n = -l$

The direction cosines are proportional to $(l, 0, -l)$.

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

$l^2 + 0^2 + (-l)^2 = 1$

$l^2 + l^2 = 1$

$2l^2 = 1$

$l^2 = \frac{1}{2}$

$l = \pm \frac{1}{\sqrt{2}}$

If $l = \frac{1}{\sqrt{2}}$, then $n = -l = -\frac{1}{\sqrt{2}}$. The direction cosines for the second line are $\left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$.

If $l = -\frac{1}{\sqrt{2}}$, then $n = -l = \frac{1}{\sqrt{2}}$. The direction cosines for the same line (opposite direction) are $\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$.

Let the direction cosines of the second line be $l_2 = \frac{1}{\sqrt{2}}, m_2 = 0, n_2 = -\frac{1}{\sqrt{2}}$.

The angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by:

$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$

Substitute the direction cosines we found:

$\cos \theta = \left|\left(0\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(0\right) + \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)\right|$

$\cos \theta = \left|0 + 0 + \left(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\right)\right|$

$\cos \theta = \left|0 + 0 + \frac{1}{2}\right|$

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

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

$\theta = \cos^{-1}\left(\frac{1}{2}\right)$

The principal value of $\cos^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{3}$.

$\theta = \frac{\pi}{3}$

Conclusion:

The angle between the two lines is $\frac{\pi}{3}$ or $60^\circ$.

Solution:

Given:

Direction cosines of the line in the first position are $(l, m, n)$.

Direction cosines of the line in an adjacent position are $(l + \delta l, m + \delta m, n + \delta n)$.

$\delta\theta$ is the small angle between the two positions of the line.

To Prove:

$\delta\theta^2 = \delta l^2 + \delta m^2 + \delta n^2$ (for small $\delta\theta$)

Proof:

Since $(l, m, n)$ are the direction cosines of a line, they satisfy the relation:

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

Similarly, since $(l + \delta l, m + \delta m, n + \delta n)$ are also direction cosines of a line (in an adjacent position), they must satisfy:

$(l + \delta l)^2 + (m + \delta m)^2 + (n + \delta n)^2 = 1$

Expand the terms:

$(l^2 + 2l\delta l + (\delta l)^2) + (m^2 + 2m\delta m + (\delta m)^2) + (n^2 + 2n\delta n + (\delta n)^2) = 1$

Rearrange the terms:

$(l^2 + m^2 + n^2) + 2(l\delta l + m\delta m + n\delta n) + ((\delta l)^2 + (\delta m)^2 + (\delta n)^2) = 1$

Substitute $l^2 + m^2 + n^2 = 1$ into the equation:

$1 + 2(l\delta l + m\delta m + n\delta n) + (\delta l)^2 + (\delta m)^2 + (\delta n)^2 = 1$

Subtract 1 from both sides:

Let $\vec{d}_1 = l\hat{i} + m\hat{j} + n\hat{k}$ and $\vec{d}_2 = (l + \delta l)\hat{i} + (m + \delta m)\hat{j} + (n + \delta n)\hat{k}$ be the direction vectors of the line in the two positions. The magnitudes of these vectors are $|\vec{d}_1| = \sqrt{l^2 + m^2 + n^2} = 1$ and $|\vec{d}_2| = \sqrt{(l + \delta l)^2 + (m + \delta m)^2 + (n + \delta n)^2} = 1$.

The angle $\delta\theta$ between the two lines is the angle between their direction vectors. Using the dot product formula for the angle between two vectors:

$\cos(\delta\theta) = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1| |\vec{d}_2|}$

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

$\vec{d}_1 \cdot \vec{d}_2 = (l\hat{i} + m\hat{j} + n\hat{k}) \cdot ((l + \delta l)\hat{i} + (m + \delta m)\hat{j} + (n + \delta n)\hat{k})$

$\vec{d}_1 \cdot \vec{d}_2 = l(l + \delta l) + m(m + \delta m) + n(n + \delta n)$

$\vec{d}_1 \cdot \vec{d}_2 = l^2 + l\delta l + m^2 + m\delta m + n^2 + n\delta n$

$\vec{d}_1 \cdot \vec{d}_2 = (l^2 + m^2 + n^2) + (l\delta l + m\delta m + n\delta n)$

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

$\vec{d}_1 \cdot \vec{d}_2 = 1 + l\delta l + m\delta m + n\delta n$

Now, substitute this into the cosine formula, along with $|\vec{d}_1|=1$ and $|\vec{d}_2|=1$:

$\cos(\delta\theta) = \frac{1 + l\delta l + m\delta m + n\delta n}{1 \times 1}$

From equation (1), we can express $l\delta l + m\delta m + n\delta n$:

$2(l\delta l + m\delta m + n\delta n) = - ((\delta l)^2 + (\delta m)^2 + (\delta n)^2)$

$l\delta l + m\delta m + n\delta n = -\frac{1}{2} ((\delta l)^2 + (\delta m)^2 + (\delta n)^2)$

Substitute this into equation (2):

For a small angle $\delta\theta$, we use the Taylor series expansion of $\cos(\delta\theta)$ around 0:

$\cos(\delta\theta) = 1 - \frac{(\delta\theta)^2}{2!} + \frac{(\delta\theta)^4}{4!} - \dots$

For a sufficiently small angle $\delta\theta$, we can approximate $\cos(\delta\theta)$ by the first two terms:

Equating the two expressions for $\cos(\delta\theta)$ from equation (3) and the small angle approximation:

$1 - \frac{(\delta\theta)^2}{2} \approx 1 - \frac{1}{2} ((\delta l)^2 + (\delta m)^2 + (\delta n)^2)$

Subtract 1 from both sides:

$-\frac{(\delta\theta)^2}{2} \approx -\frac{1}{2} ((\delta l)^2 + (\delta m)^2 + (\delta n)^2)$

Multiply both sides by -2:

$(\delta\theta)^2 \approx (\delta l)^2 + (\delta m)^2 + (\delta n)^2$

Since $\delta l$, $\delta m$, and $\delta n$ represent small changes in direction cosines corresponding to a small angle $\delta\theta$, this relationship holds true for small $\delta\theta$. The higher-order terms in the Taylor expansion and the original equation become negligible as $\delta\theta$, $\delta l$, $\delta m$, $\delta n$ approach zero.

Conclusion:

Based on the properties of direction cosines and the small angle approximation for cosine, we have shown that the square of the small angle between the two positions of the line is given by $\delta\theta^2 = \delta l^2 + \delta m^2 + \delta n^2$.

Solution:

Given:

Origin O is the point (0, 0, 0).

Point A is (a, b, c).

To Find:

1. The direction cosines of the line OA.

2. The equation of the plane through A at right angle to OA.

Calculation:

Part 1: Direction Cosines of the line OA

The line OA passes through the origin O (0, 0, 0) and point A (a, b, c).

The direction ratios of the line OA are the components of the vector $\overrightarrow{OA}$.

$\overrightarrow{OA} = \vec{A} - \vec{O} = (a\hat{i} + b\hat{j} + c\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k})$

$\overrightarrow{OA} = a\hat{i} + b\hat{j} + c\hat{k}$

The direction ratios of OA are $(a, b, c)$.

To find the direction cosines, we need to divide the direction ratios by the magnitude of the vector $\overrightarrow{OA}$.

The magnitude of $\overrightarrow{OA}$ is $|\overrightarrow{OA}| = \sqrt{a^2 + b^2 + c^2}$.

Let $d = \sqrt{a^2 + b^2 + c^2}$. Note that $d$ is the distance of point A from the origin.

The direction cosines $(l, m, n)$ of the line OA are given by:

$l = \frac{a}{|\overrightarrow{OA}|} = \frac{a}{d}$

$m = \frac{b}{|\overrightarrow{OA}|} = \frac{b}{d}$

$n = \frac{c}{|\overrightarrow{OA}|} = \frac{c}{d}$

Thus, the direction cosines of the line OA are $\left(\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\right)$.

Part 2: Equation of the plane through A at right angle to OA

The plane passes through the point A (a, b, c).

The plane is at a right angle to the line OA. This means the line OA is normal to the plane.

The direction vector of the line OA is $\overrightarrow{OA} = a\hat{i} + b\hat{j} + c\hat{k}$.

This vector $\overrightarrow{OA}$ serves as the normal vector $\vec{n}$ to the plane. So, $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.

The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $a'\hat{i} + b'\hat{j} + c'\hat{k}$ is given by $a'(x - x_0) + b'(y - y_0) + c'(z - z_0) = 0$.

Here, the point is A $(x_0, y_0, z_0) = (a, b, c)$, and the normal vector has coefficients $(a', b', c') = (a, b, c)$.

Substitute these values into the equation:

$a(x - a) + b(y - b) + c(z - c) = 0$

Expand the terms:

$ax - a^2 + by - b^2 + cz - c^2 = 0$

Rearrange the terms:

$ax + by + cz = a^2 + b^2 + c^2$

Conclusion:

The direction cosines of the line OA are $\left(\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\right)$.

The equation of the plane through A at right angle to OA is $ax + by + cz = a^2 + b^2 + c^2$.

Solution:

Given:

Two systems of rectangular axes with the same origin.

A plane intersects the axes of the first system (say, OXYZ) at distances a, b, c from the origin along the x, y, and z axes, respectively.

The same plane intersects the axes of the second system (say, OX'Y'Z') at distances a', b', c' from the origin along the x', y', and z' axes, respectively.

To Prove:

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a’^2} + \frac{1}{b’^2} + \frac{1}{c’^2}$.

Proof:

Consider the first system of rectangular axes OXYZ. The plane cuts the axes at (a, 0, 0), (0, b, 0), and (0, 0, c). The equation of the plane in the intercept form is:

The distance of this plane from the origin (0, 0, 0) is given by the formula $D = \frac{|Ax_0 + By_0 + Cz_0 - d|}{\sqrt{A^2 + B^2 + C^2}}$, where the equation of the plane is $Ax + By + Cz - d = 0$.

From equation (1), we can write the equation of the plane as $\frac{1}{a}x + \frac{1}{b}y + \frac{1}{c}z - 1 = 0$.

Comparing this with $Ax + By + Cz - d = 0$, we have $A = \frac{1}{a}$, $B = \frac{1}{b}$, $C = \frac{1}{c}$, and $d = 1$. The origin is $(x_0, y_0, z_0) = (0, 0, 0)$.

The distance $D$ of the plane from the origin is:

$D = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) + \frac{1}{c}(0) - 1|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2}}$

$D = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$

$D = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$

Squaring both sides:

Rearranging equation (2):

Now, consider the second system of rectangular axes OX'Y'Z' with the same origin. The same plane cuts these axes at distances a', b', c' from the origin along the x', y', and z' axes, respectively.

The equation of the plane in the intercept form in this coordinate system is:

The distance of this plane from the origin (0, 0, 0) is given by:

$D' = \frac{1}{\sqrt{\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}}}$

Squaring both sides:

Rearranging equation (5):

Since the two rectangular axis systems have the same origin, and it is the *same* plane in both cases, the perpendicular distance of the plane from the origin must be the same in both coordinate systems.

Therefore, $D = D'$.

Squaring both sides, we get $D^2 = D'^2$.

Equating the right-hand sides of equations (3) and (6):

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{D^2}$

$\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2} = \frac{1}{D'^2}$

Since $D^2 = D'^2$, we have:

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a’^2} + \frac{1}{b’^2} + \frac{1}{c’^2}$

This proves the required result.

Conclusion:

By expressing the equation of the plane in the intercept form for both coordinate systems and using the formula for the distance of a plane from the origin, we showed that the reciprocal of the square of the distance from the origin is equal to the sum of the reciprocals of the squares of the intercepts squared. Since the distance of the plane from the origin is independent of the coordinate system (as long as the origin is the same), the sum of the reciprocals of the squares of the intercepts must be the same for both systems.

Solution:

Given:

Point P (2, 3, –8).

Equation of the line $L$: $\frac{4 − x}{2} = \frac{y}{6} = \frac{1 − z}{3}$.

To Find:

1. The coordinates of the foot of the perpendicular from P to the line.

2. The perpendicular distance from P to the line.

Calculation:

First, let's rewrite the equation of the line $L$ in the standard symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

The given equation is $\frac{4 − x}{2} = \frac{y}{6} = \frac{1 − z}{3}$.

We can rewrite the terms to have $(x - x_0)$, $(y - y_0)$, and $(z - z_0)$ in the numerator:

$\frac{-(x - 4)}{2} = \frac{y - 0}{6} = \frac{-(z - 1)}{3}$

$\frac{x - 4}{-2} = \frac{y - 0}{6} = \frac{z - 1}{-3}$

This is the standard form of the line. It passes through the point A (4, 0, 1) and has a direction vector $\vec{b} = -2\hat{i} + 6\hat{j} - 3\hat{k}$.

Let Q be the foot of the perpendicular from the point P (2, 3, –8) to the line $L$. Since Q lies on the line $L$, its coordinates can be expressed in terms of a parameter, say $t$, from the equation of the line:

$\frac{x_Q - 4}{-2} = \frac{y_Q - 0}{6} = \frac{z_Q - 1}{-3} = t$

So, the coordinates of Q are:

$x_Q = -2t + 4$

$y_Q = 6t$

$z_Q = -3t + 1$

Thus, Q = $(-2t + 4, 6t, -3t + 1)$.

The vector $\overrightarrow{PQ}$ connects point P (2, 3, –8) to point Q $(-2t + 4, 6t, -3t + 1)$.

$\overrightarrow{PQ} = (x_Q - x_P)\hat{i} + (y_Q - y_P)\hat{j} + (z_Q - z_P)\hat{k}$

$\overrightarrow{PQ} = ((-2t + 4) - 2)\hat{i} + (6t - 3)\hat{j} + ((-3t + 1) - (-8))\hat{k}$

$\overrightarrow{PQ} = (-2t + 2)\hat{i} + (6t - 3)\hat{j} + (-3t + 9)\hat{k}$

Since PQ is the perpendicular from P to the line $L$, the vector $\overrightarrow{PQ}$ must be orthogonal (perpendicular) to the direction vector of the line $L$, which is $\vec{b} = -2\hat{i} + 6\hat{j} - 3\hat{k}$.

The dot product of two orthogonal vectors is zero: $\overrightarrow{PQ} \cdot \vec{b} = 0$.

$(-2t + 2)(-2) + (6t - 3)(6) + (-3t + 9)(-3) = 0$

Expand and simplify the equation:

$4t - 4 + 36t - 18 + 9t - 27 = 0$

Combine the terms with $t$ and the constant terms:

$49t - 49 = 0$

$49t = 49$

$t = \frac{49}{49}$

$t = 1$

Now, substitute the value of $t = 1$ back into the coordinates of Q to find the foot of the perpendicular:

$x_Q = -2(1) + 4 = -2 + 4 = 2$

$y_Q = 6(1) = 6$

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

The coordinates of the foot of the perpendicular are (2, 6, –2).

Next, we need to find the perpendicular distance from P to the line, which is the distance between P and the foot of the perpendicular Q.

The distance PQ is the magnitude of the vector $\overrightarrow{PQ}$ when $t=1$.

Substitute $t=1$ into the vector $\overrightarrow{PQ} = (-2t + 2)\hat{i} + (6t - 3)\hat{j} + (-3t + 9)\hat{k}$:

$\overrightarrow{PQ} = (-2(1) + 2)\hat{i} + (6(1) - 3)\hat{j} + (-3(1) + 9)\hat{k}$

$\overrightarrow{PQ} = (0)\hat{i} + (3)\hat{j} + (6)\hat{k}$

$\overrightarrow{PQ} = 0\hat{i} + 3\hat{j} + 6\hat{k}$

The magnitude of $\overrightarrow{PQ}$ is $|\overrightarrow{PQ}| = \sqrt{0^2 + 3^2 + 6^2}$.

$|\overrightarrow{PQ}| = \sqrt{0 + 9 + 36}$

$|\overrightarrow{PQ}| = \sqrt{45}$

Simplify the square root:

$|\overrightarrow{PQ}| = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$

The perpendicular distance from P to the line is $3\sqrt{5}$ units.

Conclusion:

The foot of the perpendicular from the point (2, 3, –8) to the given line is (2, 6, –2).

The perpendicular distance from the point (2, 3, –8) to the given line is $3\sqrt{5}$ units.

Solution:

Given:

The point P is (2, 4, –1).

The equation of the line $L$ is $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z − 6}{−9}$.

To Find:

The distance of the point P from the line $L$.

Calculation:

The equation of the line $L$ is given in the standard symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

From the equation $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z − 6}{−9}$, we can identify that the line passes through the point A (–5, –3, 6) and has a direction vector $\vec{b} = 1\hat{i} + 4\hat{j} - 9\hat{k}$.

Let Q be the foot of the perpendicular from the point P (2, 4, –1) to the line $L$. Since Q lies on the line $L$, its coordinates can be expressed in terms of a parameter. Let:

$\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z − 6}{−9} = t$

The coordinates of any point Q on the line are $(x, y, z) = (t - 5, 4t - 3, -9t + 6)$.

So, Q = $(t - 5, 4t - 3, -9t + 6)$.

The vector $\overrightarrow{PQ}$ connects the point P (2, 4, –1) to the point Q $(t - 5, 4t - 3, -9t + 6)$.

$\overrightarrow{PQ} = (x_Q - x_P)\hat{i} + (y_Q - y_P)\hat{j} + (z_Q - z_P)\hat{k}$

$\overrightarrow{PQ} = ((t - 5) - 2)\hat{i} + ((4t - 3) - 4)\hat{j} + ((-9t + 6) - (-1))\hat{k}$

$\overrightarrow{PQ} = (t - 7)\hat{i} + (4t - 7)\hat{j} + (-9t + 7)\hat{k}$

Since PQ is the perpendicular segment from P to the line $L$, the vector $\overrightarrow{PQ}$ must be orthogonal (perpendicular) to the direction vector of the line $L$, $\vec{b} = \hat{i} + 4\hat{j} - 9\hat{k}$.

The dot product of two orthogonal vectors is zero:

$\overrightarrow{PQ} \cdot \vec{b} = 0$

$(t - 7)(1) + (4t - 7)(4) + (-9t + 7)(-9) = 0$

Expand and simplify the equation:

$t - 7 + 16t - 28 + 81t - 63 = 0$

Combine the terms:

$(t + 16t + 81t) + (-7 - 28 - 63) = 0$

$98t - 98 = 0$

$98t = 98$

$t = 1$

Now, substitute the value of $t = 1$ back into the coordinates of Q to find the coordinates of the foot of the perpendicular:

$x_Q = 1 - 5 = -4$

$y_Q = 4(1) - 3 = 4 - 3 = 1$

$z_Q = -9(1) + 6 = -9 + 6 = -3$

The foot of the perpendicular Q is (–4, 1, –3).

The distance of the point P from the line $L$ is the distance between P and the foot of the perpendicular Q. We can find this by calculating the magnitude of the vector $\overrightarrow{PQ}$ (when $t=1$) or by using the distance formula between points P(2, 4, –1) and Q(–4, 1, –3).

Using the distance formula:

Distance $PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2}$

Distance $PQ = \sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2}$

Distance $PQ = \sqrt{(-6)^2 + (-3)^2 + (-2)^2}$

Distance $PQ = \sqrt{36 + 9 + 4}$

Distance $PQ = \sqrt{49}$

Distance $PQ = 7$

Conclusion:

The distance of the point (2, 4, –1) from the given line is 7 units.

Solution:

Given:

The point P is $\left( 1,\; \frac{3}{2}, \; 2 \right)$.

The equation of the plane is $2x - 2y + 4z + 5 = 0$.

To Find:

1. The coordinates of the foot of the perpendicular from point P to the plane.

2. The length of the perpendicular from point P to the plane.

Calculation:

Let the given point be P $\left( 1,\; \frac{3}{2}, \; 2 \right)$.

The equation of the plane is $2x - 2y + 4z + 5 = 0$.

The coefficients of $x$, $y$, and $z$ in the plane equation are the direction ratios of the normal vector to the plane. So, the normal vector is $\vec{n} = 2\hat{i} - 2\hat{j} + 4\hat{k}$. The direction ratios of the normal are $(2, -2, 4)$.

The line passing through the point P and perpendicular to the plane will be parallel to the normal vector $\vec{n}$.

The equation of the line passing through P $\left( 1,\; \frac{3}{2}, \; 2 \right)$ and having direction ratios $(2, -2, 4)$ is given by the symmetric form:

$\frac{x - 1}{2} = \frac{y - \frac{3}{2}}{-2} = \frac{z - 2}{4}$

Let Q be the foot of the perpendicular from P to the plane. Q is the point where the line intersects the plane. Any point on the line can be represented parametrically. Let:

$\frac{x - 1}{2} = \frac{y - \frac{3}{2}}{-2} = \frac{z - 2}{4} = \lambda$

The coordinates of any point Q on this line are:

$x_Q = 2\lambda + 1$

$y_Q = -2\lambda + \frac{3}{2}$

$z_Q = 4\lambda + 2$

Since Q lies on the plane $2x - 2y + 4z + 5 = 0$, the coordinates of Q must satisfy the plane equation. Substitute the coordinates of Q into the equation of the plane:

$2(2\lambda + 1) - 2(-2\lambda + \frac{3}{2}) + 4(4\lambda + 2) + 5 = 0$

Expand and simplify the equation:

$4\lambda + 2 + 4\lambda - 3 + 16\lambda + 8 + 5 = 0$

Combine terms with $\lambda$ and constant terms:

$(4\lambda + 4\lambda + 16\lambda) + (2 - 3 + 8 + 5) = 0$

$24\lambda + 12 = 0$

$24\lambda = -12$

$\lambda = \frac{-12}{24} = -\frac{1}{2}$

Substitute the value of $\lambda = -\frac{1}{2}$ back into the coordinates of Q to find the foot of the perpendicular:

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

$y_Q = -2(-\frac{1}{2}) + \frac{3}{2} = 1 + \frac{3}{2} = \frac{2+3}{2} = \frac{5}{2}$

$z_Q = 4(-\frac{1}{2}) + 2 = -2 + 2 = 0$

The coordinates of the foot of the perpendicular Q are $\left(0, \frac{5}{2}, 0\right)$.

Now, we find the length of the perpendicular, which is the distance between point P $\left( 1,\; \frac{3}{2}, \; 2 \right)$ and the foot of the perpendicular Q $\left(0, \frac{5}{2}, 0\right)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

$d = \sqrt{(0 - 1)^2 + \left(\frac{5}{2} - \frac{3}{2}\right)^2 + (0 - 2)^2}$

$d = \sqrt{(-1)^2 + \left(\frac{2}{2}\right)^2 + (-2)^2}$

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

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

$d = \sqrt{6}$

Alternatively, the length of the perpendicular from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by:

$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Here, $(x_1, y_1, z_1) = \left( 1,\; \frac{3}{2}, \; 2 \right)$, and the plane is $2x - 2y + 4z + 5 = 0$, so $A=2$, $B=-2$, $C=4$, $D=5$.

$d = \frac{\left|2(1) + (-2)\left(\frac{3}{2}\right) + 4(2) + 5\right|}{\sqrt{2^2 + (-2)^2 + 4^2}}$

$d = \frac{|2 - 3 + 8 + 5|}{\sqrt{4 + 4 + 16}}$

$d = \frac{|-1 + 8 + 5|}{\sqrt{24}}$

$d = \frac{|7 + 5|}{\sqrt{24}}$

$d = \frac{|12|}{\sqrt{24}}$

$d = \frac{12}{\sqrt{4 \times 6}} = \frac{12}{2\sqrt{6}} = \frac{6}{\sqrt{6}}$

Rationalize the denominator:

$d = \frac{6}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{6\sqrt{6}}{6} = \sqrt{6}$

Both methods give the same perpendicular distance.

Conclusion:

The foot of the perpendicular from the point $\left( 1,\; \frac{3}{2}, \; 2 \right)$ to the plane $2x - 2y + 4z + 5 = 0$ is $\left(0, \frac{5}{2}, 0\right)$.

The length of the perpendicular from the point $\left( 1,\; \frac{3}{2}, \; 2 \right)$ to the plane $2x - 2y + 4z + 5 = 0$ is $\sqrt{6}$ units.

Solution:

Given:

The line passes through the point P (3, 0, 1).

The line is parallel to the plane $P_1$: $x + 2y = 0$.

The line is parallel to the plane $P_2$: $3y - z = 0$.

To Find:

The equations of the line.

Calculation:

Let the equation of the required line be $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$, where $(x_0, y_0, z_0)$ is a point on the line and $(a, b, c)$ are the direction ratios of the line.

We are given that the line passes through the point (3, 0, 1). So, $(x_0, y_0, z_0) = (3, 0, 1)$.

The equation of the line is $\frac{x - 3}{a} = \frac{y - 0}{b} = \frac{z - 1}{c}$, or $\frac{x - 3}{a} = \frac{y}{b} = \frac{z - 1}{c}$.

If a line is parallel to a plane, its direction vector is perpendicular to the normal vector of the plane. The dot product of their direction ratios is zero.

The equation of the first plane $P_1$ is $x + 2y + 0z = 0$. The normal vector to this plane is $\vec{n}_1 = 1\hat{i} + 2\hat{j} + 0\hat{k}$. The direction ratios of $\vec{n}_1$ are $(1, 2, 0)$.

Since the line is parallel to $P_1$, the direction vector of the line $(a, b, c)$ is perpendicular to the normal vector of $P_1$ $(1, 2, 0)$.

Their dot product is zero:

$(a)(1) + (b)(2) + (c)(0) = 0$

The equation of the second plane $P_2$ is $0x + 3y - z = 0$. The normal vector to this plane is $\vec{n}_2 = 0\hat{i} + 3\hat{j} - 1\hat{k}$. The direction ratios of $\vec{n}_2$ are $(0, 3, -1)$.

Since the line is parallel to $P_2$, the direction vector of the line $(a, b, c)$ is perpendicular to the normal vector of $P_2$ $(0, 3, -1)$.

Their dot product is zero:

$(a)(0) + (b)(3) + (c)(-1) = 0$

We have a system of two linear equations in terms of the direction ratios $a$, $b$, and $c$:

$a + 2b = 0$

$3b - c = 0$

From equation (1), $a = -2b$.

From equation (2), $c = 3b$.

The direction ratios are proportional to $(-2b, b, 3b)$. Since $b$ can be any non-zero scalar, we can choose a simple value for $b$ to find a set of direction ratios. Let's choose $b=1$.

If $b = 1$, then $a = -2(1) = -2$ and $c = 3(1) = 3$.

So, the direction ratios of the line are $(-2, 1, 3)$.

The direction vector of the line is $\vec{d} = -2\hat{i} + \hat{j} + 3\hat{k}$.

Now we can write the equations of the line passing through P (3, 0, 1) with direction ratios $(-2, 1, 3)$.

Cartesian Equation:

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

$\frac{x - 3}{-2} = \frac{y - 0}{1} = \frac{z - 1}{3}$

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

Vector Equation:

The position vector of the point P is $\vec{r}_0 = 3\hat{i} + 0\hat{j} + 1\hat{k} = 3\hat{i} + \hat{k}$.

The direction vector of the line is $\vec{d} = -2\hat{i} + \hat{j} + 3\hat{k}$.

The vector equation of the line is $\vec{r} = \vec{r}_0 + \lambda \vec{d}$, where $\lambda$ is a scalar parameter.

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

Conclusion:

The equations of the line passing through the point (3, 0, 1) and parallel to the given planes are:

Cartesian form: $\frac{x - 3}{-2} = \frac{y}{1} = \frac{z - 1}{3}$

Vector form: $\vec{r} = 3\hat{i} + \hat{k} + \lambda (-2\hat{i} + \hat{j} + 3\hat{k})$, where $\lambda$ is a scalar.

Solution:

Given:

The required plane passes through two points: P$_1$ (2, 1, –1) and P$_2$ (–1, 3, 4).

The required plane is perpendicular to the plane $P_3$: $x – 2y + 4z = 10$.

To Find:

The equation of the plane passing through P$_1$ and P$_2$ and perpendicular to $P_3$.

Calculation:

Method 1: Using the general equation of a plane.

Let the equation of the required plane be $Ax + By + Cz + D = 0$.

Since the points P$_1$(2, 1, –1) and P$_2$(–1, 3, 4) lie on the plane, their coordinates must satisfy the equation:

For P$_1$(2, 1, –1):

$A(2) + B(1) + C(-1) + D = 0$

For P$_2$(–1, 3, 4):

$A(-1) + B(3) + C(4) + D = 0$

The normal vector to the required plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$, with direction ratios $(A, B, C)$.

The equation of the plane $P_3$ is $x - 2y + 4z = 10$. The normal vector to this plane is $\vec{n}_3 = 1\hat{i} - 2\hat{j} + 4\hat{k}$, with direction ratios $(1, -2, 4)$.

Since the required plane is perpendicular to $P_3$, their normal vectors are perpendicular. The dot product of their direction ratios is zero:

$(A)(1) + (B)(-2) + (C)(4) = 0$

Subtract equation (2) from equation (1) to eliminate D:

$(2A + B - C + D) - (-A + 3B + 4C + D) = 0$

$2A + B - C + D + A - 3B - 4C - D = 0$

Now we have a system of two equations (3 and 4) in terms of A, B, and C:

$A - 2B + 4C = 0$

$3A - 2B - 5C = 0$

Subtract equation (3) from equation (4):

$(3A - 2B - 5C) - (A - 2B + 4C) = 0$

$3A - 2B - 5C - A + 2B - 4C = 0$

$2A - 9C = 0 \implies 2A = 9C \implies A = \frac{9}{2}C$

Substitute $A = \frac{9}{2}C$ into equation (3):

$\frac{9}{2}C - 2B + 4C = 0$

$\frac{9}{2}C + \frac{8}{2}C = 2B$

$\frac{17}{2}C = 2B \implies B = \frac{17}{4}C$

Now substitute the values of A and B (in terms of C) into equation (1) to find D in terms of C:

$2\left(\frac{9}{2}C\right) + \left(\frac{17}{4}C\right) - C + D = 0$

$9C + \frac{17}{4}C - C + D = 0$

$8C + \frac{17}{4}C + D = 0$

$\frac{32C + 17C}{4} + D = 0$

$\frac{49C}{4} + D = 0 \implies D = -\frac{49}{4}C$

The direction ratios $(A, B, C)$ and the constant $D$ are proportional to $\left(\frac{9}{2}C, \frac{17}{4}C, C, -\frac{49}{4}C\right)$. We can choose $C=4$ to get integer values:

$A = \frac{9}{2}(4) = 18$

$B = \frac{17}{4}(4) = 17$

$C = 4$

$D = -\frac{49}{4}(4) = -49$

Substituting these values into the general equation $Ax + By + Cz + D = 0$:

$18x + 17y + 4z - 49 = 0$

$18x + 17y + 4z = 49$

Method 2: Using vectors.

Let the required plane be $P$. It passes through P$_1$(2, 1, –1) and P$_2$(–1, 3, 4).

The vector $\overrightarrow{P_1P_2}$ lies in the plane $P$.

$\overrightarrow{P_1P_2} = \vec{P_2} - \vec{P_1} = (-1-2)\hat{i} + (3-1)\hat{j} + (4-(-1))\hat{k} = -3\hat{i} + 2\hat{j} + 5\hat{k}$.

The plane $P$ is perpendicular to the plane $P_3$: $x - 2y + 4z = 10$. The normal vector to $P_3$ is $\vec{n}_3 = 1\hat{i} - 2\hat{j} + 4\hat{k}$. This vector $\vec{n}_3$ is parallel to the required plane $P$ (it lies in a direction parallel to the plane).

The normal vector $\vec{n}$ of the required plane $P$ is perpendicular to both $\overrightarrow{P_1P_2}$ (a vector in the plane) and $\vec{n}_3$ (a vector parallel to the plane). Therefore, $\vec{n}$ is proportional to the cross product of $\overrightarrow{P_1P_2}$ and $\vec{n}_3$.

$\vec{n} = \overrightarrow{P_1P_2} \times \vec{n}_3$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 5 \\ 1 & -2 & 4 \end{vmatrix}$

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

$\vec{n} = \hat{i}(8 + 10) - \hat{j}(-12 - 5) + \hat{k}(6 - 2)$

$\vec{n} = 18\hat{i} + 17\hat{j} + 4\hat{k}$

The normal vector to the plane is $\vec{n} = 18\hat{i} + 17\hat{j} + 4\hat{k}$. The direction ratios are $(18, 17, 4)$.

The plane passes through the point P$_1$ (2, 1, –1). The equation of a plane with normal vector $A\hat{i} + B\hat{j} + C\hat{k}$ passing through $(x_0, y_0, z_0)$ is $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.

Using P$_1$(2, 1, –1) and normal vector $(18, 17, 4)$:

$18(x - 2) + 17(y - 1) + 4(z - (-1)) = 0$

$18(x - 2) + 17(y - 1) + 4(z + 1) = 0$

$18x - 36 + 17y - 17 + 4z + 4 = 0$

$18x + 17y + 4z - 36 - 17 + 4 = 0$

$18x + 17y + 4z - 49 = 0$

$18x + 17y + 4z = 49$

Conclusion:

The equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane $x – 2y + 4z = 10$ is $18x + 17y + 4z = 49$.

Solution:

Given:

Equation of the first line ($L_1$): $\vec{r}= (8 + 3λ)\hat{i}− (9 + 16λ)\hat{j} + (10 + 7λ)\hat{k}$

Equation of the second line ($L_2$): $\vec{r}= 15\hat{i}+ 29\hat{j}+ 5\hat{k}+ μ(3\hat{i}+ 8\hat{j}− 5\hat{k})$

To Find:

The shortest distance between the two lines.

Calculation:

We can write the equations of the lines in the standard form $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, where $\vec{a}_1$ and $\vec{a}_2$ are the position vectors of points on the lines, and $\vec{b}_1$ and $\vec{b}_2$ are the direction vectors of the lines.

For the first line $L_1$, $\vec{r}= (8 + 3λ)\hat{i}− (9 + 16λ)\hat{j} + (10 + 7λ)\hat{k}$:

$\vec{r} = (8\hat{i} - 9\hat{j} + 10\hat{k}) + \lambda (3\hat{i} - 16\hat{j} + 7\hat{k})$

So, $\vec{a}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k}$

And $\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}$

For the second line $L_2$, $\vec{r}= 15\hat{i}+ 29\hat{j}+ 5\hat{k}+ μ(3\hat{i}+ 8\hat{j}− 5\hat{k})$:

So, $\vec{a}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k}$

And $\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$

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 the formula:

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

Calculate $\vec{a}_2 - \vec{a}_1$:

$\vec{a}_2 - \vec{a}_1 = (15\hat{i} + 29\hat{j} + 5\hat{k}) - (8\hat{i} - 9\hat{j} + 10\hat{k})$

$\vec{a}_2 - \vec{a}_1 = (15 - 8)\hat{i} + (29 - (-9))\hat{j} + (5 - 10)\hat{k}$

$\vec{a}_2 - \vec{a}_1 = 7\hat{i} + 38\hat{j} - 5\hat{k}$

Calculate $\vec{b}_1 \times \vec{b}_2$:

$\vec{b}_1 \times \vec{b}_2 = (3\hat{i} - 16\hat{j} + 7\hat{k}) \times (3\hat{i} + 8\hat{j} - 5\hat{k})$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}[(-16)(-5) - (7)(8)] - \hat{j}[(3)(-5) - (7)(3)] + \hat{k}[(3)(8) - (-16)(3)]$

$\vec{b}_1 \times \vec{b}_2 = \hat{i}[80 - 56] - \hat{j}[-15 - 21] + \hat{k}[24 + 48]$

$\vec{b}_1 \times \vec{b}_2 = 24\hat{i} + 36\hat{j} + 72\hat{k}$

Calculate the scalar triple product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (7\hat{i} + 38\hat{j} - 5\hat{k}) \cdot (24\hat{i} + 36\hat{j} + 72\hat{k})$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (7)(24) + (38)(36) + (-5)(72)$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 168 + 1368 - 360$

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 1536 - 360 = 1176$

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

$|\vec{b}_1 \times \vec{b}_2| = |24\hat{i} + 36\hat{j} + 72\hat{k}|$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{24^2 + 36^2 + 72^2}$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{576 + 1296 + 5184}$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{7056}$

$|\vec{b}_1 \times \vec{b}_2| = 84$

Substitute the values into the shortest distance formula:

$D = \frac{|1176|}{84}$

$D = \frac{1176}{84}$

$D = 14$

Conclusion:

The shortest distance between the given lines is 14 units.

Solution:

Given:

The required plane is perpendicular to the plane $P_3$: $5x + 3y + 6z + 8 = 0$.

The required plane contains the line of intersection of the planes $P_1$: $x + 2y + 3z – 4 = 0$ and $P_2$: $2x + y – z + 5 = 0$.

To Find:

The equation of the required plane.

Calculation:

The equation of a plane that contains the line of intersection of two planes $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter.

In this case, the equation of the required plane is:

$(x + 2y + 3z – 4) + \lambda (2x + y – z + 5) = 0$

Rearrange this equation to the general form $Ax + By + Cz + D = 0$:

$x + 2y + 3z – 4 + 2\lambda x + \lambda y – \lambda z + 5\lambda = 0$

$(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$

The normal vector to this plane is $\vec{n} = (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k}$. The direction ratios of the normal are $(1 + 2\lambda, 2 + \lambda, 3 - \lambda)$.

The required plane is perpendicular to the plane $P_3$: $5x + 3y + 6z + 8 = 0$. The normal vector to $P_3$ is $\vec{n}_3 = 5\hat{i} + 3\hat{j} + 6\hat{k}$. The direction ratios of $\vec{n}_3$ are $(5, 3, 6)$.

Since the two planes are perpendicular, their normal vectors are orthogonal. The dot product of their direction ratios is zero:

$\vec{n} \cdot \vec{n}_3 = 0$

$(1 + 2\lambda)(5) + (2 + \lambda)(3) + (3 - \lambda)(6) = 0$

Expand and simplify the equation to find the value of $\lambda$:

$5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0$

Combine terms with $\lambda$ and constant terms:

$(10\lambda + 3\lambda - 6\lambda) + (5 + 6 + 18) = 0$

$7\lambda + 29 = 0$

$7\lambda = -29$

$\lambda = -\frac{29}{7}$

Substitute the value of $\lambda = -\frac{29}{7}$ back into the equation of the required plane:

$(x + 2y + 3z – 4) + \left(-\frac{29}{7}\right) (2x + y – z + 5) = 0$

Multiply the entire equation by 7 to eliminate the fraction:

$7(x + 2y + 3z – 4) - 29(2x + y – z + 5) = 0$

$7x + 14y + 21z – 28 - 58x - 29y + 29z - 145 = 0$

Combine like terms:

$(7x - 58x) + (14y - 29y) + (21z + 29z) + (-28 - 145) = 0$

$-51x - 15y + 50z - 173 = 0$

Multiply by -1 to make the coefficient of x positive (optional, but common practice):

$51x + 15y - 50z + 173 = 0$

Conclusion:

The equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z – 4 = 0$ and $2x + y – z + 5 = 0$ is $51x + 15y - 50z + 173 = 0$.

Solution:

Given:

Original plane $P_1$: $ax + by = 0$.

Second plane $P_2$: $z = 0$.

The plane $P_1$ is rotated about the line of intersection of $P_1$ and $P_2$ through an angle $\alpha$.

To Prove:

The equation of the plane in its new position is $ax + by ± \left( \sqrt{a^2+b^2} \tan α \right) z = 0$.

Proof:

The line of intersection of the planes $P_1: ax + by = 0$ and $P_2: z = 0$ is given by the system of equations:

Any plane passing through the line of intersection of $P_1$ and $P_2$ has the equation of the form $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter.

So, the equation of the new plane (after rotation) can be written as:

$ax + by + \lambda z = 0$.

The normal vector to the original plane $P_1: ax + by + 0z = 0$ is $\vec{n}_1 = a\hat{i} + b\hat{j} + 0\hat{k}$.

The normal vector to the new plane $P'$ (equation (iii)) is $\vec{n}' = a\hat{i} + b\hat{j} + \lambda\hat{k}$.

The angle between the two planes $P_1$ and $P'$ is equal to the angle between their normal vectors $\vec{n}_1$ and $\vec{n}'$. We are given that the angle of rotation is $\alpha$, so the angle between the original plane and the new plane is $\alpha$.

The cosine of the angle between two vectors $\vec{v}_1$ and $\vec{v}_2$ is given by $\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$.

Here, $\theta = \alpha$, $\vec{v}_1 = \vec{n}_1 = a\hat{i} + b\hat{j}$, and $\vec{v}_2 = \vec{n}' = a\hat{i} + b\hat{j} + \lambda\hat{k}$.

Calculate the dot product $\vec{n}_1 \cdot \vec{n}'$:

$\vec{n}_1 \cdot \vec{n}' = (a\hat{i} + b\hat{j}) \cdot (a\hat{i} + b\hat{j} + \lambda\hat{k})$

$\vec{n}_1 \cdot \vec{n}' = (a)(a) + (b)(b) + (0)(\lambda) = a^2 + b^2$

Calculate the magnitudes of the normal vectors:

$|\vec{n}_1| = |a\hat{i} + b\hat{j}| = \sqrt{a^2 + b^2 + 0^2} = \sqrt{a^2 + b^2}$

$|\vec{n}'| = |a\hat{i} + b\hat{j} + \lambda\hat{k}| = \sqrt{a^2 + b^2 + \lambda^2}$

Substitute these into the cosine formula:

$\cos \alpha = \frac{|a^2 + b^2|}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + \lambda^2}}$

Since $a^2 + b^2$ is always non-negative, $|a^2 + b^2| = a^2 + b^2$. Assume $a$ and $b$ are not both zero, so $\sqrt{a^2+b^2} > 0$.

$\cos \alpha = \frac{a^2 + b^2}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + \lambda^2}}$

$\cos \alpha = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2 + \lambda^2}}$

Square both sides:

$\cos^2 \alpha = \frac{a^2 + b^2}{a^2 + b^2 + \lambda^2}$

Rearrange the equation to solve for $\lambda^2$:

$(a^2 + b^2 + \lambda^2) \cos^2 \alpha = a^2 + b^2$

$a^2 \cos^2 \alpha + b^2 \cos^2 \alpha + \lambda^2 \cos^2 \alpha = a^2 + b^2$

$\lambda^2 \cos^2 \alpha = a^2(1 - \cos^2 \alpha) + b^2(1 - \cos^2 \alpha)$

$\lambda^2 \cos^2 \alpha = a^2 \sin^2 \alpha + b^2 \sin^2 \alpha$

$\lambda^2 \cos^2 \alpha = (a^2 + b^2) \sin^2 \alpha$

Assuming $\cos \alpha \neq 0$ (i.e., $\alpha \neq \frac{\pi}{2} + n\pi$), we can divide by $\cos^2 \alpha$:

$\lambda^2 = (a^2 + b^2) \frac{\sin^2 \alpha}{\cos^2 \alpha}$

$\lambda^2 = (a^2 + b^2) \tan^2 \alpha$

Take the square root of both sides:

$\lambda = \pm \sqrt{(a^2 + b^2) \tan^2 \alpha}$

$\lambda = \pm \sqrt{a^2 + b^2} |\tan \alpha|$

Since the problem statement uses $\tan \alpha$ without absolute value, and angle $\alpha$ is mentioned as the angle of rotation (typically taken as $0 \leq \alpha < \pi$), we can assume $\tan \alpha$ can be positive or negative depending on $\alpha$. The $\pm$ sign indicates the two possible directions of rotation (clockwise or counter-clockwise) from the original plane, leading to two possible new planes (unless $\alpha = 0$ or $\alpha = \pi$).

So, we can write $\lambda = \pm \sqrt{a^2 + b^2} \tan \alpha$.

Substitute this value of $\lambda$ back into the equation of the new plane (iii):

$ax + by + \lambda z = 0$

$ax + by + \left(\pm \sqrt{a^2 + b^2} \tan \alpha\right) z = 0$

$ax + by \pm \left( \sqrt{a^2+b^2} \tan α \right) z = 0$

This is the required equation of the plane in its new position.

If $\alpha = \frac{\pi}{2}$, then $\cos \alpha = 0$. In this case, from $\lambda^2 \cos^2 \alpha = (a^2 + b^2) \sin^2 \alpha$, we get $0 = (a^2 + b^2) \sin^2(\frac{\pi}{2}) = a^2 + b^2$. This implies $a=0$ and $b=0$. If $a=0$ and $b=0$, the original plane $ax+by=0$ is $0=0$, which is not a plane. If $a$ or $b$ is non-zero, $\cos \alpha \neq 0$ and the derivation holds.

Conclusion:

We have shown that by using the property that the angle between the original plane and the new plane is the angle of rotation, and expressing the new plane's equation through the line of intersection, the resulting equation is $ax + by ± \left( \sqrt{a^2+b^2} \tan α \right) z = 0$.

Solution:

Given:

Two planes: $P_1: \vec{r}. (\hat{i}+ 3\hat{j}) − 6 = 0$ and $P_2: \vec{r}. (3\hat{i}− \hat{j}− 4\hat{k}) = 0$.

The required plane passes through the line of intersection of $P_1$ and $P_2$.

The perpendicular distance of the required plane from the origin is 1 unit.

To Find:

The equation of the required plane.

Calculation:

The equation of a plane passing through the line of intersection of two planes $\vec{r} \cdot \vec{n}_1 - d_1 = 0$ and $\vec{r} \cdot \vec{n}_2 - d_2 = 0$ is given by $(\vec{r} \cdot \vec{n}_1 - d_1) + \lambda (\vec{r} \cdot \vec{n}_2 - d_2) = 0$, where $\lambda$ is a scalar parameter.

The equations of the given planes are:

$P_1: \vec{r}. (\hat{i}+ 3\hat{j}) − 6 = 0$

$P_2: \vec{r}. (3\hat{i}− \hat{j}− 4\hat{k}) + 0 = 0$ (Rewriting $P_2$ in the form $\vec{r} \cdot \vec{n}_2 - d_2 = 0$)

So, for $P_1$, $\vec{n}_1 = \hat{i}+ 3\hat{j}$ and $d_1 = 6$.

And for $P_2$, $\vec{n}_2 = 3\hat{i}− \hat{j}− 4\hat{k}$ and $d_2 = 0$.

The equation of the required plane is:

$(\vec{r}. (\hat{i}+ 3\hat{j}) − 6) + \lambda (\vec{r}. (3\hat{i}− \hat{j}− 4\hat{k})) = 0$

Rearrange the terms:

$\vec{r} \cdot (\hat{i}+ 3\hat{j}) - 6 + \lambda (\vec{r} \cdot (3\hat{i}− \hat{j}− 4\hat{k})) = 0$

$\vec{r} \cdot (\hat{i}+ 3\hat{j} + \lambda (3\hat{i}− \hat{j}− 4\hat{k})) - 6 = 0$

$\vec{r} \cdot (\hat{i}+ 3\hat{j} + 3\lambda\hat{i} - \lambda\hat{j} - 4\lambda\hat{k}) - 6 = 0$

This is the equation of the required plane in vector form. Let the normal vector to this plane be $\vec{n} = (1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}$.

The general equation of a plane is $\vec{r} \cdot \vec{n} = d$. In equation (1), the distance term is on the right side if we write $\vec{r} \cdot \vec{n} = 6$. So, $d = 6$.

The perpendicular distance of a plane $\vec{r} \cdot \vec{n} = d$ from the origin is given by $\frac{|d|}{|\vec{n}|}$.

We are given that the perpendicular distance from the origin is 1 unit. So, $\frac{|6|}{|\vec{n}|} = 1$.

$|\vec{n}| = 6$

Calculate the magnitude of the normal vector $\vec{n} = (1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}$:

$|\vec{n}| = \sqrt{(1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2}$

$|\vec{n}|^2 = (1 + 3\lambda)^2 + (3 - \lambda)^2 + (-4\lambda)^2$

$|\vec{n}|^2 = (1 + 6\lambda + 9\lambda^2) + (9 - 6\lambda + \lambda^2) + (16\lambda^2)$

$|\vec{n}|^2 = 9\lambda^2 + \lambda^2 + 16\lambda^2 + 6\lambda - 6\lambda + 1 + 9$

$|\vec{n}|^2 = 26\lambda^2 + 10$

We have $|\vec{n}| = 6$, so $|\vec{n}|^2 = 6^2 = 36$.

Equate the expressions for $|\vec{n}|^2$:

$26\lambda^2 + 10 = 36$

$26\lambda^2 = 36 - 10$

$26\lambda^2 = 26$

$\lambda^2 = 1$

$\lambda = \pm 1$

Substitute the values of $\lambda$ back into the equation of the plane (1) to find the equations of the two possible planes.

Case 1: $\lambda = 1$

Substitute $\lambda = 1$ into $\vec{r} \cdot ((1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}) - 6 = 0$:

$\vec{r} \cdot ((1 + 3(1))\hat{i} + (3 - 1)\hat{j} - 4(1)\hat{k}) - 6 = 0$

$\vec{r} \cdot (4\hat{i} + 2\hat{j} - 4\hat{k}) - 6 = 0$

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

This is the vector equation. In Cartesian form (let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$):

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

$4x + 2y - 4z = 6$

Divide by 2 to simplify:

$2x + y - 2z = 3$

$2x + y - 2z - 3 = 0$

Case 2: $\lambda = -1$

Substitute $\lambda = -1$ into $\vec{r} \cdot ((1 + 3\lambda)\hat{i} + (3 - \lambda)\hat{j} - 4\lambda\hat{k}) - 6 = 0$:

$\vec{r} \cdot ((1 + 3(-1))\hat{i} + (3 - (-1))\hat{j} - 4(-1)\hat{k}) - 6 = 0$

$\vec{r} \cdot ((1 - 3)\hat{i} + (3 + 1)\hat{j} + 4\hat{k}) - 6 = 0$

$\vec{r} \cdot (-2\hat{i} + 4\hat{j} + 4\hat{k}) - 6 = 0$

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

This is the vector equation. In Cartesian form:

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

$-2x + 4y + 4z = 6$

Divide by -2 to simplify and make the coefficient of x positive:

$x - 2y - 2z = -3$

$x - 2y - 2z + 3 = 0$

Both planes $2x + y - 2z - 3 = 0$ and $x - 2y - 2z + 3 = 0$ pass through the line of intersection of the given planes and are at a perpendicular distance of 1 unit from the origin.

Let's verify the distance for $2x + y - 2z - 3 = 0$ from origin (0,0,0):

$d = \frac{|2(0) + 1(0) - 2(0) - 3|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|-3|}{\sqrt{4 + 1 + 4}} = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1$.

Let's verify the distance for $x - 2y - 2z + 3 = 0$ from origin (0,0,0):

$d = \frac{|1(0) + (-2)(0) + (-2)(0) + 3|}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \frac{|3|}{\sqrt{1 + 4 + 4}} = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1$.

Both equations are correct.

Conclusion:

The equations of the planes satisfying the given conditions are $2x + y - 2z - 3 = 0$ and $x - 2y - 2z + 3 = 0$.

Solution:

Given:

Point P with position vector $\vec{p} = \hat{i}− \hat{j}+ 3\hat{k}$.

Point Q with position vector $\vec{q} = 3(\hat{i}+ \hat{j}+ \hat{k}) = 3\hat{i}+ 3\hat{j}+ 3\hat{k}$.

Equation of the plane: $\vec{r}. (5\hat{i}+ 2\hat{j}− 7\hat{k}) + 9 = 0$.

To Show:

1. The points P and Q are equidistant from the plane.

2. The points P and Q lie on opposite sides of the plane.

Proof:

The equation of the plane is given in vector form $\vec{r} \cdot \vec{n} + D = 0$, where $\vec{n} = 5\hat{i}+ 2\hat{j}− 7\hat{k}$ and $D = 9$.

The Cartesian equation of the plane is $5x + 2y - 7z + 9 = 0$. The normal vector is $(5, 2, -7)$.

The coordinates of point P are $(1, -1, 3)$.

The coordinates of point Q are $(3, 3, 3)$.

The perpendicular distance of a point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Alternatively, using the vector form, the distance of a point with position vector $\vec{p}_0$ from the plane $\vec{r} \cdot \vec{n} + D = 0$ is $\frac{|\vec{p}_0 \cdot \vec{n} + D|}{|\vec{n}|}$.

Here, the normal vector is $\vec{n} = 5\hat{i}+ 2\hat{j}− 7\hat{k}$. Its magnitude is $|\vec{n}| = \sqrt{5^2 + 2^2 + (-7)^2} = \sqrt{25 + 4 + 49} = \sqrt{78}$.

The constant term in the plane equation is $D=9$.

Calculate the distance of point P $(\hat{i}− \hat{j}+ 3\hat{k})$ from the plane:

$\vec{p} \cdot \vec{n} = (\hat{i}− \hat{j}+ 3\hat{k}) \cdot (5\hat{i}+ 2\hat{j}− 7\hat{k}) = (1)(5) + (-1)(2) + (3)(-7) = 5 - 2 - 21 = -18$.

The term $\vec{p} \cdot \vec{n} + D = -18 + 9 = -9$.

The distance of P from the plane is $d_P = \frac{|\vec{p} \cdot \vec{n} + D|}{|\vec{n}|} = \frac{|-9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.

Calculate the distance of point Q $(3\hat{i}+ 3\hat{j}+ 3\hat{k})$ from the plane:

$\vec{q} \cdot \vec{n} = (3\hat{i}+ 3\hat{j}+ 3\hat{k}) \cdot (5\hat{i}+ 2\hat{j}− 7\hat{k}) = (3)(5) + (3)(2) + (3)(-7) = 15 + 6 - 21 = 21 - 21 = 0$.

The term $\vec{q} \cdot \vec{n} + D = 0 + 9 = 9$.

The distance of Q from the plane is $d_Q = \frac{|\vec{q} \cdot \vec{n} + D|}{|\vec{n}|} = \frac{|9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.

Since $d_P = \frac{9}{\sqrt{78}}$ and $d_Q = \frac{9}{\sqrt{78}}$, the points P and Q are equidistant from the plane.

To determine if the points lie on opposite sides of the plane, we examine the sign of the expression $Ax_0 + By_0 + Cz_0 + D$ (or $\vec{p}_0 \cdot \vec{n} + D$) when the coordinates (or position vector) of the points are substituted. If the signs are opposite, the points lie on opposite sides of the plane.

For point P, the value of $\vec{p} \cdot \vec{n} + D$ is $-18 + 9 = -9$.

For point Q, the value of $\vec{q} \cdot \vec{n} + D$ is $0 + 9 = 9$.

Since the value of the expression is $-9$ for point P and $9$ for point Q, the signs are opposite (one is negative, the other is positive). This indicates that the points P and Q lie on opposite sides of the plane.

Conclusion:

We have calculated the perpendicular distances of points P and Q from the given plane and found them to be equal ($\frac{9}{\sqrt{78}}$ units), showing that the points are equidistant from the plane. By evaluating the expression $\vec{r} \cdot \vec{n} + D$ for each point and observing that the signs are opposite, we have shown that the points lie on opposite sides of the plane.

Solution:

Given:

Direction vector of line AB: $\vec{v}_1 = \overrightarrow{AB} = 3\hat{i}− \hat{j}+ \hat{k}$.

Direction vector of line CD: $\vec{v}_2 = \overrightarrow{CD} = −3\hat{i}+ 2\hat{j}+ 4\hat{k}$.

Position vector of point A: $\vec{a} = 6\hat{i}+ 7\hat{j}+ 4\hat{k}$. Point A lies on line AB.

Position vector of point C: $\vec{c} = −9\hat{j} + 2\hat{k}$. Point C lies on line CD.

P is a point on line AB and Q is a point on line CD.

$\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{CD}$ both.

To Find:

The position vector of point P and the position vector of point Q.

Calculation:

The equation of the line AB passing through point A with position vector $\vec{a}$ and having direction vector $\vec{v}_1$ is $\vec{r} = \vec{a} + \lambda \vec{v}_1$, where $\lambda$ is a scalar parameter.

The position vector of any point P on line AB is $\vec{p} = \vec{a} + \lambda \vec{v}_1 = (6\hat{i}+ 7\hat{j}+ 4\hat{k}) + \lambda (3\hat{i}− \hat{j}+ \hat{k})$

$\vec{p} = (6 + 3\lambda)\hat{i} + (7 - \lambda)\hat{j} + (4 + \lambda)\hat{k}$.

The equation of the line CD passing through point C with position vector $\vec{c}$ and having direction vector $\vec{v}_2$ is $\vec{r} = \vec{c} + \mu \vec{v}_2$, where $\mu$ is a scalar parameter.

The position vector of any point Q on line CD is $\vec{q} = \vec{c} + \mu \vec{v}_2 = (−9\hat{j} + 2\hat{k}) + \mu (−3\hat{i}+ 2\hat{j}+ 4\hat{k})$

$\vec{q} = -3\mu\hat{i} + (-9 + 2\mu)\hat{j} + (2 + 4\mu)\hat{k}$.

The vector $\overrightarrow{PQ}$ is the vector from P to Q:

$\overrightarrow{PQ} = \vec{q} - \vec{p}$

$\overrightarrow{PQ} = (-3\mu\hat{i} + (-9 + 2\mu)\hat{j} + (2 + 4\mu)\hat{k}) - ((6 + 3\lambda)\hat{i} + (7 - \lambda)\hat{j} + (4 + \lambda)\hat{k})$

$\overrightarrow{PQ} = (-3\mu - (6 + 3\lambda))\hat{i} + ((-9 + 2\mu) - (7 - \lambda))\hat{j} + ((2 + 4\mu) - (4 + \lambda))\hat{k}$

$\overrightarrow{PQ} = (-3\mu - 6 - 3\lambda)\hat{i} + (-9 + 2\mu - 7 + \lambda)\hat{j} + (2 + 4\mu - 4 - \lambda)\hat{k}$

$\overrightarrow{PQ} = (-3\lambda - 3\mu - 6)\hat{i} + (\lambda + 2\mu - 16)\hat{j} + (-\lambda + 4\mu - 2)\hat{k}$.

We are given that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{CD}$. This means the dot product of $\overrightarrow{PQ}$ with $\vec{v}_1$ and $\vec{v}_2$ is zero.

$\overrightarrow{PQ} \cdot \vec{v}_1 = 0$

$(-3\lambda - 3\mu - 6)(3) + (\lambda + 2\mu - 16)(-1) + (-\lambda + 4\mu - 2)(1) = 0$

$-9\lambda - 9\mu - 18 - \lambda - 2\mu + 16 - \lambda + 4\mu - 2 = 0$

Combine terms with $\lambda$ and $\mu$ and constant terms:

$(-9\lambda - \lambda - \lambda) + (-9\mu - 2\mu + 4\mu) + (-18 + 16 - 2) = 0$

$-11\lambda - 7\mu - 4 = 0$

$\overrightarrow{PQ} \cdot \vec{v}_2 = 0$

$(-3\lambda - 3\mu - 6)(-3) + (\lambda + 2\mu - 16)(2) + (-\lambda + 4\mu - 2)(4) = 0$

$9\lambda + 9\mu + 18 + 2\lambda + 4\mu - 32 - 4\lambda + 16\mu - 8 = 0$

Combine terms with $\lambda$ and $\mu$ and constant terms:

$(9\lambda + 2\lambda - 4\lambda) + (9\mu + 4\mu + 16\mu) + (18 - 32 - 8) = 0$

$7\lambda + 29\mu - 22 = 0$

We have a system of two linear equations in $\lambda$ and $\mu$:

$11\lambda + 7\mu = -4$

$7\lambda + 29\mu = 22$

Multiply the first equation by 7 and the second equation by 11 to eliminate $\lambda$:

Subtract equation (3) from equation (4):

$(77\lambda + 319\mu) - (77\lambda + 49\mu) = 242 - (-28)$

$270\mu = 242 + 28$

$270\mu = 270$

$\mu = 1$

Substitute $\mu = 1$ into equation (1):

$11\lambda + 7(1) + 4 = 0$

$11\lambda + 11 = 0$

$11\lambda = -11$

$\lambda = -1$

Now, substitute the values of $\lambda = -1$ and $\mu = 1$ into the position vectors of P and Q.

Position vector of P:

$\vec{p} = (6 + 3\lambda)\hat{i} + (7 - \lambda)\hat{j} + (4 + \lambda)\hat{k}$

$\vec{p} = (6 + 3(-1))\hat{i} + (7 - (-1))\hat{j} + (4 + (-1))\hat{k}$

$\vec{p} = (6 - 3)\hat{i} + (7 + 1)\hat{j} + (4 - 1)\hat{k}$

$\vec{p} = 3\hat{i} + 8\hat{j} + 3\hat{k}$

Position vector of Q:

$\vec{q} = -3\mu\hat{i} + (-9 + 2\mu)\hat{j} + (2 + 4\mu)\hat{k}$

$\vec{q} = -3(1)\hat{i} + (-9 + 2(1))\hat{j} + (2 + 4(1))\hat{k}$

$\vec{q} = -3\hat{i} + (-9 + 2)\hat{j} + (2 + 4)\hat{k}$

$\vec{q} = -3\hat{i} - 7\hat{j} + 6\hat{k}$

The vector $\overrightarrow{PQ}$ connecting these points is $\vec{q} - \vec{p} = (-3\hat{i} - 7\hat{j} + 6\hat{k}) - (3\hat{i} + 8\hat{j} + 3\hat{k}) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.

Check perpendicularity:

$\overrightarrow{PQ} \cdot \vec{v}_1 = (-6\hat{i} - 15\hat{j} + 3\hat{k}) \cdot (3\hat{i}− \hat{j}+ \hat{k}) = (-6)(3) + (-15)(-1) + (3)(1) = -18 + 15 + 3 = 0$. (Perpendicular to $\overrightarrow{AB}$)

$\overrightarrow{PQ} \cdot \vec{v}_2 = (-6\hat{i} - 15\hat{j} + 3\hat{k}) \cdot (−3\hat{i}+ 2\hat{j}+ 4\hat{k}) = (-6)(-3) + (-15)(2) + (3)(4) = 18 - 30 + 12 = 0$. (Perpendicular to $\overrightarrow{CD}$)

The calculations are correct.

Conclusion:

The position vector of point P on the line AB is $3\hat{i} + 8\hat{j} + 3\hat{k}$.

The position vector of point Q on the line CD is $-3\hat{i} - 7\hat{j} + 6\hat{k}$.

Solution:

Given:

The direction cosines $(l, m, n)$ of the lines satisfy the following two equations:

To Show:

The straight lines whose direction cosines satisfy these conditions are at right angles.

Proof:

From equation (1), we can express $n$ in terms of $l$ and $m$:

Substitute this expression for $n$ into equation (2):

$m(2l + 2m) + (2l + 2m)l + lm = 0$

Expand the terms:

$2lm + 2m^2 + 2l^2 + 2lm + lm = 0$

Combine the like terms:

Equation (4) is a quadratic relation between $l$ and $m$. Since the direction cosines of the two lines satisfy the given equations, this quadratic will give us the relationship between the direction ratios for each line.

We can factor this quadratic equation in $l$ and $m$. Treating it as a quadratic in $l$ with variable $m$, or vice-versa, or simply factoring:

$2l^2 + 4lm + lm + 2m^2 = 0$

$2l(l + 2m) + m(l + 2m) = 0$

$(2l + m)(l + 2m) = 0$

This equation gives two possibilities:

Case 1: $2l + m = 0 \implies m = -2l$

Case 2: $l + 2m = 0 \implies l = -2m$

Now, we find the direction cosines (or ratios) for each case using equation (3):

Case 1: $m = -2l$

Substitute $m = -2l$ into $n = 2l + 2m$:

$n = 2l + 2(-2l) = 2l - 4l = -2l$

The direction cosines $(l, m, n)$ are proportional to $(l, -2l, -2l) = l(1, -2, -2)$.

Let the direction ratios of the first line be proportional to $(a_1, b_1, c_1) = (1, -2, -2)$.

The actual direction cosines $(l_1, m_1, n_1)$ are obtained by dividing the direction ratios by the magnitude $\sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

So, $(l_1, m_1, n_1) = \left(\frac{1}{3}, \frac{-2}{3}, \frac{-2}{3}\right)$.

Case 2: $l = -2m$

Substitute $l = -2m$ into $n = 2l + 2m$:

$n = 2(-2m) + 2m = -4m + 2m = -2m$

The direction cosines $(l, m, n)$ are proportional to $(-2m, m, -2m) = m(-2, 1, -2)$.

Let the direction ratios of the second line be proportional to $(a_2, b_2, c_2) = (-2, 1, -2)$.

The actual direction cosines $(l_2, m_2, n_2)$ are obtained by dividing the direction ratios by the magnitude $\sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

So, $(l_2, m_2, n_2) = \left(\frac{-2}{3}, \frac{1}{3}, \frac{-2}{3}\right)$.

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

Calculate the sum of the products of the corresponding direction cosines for the two cases we found:

$l_1 l_2 + m_1 m_2 + n_1 n_2 = \left(\frac{1}{3}\right)\left(\frac{-2}{3}\right) + \left(\frac{-2}{3}\right)\left(\frac{1}{3}\right) + \left(\frac{-2}{3}\right)\left(\frac{-2}{3}\right)$

$l_1 l_2 + m_1 m_2 + n_1 n_2 = -\frac{2}{9} - \frac{2}{9} + \frac{4}{9}$

$l_1 l_2 + m_1 m_2 + n_1 n_2 = \frac{-2 - 2 + 4}{9} = \frac{0}{9} = 0$

Since the sum of the products of the corresponding direction cosines is 0, the two lines are at right angles.

Conclusion:

The two straight lines whose direction cosines satisfy the given equations $2l + 2m – n = 0$ and $mn + nl + lm = 0$ are indeed at right angles.

Solution:

Given:

Three lines with direction cosines $(l_1, m_1, n_1)$, $(l_2, m_2, n_2)$, and $(l_3, m_3, n_3)$.

These three lines are mutually perpendicular.

A new line whose direction cosines are proportional to $(l_1 + l_2 + l_3, m_1 + m_2 + m_3, n_1 + n_2 + n_3)$.

To Prove:

The new line makes equal angles with the three mutually perpendicular lines.

Proof:

Since $(l_1, m_1, n_1)$, $(l_2, m_2, n_2)$, and $(l_3, m_3, n_3)$ are the direction cosines of three mutually perpendicular lines, they satisfy the following orthogonality and normalization conditions:

Normalization:

Orthogonality (mutually perpendicular):

Let the direction ratios of the new line be proportional to $L = l_1 + l_2 + l_3$, $M = m_1 + m_2 + m_3$, and $N = n_1 + n_2 + n_3$.

The direction cosines of the new line are $(l, m, n) = \left(\frac{L}{R}, \frac{M}{R}, \frac{N}{R}\right)$, where $R = \sqrt{L^2 + M^2 + N^2}$ is the magnitude of the direction vector $(L, M, N)$.

Calculate $R^2 = L^2 + M^2 + N^2$:

$R^2 = (l_1 + l_2 + l_3)^2 + (m_1 + m_2 + m_3)^2 + (n_1 + n_2 + n_3)^2$

Expand each squared term:

$(l_1 + l_2 + l_3)^2 = l_1^2 + l_2^2 + l_3^2 + 2l_1 l_2 + 2l_1 l_3 + 2l_2 l_3$

$(m_1 + m_2 + m_3)^2 = m_1^2 + m_2^2 + m_3^2 + 2m_1 m_2 + 2m_1 m_3 + 2m_2 m_3$

$(n_1 + n_2 + n_3)^2 = n_1^2 + n_2^2 + n_3^2 + 2n_1 n_2 + 2n_1 n_3 + 2n_2 n_3$

Summing these expansions for $R^2$:

$R^2 = (l_1^2 + m_1^2 + n_1^2) + (l_2^2 + m_2^2 + n_2^2) + (l_3^2 + m_3^2 + n_3^2) + 2(l_1 l_2 + m_1 m_2 + n_1 n_2) + 2(l_1 l_3 + m_1 m_3 + n_1 n_3) + 2(l_2 l_3 + m_2 m_3 + n_2 n_3)$

Using the normalization conditions (1a), (1b), (1c) and orthogonality conditions (2a), (2b), (2c):

$R^2 = (1) + (1) + (1) + 2(0) + 2(0) + 2(0)$

$R^2 = 3$

$R = \sqrt{3}$

The direction cosines of the new line are $(l, m, n) = \left(\frac{l_1 + l_2 + l_3}{\sqrt{3}}, \frac{m_1 + m_2 + m_3}{\sqrt{3}}, \frac{n_1 + n_2 + n_3}{\sqrt{3}}\right)$.

Let $\theta_1$ be the angle between the new line and the first line (with direction cosines $l_1, m_1, n_1$). The cosine of the angle is given by the sum of the products of the corresponding direction cosines:

$\cos \theta_1 = l l_1 + m m_1 + n n_1$

$\cos \theta_1 = \left(\frac{l_1 + l_2 + l_3}{\sqrt{3}}\right)l_1 + \left(\frac{m_1 + m_2 + m_3}{\sqrt{3}}\right)m_1 + \left(\frac{n_1 + n_2 + n_3}{\sqrt{3}}\right)n_1$

$\cos \theta_1 = \frac{1}{\sqrt{3}} [l_1(l_1 + l_2 + l_3) + m_1(m_1 + m_2 + m_3) + n_1(n_1 + n_2 + n_3)]$

$\cos \theta_1 = \frac{1}{\sqrt{3}} [l_1^2 + l_1 l_2 + l_1 l_3 + m_1^2 + m_1 m_2 + m_1 m_3 + n_1^2 + n_1 n_2 + n_1 n_3]$

Rearrange terms:

$\cos \theta_1 = \frac{1}{\sqrt{3}} [(l_1^2 + m_1^2 + n_1^2) + (l_1 l_2 + m_1 m_2 + n_1 n_2) + (l_1 l_3 + m_1 m_3 + n_1 n_3)]$

Using the normalization condition (1a) and orthogonality conditions (2a), (2c):

$\cos \theta_1 = \frac{1}{\sqrt{3}} [(1) + (0) + (0)]$

Let $\theta_2$ be the angle between the new line and the second line (with direction cosines $l_2, m_2, n_2$).

$\cos \theta_2 = l l_2 + m m_2 + n n_2$

$\cos \theta_2 = \left(\frac{l_1 + l_2 + l_3}{\sqrt{3}}\right)l_2 + \left(\frac{m_1 + m_2 + m_3}{\sqrt{3}}\right)m_2 + \left(\frac{n_1 + n_2 + n_3}{\sqrt{3}}\right)n_2$

$\cos \theta_2 = \frac{1}{\sqrt{3}} [l_2(l_1 + l_2 + l_3) + m_2(m_1 + m_2 + m_3) + n_2(n_1 + n_2 + n_3)]$

$\cos \theta_2 = \frac{1}{\sqrt{3}} [l_1 l_2 + l_2^2 + l_2 l_3 + m_1 m_2 + m_2^2 + m_2 m_3 + n_1 n_2 + n_2^2 + n_2 n_3]$

Rearrange terms:

$\cos \theta_2 = \frac{1}{\sqrt{3}} [(l_2^2 + m_2^2 + n_2^2) + (l_1 l_2 + m_1 m_2 + n_1 n_2) + (l_2 l_3 + m_2 m_3 + n_2 n_3)]$

Using the normalization condition (1b) and orthogonality conditions (2a), (2b):

$\cos \theta_2 = \frac{1}{\sqrt{3}} [(1) + (0) + (0)]$

Let $\theta_3$ be the angle between the new line and the third line (with direction cosines $l_3, m_3, n_3$).

$\cos \theta_3 = l l_3 + m m_3 + n n_3$

$\cos \theta_3 = \left(\frac{l_1 + l_2 + l_3}{\sqrt{3}}\right)l_3 + \left(\frac{m_1 + m_2 + m_3}{\sqrt{3}}\right)m_3 + \left(\frac{n_1 + n_2 + n_3}{\sqrt{3}}\right)n_3$

$\cos \theta_3 = \frac{1}{\sqrt{3}} [l_3(l_1 + l_2 + l_3) + m_3(m_1 + m_2 + m_3) + n_3(n_1 + n_2 + n_3)]$

$\cos \theta_3 = \frac{1}{\sqrt{3}} [l_1 l_3 + l_2 l_3 + l_3^2 + m_1 m_3 + m_2 m_3 + m_3^2 + n_1 n_3 + n_2 n_3 + n_3^2]$

Rearrange terms:

$\cos \theta_3 = \frac{1}{\sqrt{3}} [(l_3^2 + m_3^2 + n_3^2) + (l_1 l_3 + m_1 m_3 + n_1 n_3) + (l_2 l_3 + m_2 m_3 + n_2 n_3)]$

Using the normalization condition (1c) and orthogonality conditions (2b), (2c):

$\cos \theta_3 = \frac{1}{\sqrt{3}} [(1) + (0) + (0)]$

From (3a), (3b), and (3c), we have $\cos \theta_1 = \cos \theta_2 = \cos \theta_3 = \frac{1}{\sqrt{3}}$.

Since the cosines of the angles are equal, and assuming the angles are in $[0, \pi]$ (which they are, as angles between lines are usually considered), the angles themselves are equal: $\theta_1 = \theta_2 = \theta_3$.

Conclusion:

The line whose direction cosines are proportional to $l_1 + l_2 + l_3, m_1 + m_2 + m_3, n_1 + n_2 + n_3$ makes equal angles with the three mutually perpendicular lines.

Solution:

Given:

The point P has coordinates $(\alpha, \beta, \gamma)$.

To Find:

The distance of the point $(\alpha, \beta, \gamma)$ from the y-axis.

Calculation:

The y-axis is the line passing through the origin (0, 0, 0) with direction ratios (0, 1, 0).

The equation of the y-axis can be written as $\frac{x-0}{0} = \frac{y-0}{1} = \frac{z-0}{0}$ (interpreting divisions by zero appropriately, it means $x=0$ and $z=0$).

Any point on the y-axis has coordinates of the form (0, $k$, 0) for some real number $k$.

Let Q be the foot of the perpendicular from the point P$(\alpha, \beta, \gamma)$ to the y-axis.

Q must be a point on the y-axis, so its coordinates are $(0, k, 0)$.

The vector $\overrightarrow{PQ}$ connects P$(\alpha, \beta, \gamma)$ to Q$(0, k, 0)$.

$\overrightarrow{PQ} = (0 - \alpha)\hat{i} + (k - \beta)\hat{j} + (0 - \gamma)\hat{k} = -\alpha\hat{i} + (k - \beta)\hat{j} - \gamma\hat{k}$.

Since $\overrightarrow{PQ}$ is perpendicular to the y-axis, it must be orthogonal to the direction vector of the y-axis, which is $\hat{j}$ (or $0\hat{i} + 1\hat{j} + 0\hat{k}$).

$\overrightarrow{PQ} \cdot \hat{j} = 0$

$(-\alpha)(0) + (k - \beta)(1) + (-\gamma)(0) = 0$

$k - \beta = 0$

$k = \beta$

So, the foot of the perpendicular from P to the y-axis is Q$(0, \beta, 0)$.

The distance of the point P$(\alpha, \beta, \gamma)$ from the y-axis is the distance between P and Q.

Using the distance formula for points P$(\alpha, \beta, \gamma)$ and Q$(0, \beta, 0)$:

Distance $PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2}$

Distance $PQ = \sqrt{(0 - \alpha)^2 + (\beta - \beta)^2 + (0 - \gamma)^2}$

Distance $PQ = \sqrt{(-\alpha)^2 + (0)^2 + (-\gamma)^2}$

Distance $PQ = \sqrt{\alpha^2 + 0 + \gamma^2}$

Distance $PQ = \sqrt{\alpha^2 + \gamma^2}$

Comparing the result with the given options:

(A) $\beta$

(B) $|\beta|$

(C) $|\beta|+|\gamma|$

(D) $\sqrt{α^2 + γ^2}$

The calculated distance matches option (D).

Conclusion:

The distance of the point $(\alpha, \beta, \gamma)$ from the y-axis is $\sqrt{\alpha^2 + \gamma^2}$.

The correct option is (D).

Solution:

Given:

The direction cosines of a line are $l = k$, $m = k$, and $n = k$.

To Find:

The possible value(s) of $k$.

Calculation:

The direction cosines $(l, m, n)$ of any line in three dimensions must satisfy the fundamental property that the sum of the squares of the direction cosines is equal to 1.

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

Substitute the given direction cosines into this equation:

$k^2 + k^2 + k^2 = 1$

$3k^2 = 1$

$k^2 = \frac{1}{3}$

Taking the square root of both sides:

$k = \pm \sqrt{\frac{1}{3}}$

$k = \pm \frac{1}{\sqrt{3}}$

So, the possible values for $k$ are $\frac{1}{\sqrt{3}}$ and $-\frac{1}{\sqrt{3}}$.

Comparing the result with the given options:

(A) k > 0 (This is only one possibility, not the complete solution)

(B) 0 < k < 1 (This is not necessarily true; $-\frac{1}{\sqrt{3}}$ is negative)

(C) k = 1 (Substituting k=1 gives $1^2+1^2+1^2 = 3 \neq 1$, so k=1 is incorrect)

(D) $k = \frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$ (This matches our derived values)

Conclusion:

If the direction cosines of a line are $k, k, k$, then $k$ must be $\frac{1}{\sqrt{3}}$ or $-\frac{1}{\sqrt{3}}$.

The correct option is (D).

Solution:

Given:

The equation of the plane is $\vec{r} \cdot \left( \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k} \right)= 1$.

To Find:

The distance of the plane from the origin.

Calculation:

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

The given equation of the plane is $\vec{r} \cdot \left( \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k} \right)= 1$.

Let the normal vector be $\vec{N} = \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k}$.

We need to check if this normal vector is a unit vector by calculating its magnitude:

$|\vec{N}| = \left|\frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k}\right| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2}$

$|\vec{N}| = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{4 + 9 + 36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1$

Since the magnitude of the normal vector is 1, the vector $\vec{N}$ is indeed a unit vector. Thus, $\hat{n} = \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k}$.

The given equation of the plane is in the form $\vec{r} \cdot \hat{n} = d$.

Comparing the equation $\vec{r} \cdot \left( \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k} \right)= 1$ with $\vec{r} \cdot \hat{n} = d$, we can directly identify the distance $d$ from the origin.

The perpendicular distance of the plane from the origin is 1 unit.

Comparing the result with the given options:

(A) 1

(B) 7

(C) $\frac{1}{7}$

(D) None of these

The calculated distance matches option (A).

Conclusion:

The distance of the plane $\vec{r} \left( \frac{2}{7} \hat{i}+ \frac{3}{7} \hat{j}− \frac{6}{7} \hat{k} \right)= 1$ from the origin is 1.

The correct option is (A).

Solution:

Given:

Equation of the line: $\frac{x − 2}{3} = \frac{y − 3}{5} = \frac{z − 4}{5}$.

Equation of the plane: $2x – 2y + z = 5$.

To Find:

The sine of the angle ($\theta$) between the line and the plane.

Calculation:

The direction vector of the line is given by the denominators in the symmetric form: $\vec{b} = 3\hat{i} + 5\hat{j} + 5\hat{k}$.

The normal vector to the plane is given by the coefficients of $x$, $y$, and $z$: $\vec{n} = 2\hat{i} - 2\hat{j} + 1\hat{k}$.

The sine of the angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by the formula:

$\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

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

$\vec{b} \cdot \vec{n} = (3)(2) + (5)(-2) + (5)(1) = 6 - 10 + 5 = 1$.

Calculate the magnitude of $\vec{b}$:

$|\vec{b}| = \sqrt{3^2 + 5^2 + 5^2} = \sqrt{9 + 25 + 25} = \sqrt{59}$.

Calculate the magnitude of $\vec{n}$:

$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

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

$\sin \theta = \frac{|1|}{\sqrt{59} \times 3} = \frac{1}{3\sqrt{59}}$.

Comparing the calculated result $\frac{1}{3\sqrt{59}}$ with the given options (A, B, C, D), it does not match any of them.

Conclusion:

Based on the provided equations for the line and the plane, the sine of the angle between them is $\frac{1}{3\sqrt{59}}$. This value is not present among the given options, indicating a likely error in the question statement or the provided choices.

[Note: The calculated answer based on the provided question parameters is $\frac{1}{3\sqrt{59}}$, which is not among the options. There is likely an error in the question or options.]

Solution:

Given:

The point P has coordinates $(\alpha, \beta, \gamma)$.

The plane of reflection is the xy-plane.

To Find:

The coordinates of the reflection of the point $(\alpha, \beta, \gamma)$ in the xy-plane.

Calculation:

The xy-plane is the plane where the z-coordinate is zero. Its equation is $z = 0$.

When a point is reflected in a plane, the coordinates parallel to the plane remain unchanged, while the coordinate perpendicular to the plane changes its sign.

In the case of reflection in the xy-plane:

The x-coordinate remains $\alpha$.

The y-coordinate remains $\beta$.

The z-coordinate changes sign from $\gamma$ to $-\gamma$.

Therefore, the coordinates of the reflection of the point $(\alpha, \beta, \gamma)$ in the xy-plane are $(\alpha, \beta, -\gamma)$.

Comparing this result with the given options:

(A) $(α, β, 0)$ is the projection onto the xy-plane.

(B) $(0, 0, γ)$ is a point on the z-axis.

(C) $(–α, –β, γ)$ is related to reflection through the z-axis.

(D) $(α, β, –γ)$ matches our result.

Conclusion:

The reflection of the point $(\alpha, \beta, \gamma)$ in the xy-plane is $(\alpha, \beta, -\gamma)$.

The correct option is (D).

Solution:

Given:

The vertices of the quadrilateral ABCD are A(0, 4, 1), B (2, 3, –1), C(4, 5, 0), and D (2, 6, 2).

To Find:

The area of the quadrilateral ABCD.

Calculation:

The area of a quadrilateral in space can be calculated as half of the magnitude of the cross product of its diagonals.

First, let's find the vectors representing the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$.

The coordinates of the points are A(0, 4, 1), B(2, 3, –1), C(4, 5, 0), and D(2, 6, 2).

Vector $\overrightarrow{AC}$ = $\vec{C} - \vec{A}$:

$\overrightarrow{AC} = (4-0)\hat{i} + (5-4)\hat{j} + (0-1)\hat{k} = 4\hat{i} + 1\hat{j} - 1\hat{k}$.

Vector $\overrightarrow{BD}$ = $\vec{D} - \vec{B}$:

$\overrightarrow{BD} = (2-2)\hat{i} + (6-3)\hat{j} + (2-(-1))\hat{k} = 0\hat{i} + 3\hat{j} + 3\hat{k}$.

Next, calculate the cross product of the diagonal vectors $\overrightarrow{AC} \times \overrightarrow{BD}$:

$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & -1 \\ 0 & 3 & 3 \end{vmatrix}$

$\overrightarrow{AC} \times \overrightarrow{BD} = \hat{i}((1)(3) - (-1)(3)) - \hat{j}((4)(3) - (-1)(0)) + \hat{k}((4)(3) - (1)(0))$

$\overrightarrow{AC} \times \overrightarrow{BD} = \hat{i}(3 + 3) - \hat{j}(12 - 0) + \hat{k}(12 - 0)$

$\overrightarrow{AC} \times \overrightarrow{BD} = 6\hat{i} - 12\hat{j} + 12\hat{k}$.

Now, find the magnitude of the cross product:

$|\overrightarrow{AC} \times \overrightarrow{BD}| = |6\hat{i} - 12\hat{j} + 12\hat{k}| = \sqrt{6^2 + (-12)^2 + 12^2}$

$|\overrightarrow{AC} \times \overrightarrow{BD}| = \sqrt{36 + 144 + 144} = \sqrt{324}$.

The square root of 324 is 18.

$|\overrightarrow{AC} \times \overrightarrow{BD}| = 18$.

The area of the quadrilateral ABCD is half the magnitude of the cross product of its diagonals:

Area $= \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}| = \frac{1}{2} \times 18 = 9$.

The area of the quadrilateral is 9 square units.

Comparing the calculated area with the given options:

(A) 9 sq. units

(B) 18 sq. units

(C) 27 sq. units

(D) 81 sq. Units

The calculated area matches option (A).

Conclusion:

The area of the quadrilateral ABCD is 9 sq. units.

The correct option is (A).

Solution:

Given:

The equation $xy + yz = 0$.

To Find:

The type of locus represented by the given equation in three-dimensional space.

Calculation:

The given equation is $xy + yz = 0$.

We can factor out the common term $y$ from the equation:

$y(x + z) = 0$

This equation is satisfied if and only if either $y = 0$ or $x + z = 0$ (or both).

Thus, the locus of the points $(x, y, z)$ that satisfy the equation $y(x + z) = 0$ is the set of points that lie on either the surface defined by $y = 0$ or the surface defined by $x + z = 0$.

Consider the equation $y = 0$. In a three-dimensional Cartesian coordinate system, this equation represents a plane. Specifically, it is the xz-plane.

The equation of the plane can be written as $0x + 1y + 0z = 0$. The normal vector to this plane is $\vec{n}_1 = 0\hat{i} + 1\hat{j} + 0\hat{k} = \hat{j}$.

Consider the equation $x + z = 0$. In a three-dimensional Cartesian coordinate system, this equation also represents a plane. This plane passes through the origin since $(0) + (0) = 0$.

The equation can be written as $1x + 0y + 1z = 0$. The normal vector to this plane is $\vec{n}_2 = 1\hat{i} + 0\hat{j} + 1\hat{k} = \hat{i} + \hat{k}$.

The locus represented by $xy + yz = 0$ is the union of the plane $y = 0$ and the plane $x + z = 0$. These are two distinct planes.

Now, we check if these two planes are perpendicular or parallel.

Two planes are perpendicular if their normal vectors are orthogonal (their dot product is zero).

Calculate the dot product of the normal vectors $\vec{n}_1$ and $\vec{n}_2$:

$\vec{n}_1 \cdot \vec{n}_2 = (\hat{j}) \cdot (\hat{i} + \hat{k})$

$\vec{n}_1 \cdot \vec{n}_2 = (0)(1) + (1)(0) + (0)(1)$

$\vec{n}_1 \cdot \vec{n}_2 = 0 + 0 + 0 = 0$

Since the dot product of the normal vectors is zero, the normal vectors $\vec{n}_1$ and $\vec{n}_2$ are orthogonal. Therefore, the two planes $y = 0$ and $x + z = 0$ are perpendicular to each other.

The locus represented by the equation $xy + yz = 0$ is a pair of perpendicular planes.

Comparing the result with the given options:

(A) A pair of perpendicular lines - Incorrect.

(B) A pair of parallel lines - Incorrect.

(C) A pair of parallel planes - Incorrect, the planes are perpendicular.

(D) A pair of perpendicular planes - Correct.

Conclusion:

The locus represented by the equation $xy + yz = 0$ is a pair of perpendicular planes.

The correct option is (D).

Solution:

Given:

Equation of the plane: $2x – 3y + 6z – 11 = 0$.

The angle between the plane and the x-axis is $\sin^{-1}(\alpha)$.

To Find:

The value of $\alpha$.

Calculation:

Let $\theta$ be the angle between the plane and the x-axis. We are given that $\theta = \sin^{-1}(\alpha)$, which means $\sin \theta = \alpha$.

The direction vector of the x-axis is $\vec{b} = \hat{i}$ (since the x-axis has direction ratios (1, 0, 0)). The magnitude is $|\vec{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1$.

The normal vector to the plane $2x – 3y + 6z – 11 = 0$ is $\vec{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ (the coefficients of x, y, and z). The magnitude of the normal vector is $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

The sine of the angle $\theta$ between a line (x-axis) with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by the formula:

$\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

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

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

$\vec{b} \cdot \vec{n} = (1)(2) + (0)(-3) + (0)(6) = 2 + 0 + 0 = 2$.

The absolute value is $|\vec{b} \cdot \vec{n}| = |2| = 2$.

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

$\sin \theta = \frac{2}{(1)(7)} = \frac{2}{7}$

We are given that $\sin \theta = \alpha$.

Therefore, $\alpha = \frac{2}{7}$.

Comparing the result with the given options:

(A) $\frac{\sqrt{3}}{2}$

(B) $\frac{\sqrt{2}}{3}$

(C) $\frac{2}{7}$

(D) $\frac{3}{7}$

The calculated value of $\alpha$ matches option (C).

Conclusion:

The value of $\alpha$ is $\frac{2}{7}$.

The correct option is (C).

Given:

The plane passes through the points (2, 0, 0), (0, 3, 0) and (0, 0, 4).

To Find:

The equation of the plane.

Solution:

The points through which the plane passes are (2, 0, 0), (0, 3, 0), and (0, 0, 4).

These points represent the intercepts of the plane on the x, y, and z axes, respectively.

The x-intercept is $a = 2$.

The y-intercept is $b = 3$.

The z-intercept is $c = 4$.

The equation of a plane in the intercept form is given by:

Substitute the values of a, b, and c from the given points into equation (i):

To remove the denominators, find the least common multiple (LCM) of 2, 3, and 4.

The LCM of 2, 3, and 4 is 12.

Multiply equation (ii) by 12:

Simplify equation (iii):

This simplifies to:

The equation can also be written in the general form:

The blank in the question should be filled with the equation of the plane.

Answer:

The equation of the plane is $\mathbf{6x + 4y + 3z = 12}$.

Given:

The vector is $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$.

To Find:

The direction cosines of the given vector.

Solution:

Let the given vector be $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$.

The components of the vector are $a = 2$, $b = 2$, and $c = -1$.

The magnitude of the vector $\vec{v}$ is given by $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$.

Calculate the magnitude:

The direction cosines of a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ are given by $\frac{a}{|\vec{v}|}, \frac{b}{|\vec{v}|}, \frac{c}{|\vec{v}|}$.

The direction cosine along the x-axis is $\ell = \frac{a}{|\vec{v}|} = \frac{2}{3}$.

The direction cosine along the y-axis is $m = \frac{b}{|\vec{v}|} = \frac{2}{3}$.

The direction cosine along the z-axis is $n = \frac{c}{|\vec{v}|} = \frac{-1}{3}$.

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

We can verify that the sum of the squares of the direction cosines is 1:

Answer:

The direction cosines of the vector $(2\hat{i}+ 2\hat{j}− \hat{k})$ are $\mathbf{\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}}$.

Given:

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

To Find:

The vector equation of the line.

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:

Comparing the given equation $\frac{x − 5}{3} = \frac{y + 4}{7} = \frac{z − 6}{2}$ with the general form (i), we can identify the point on the line and the direction ratios.

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

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

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.

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

The direction vector parallel to the line (formed by the direction ratios) is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

Substituting these vectors into the vector equation form:

This is the vector equation of the given line.

Answer:

The vector equation of the line is $\mathbf{\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k})}$ where $\lambda$ is a scalar.

Given:

The line passes through the points (3, 4, -7) and (1, -1, 6).

To Find:

The vector equation of the line.

Solution:

Let the two given points be P(3, 4, -7) and Q(1, -1, 6).

The position vector of point P is $\vec{a} = 3\hat{i} + 4\hat{j} - 7\hat{k}$.

The position vector of point Q is $\vec{b} = \hat{i} - \hat{j} + 6\hat{k}$.

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

where $\lambda$ is a scalar parameter.

First, calculate the vector $(\vec{b} - \vec{a})$ which represents the direction vector of the line:

$(\vec{b} - \vec{a}) = (\hat{i} - \hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} - 7\hat{k})$

$(\vec{b} - \vec{a}) = (1-3)\hat{i} + (-1-4)\hat{j} + (6 - (-7))\hat{k}$

$(\vec{b} - \vec{a}) = -2\hat{i} - 5\hat{j} + 13\hat{k}$

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

This is the vector equation of the line passing through the two given points.

Alternate Solution:

Alternatively, we can use point Q as the starting point. The vector equation is $\vec{r} = \vec{b} + \lambda (\vec{a} - \vec{b})$.

Calculate the vector $(\vec{a} - \vec{b})$:

$(\vec{a} - \vec{b}) = (3\hat{i} + 4\hat{j} - 7\hat{k}) - (\hat{i} - \hat{j} + 6\hat{k})$

$(\vec{a} - \vec{b}) = (3-1)\hat{i} + (4 - (-1))\hat{j} + (-7-6)\hat{k}$

$(\vec{a} - \vec{b}) = 2\hat{i} + 5\hat{j} - 13\hat{k}$

Substitute the position vector $\vec{b}$ and the direction vector $(\vec{a} - \vec{b})$ into the alternate form:

Both equations (ii) and (iii) represent the same line.

Answer:

The vector equation of the line is $\mathbf{\vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda (-2\hat{i} - 5\hat{j} + 13\hat{k})}$ where $\lambda$ is a scalar.

Given:

The vector equation of the plane is $\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2$.

To Find:

The Cartesian equation of the plane.

Solution:

The general form of the vector equation of a plane is $\vec{r} \cdot \vec{n} = d$, where $\vec{r}$ is the position vector of any point on the plane, $\vec{n}$ is a normal vector to the plane, and $d$ is a constant.

In the given equation $\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2$, we can identify:

The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} - \hat{k}$.

The constant is $d = 2$.

Let $\vec{r}$ be the position vector of any arbitrary point P(x, y, z) on the plane.

Substitute equation (i) into the given vector equation:

Perform the dot product of the two vectors. The dot product of two vectors $(a_1\hat{i} + b_1\hat{j} + c_1\hat{k})$ and $(a_2\hat{i} + b_2\hat{j} + c_2\hat{k})$ is $a_1a_2 + b_1b_2 + c_1c_2$.

Applying this to equation (ii):

Simplify equation (iii):

Equation (iv) is the Cartesian equation of the plane.

This can also be written as $x + y - z - 2 = 0$.

Answer:

The Cartesian equation of the plane $\vec{r}(\hat{i}+ \hat{j}− \hat{k}) = 2$ is $\mathbf{x + y - z = 2}$.

Given:

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

The claimed unit vector normal to the plane is $\hat{n}_{\text{claimed}} = \frac{1}{\sqrt{14}} \hat{i}+ \frac{2}{\sqrt{14}} \hat{j}+ \frac{3}{\sqrt{14}} \hat{k}$.

To Check:

Determine if the given statement is True or False.

Solution:

The general Cartesian equation of a plane is $Ax + By + Cz + D = 0$.

A normal vector to this plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

Comparing the given equation $x + 2y + 3z - 6 = 0$ with the general form, we have $A=1$, $B=2$, and $C=3$.

The normal vector to the plane is $\vec{n} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

To find the unit normal vector, we need to divide the normal vector by its magnitude.

The magnitude of the normal vector is $|\vec{n}| = \sqrt{A^2 + B^2 + C^2}$.

The unit normal vector $\hat{n}$ is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}|}$.

This can be written as:

Comparing the calculated unit normal vector (v) with the claimed unit normal vector in the question, we see that they are identical.

Answer:

The statement is True.

Given:

The equation of the plane is $2x - 3y + 5z + 4 = 0$.

The claimed intercepts on the coordinate axes are -2, $\frac{4}{3}$, and $-\frac{4}{5}$.

To Check:

Determine if the given statement about the intercepts is True or False.

Solution:

The equation of the plane is $2x - 3y + 5z + 4 = 0$.

To find the intercepts, we need to rewrite the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a, b, c$ are the x, y, and z intercepts, respectively.

First, move the constant term to the right side of the equation:

To make the right side equal to 1, divide the entire equation by -4:

Simplify the terms:

Rewrite the terms involving y and z to match the intercept form $\frac{y}{b}$ and $\frac{z}{c}$:

Comparing equation (iv) with the intercept form, we find the intercepts:

The x-intercept is $a = -2$.

The y-intercept is $b = \frac{4}{3}$.

The z-intercept is $c = -\frac{4}{5}$.

The intercepts made by the plane on the coordinate axes are $-2, \frac{4}{3}, -\frac{4}{5}$.

These calculated intercepts match the intercepts given in the statement.

Answer:

The statement is True.

Given:

The equation of the line is $\vec{r} = (5\hat{i}− \hat{j}− 4\hat{k}) + λ (2\hat{i}− \hat{j}+ \hat{k})$.

The equation of the plane is $\vec{r} \cdot (3\hat{i}− 4\hat{j}− \hat{k}) + 5 = 0$, which can be written as $\vec{r} \cdot (3\hat{i}− 4\hat{j}− \hat{k}) = -5$.

The claimed angle between the line and the plane is $\theta = \sin^{-1} \left( \frac{5}{2\sqrt{91}} \right)$.

To Check:

Determine if the given statement is True or False.

Solution:

The equation of the line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a} = 5\hat{i}− \hat{j}− 4\hat{k}$ is a point on the line and $\vec{b} = 2\hat{i}− \hat{j}+ \hat{k}$ is the direction vector of the line.

The equation of the plane is of the form $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector to the plane.

The angle $\theta$ between the line and the plane is given by the formula:

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

Next, calculate the magnitude of the direction vector $|\vec{b}| $:

Then, calculate the magnitude of the normal vector $|\vec{n}| $:

Substitute the values from (iii), (iv), and (v) into the formula for $\sin \theta$ (i):

Simplify the square root $\sqrt{156}$:

Substitute (viii) into (vii):

Thus, the angle $\theta$ is $\sin^{-1} \left( \frac{9}{2\sqrt{39}} \right)$.

The claimed angle in the question is $\sin^{-1} \left( \frac{5}{2\sqrt{91}} \right)$.

Comparing our calculated value $\frac{9}{2\sqrt{39}}$ with the claimed value $\frac{5}{2\sqrt{91}}$, we see they are not equal:

Answer:

The statement is False.

Given:

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

The equation of the second plane is $\vec{r} \cdot (\hat{i}− \hat{j}) = 4$, which can be written as $\vec{r} \cdot (\hat{i}− \hat{j} + 0\hat{k}) = 4$.

The claimed angle between the planes is $\theta = \cos^{-1} \left( \frac{−5}{\sqrt{58}} \right)$.

To Check:

Determine if the given statement is True or False.

Solution:

The angle $\theta$ between two planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$ is given by the angle between their normal vectors, which is calculated using the formula:

From the given equations of the planes, the normal vectors are:

First, calculate the dot product $\vec{n_1} \cdot \vec{n_2}$:

Next, calculate the magnitude of the first normal vector $|\vec{n_1}| $:

Then, calculate the magnitude of the second normal vector $|\vec{n_2}| $:

Substitute the values from (iii), (iv), and (v) into the formula for $\cos \theta$ (i):

Simplify the square root $\sqrt{28}$:

Substitute (viii) into (vii):

Thus, the angle $\theta$ between the planes is $\cos^{-1} \left( \frac{5}{2\sqrt{7}} \right)$.

The claimed angle in the question is $\cos^{-1} \left( \frac{−5}{\sqrt{58}} \right)$.

Comparing our calculated value $\frac{5}{2\sqrt{7}}$ with the claimed value $\frac{-5}{\sqrt{58}}$, we see they are not equal. Also, the cosine of the angle between two planes is usually taken as the acute angle, which corresponds to the absolute value of the dot product.

Answer:

The statement is False.

Given:

The equation of the line is $\vec{r} = 2\hat{i}− 3\hat{j}− \hat{k}+ λ (\hat{i}− \hat{j}+ 2\hat{k})$.

The equation of the plane is $\vec{r} \cdot (3\hat{i}+ \hat{j}− \hat{k}) + 2 = 0$, which can be written as $\vec{r} \cdot (3\hat{i}+ \hat{j}− \hat{k}) = -2$.

To Check:

Determine if the given statement is True or False.

Solution:

A line $\vec{r} = \vec{a} + \lambda \vec{b}$ lies in the plane $\vec{r} \cdot \vec{n} = d$ if and only if two conditions are met:

1. The direction vector of the line $\vec{b}$ is perpendicular to the normal vector of the plane $\vec{n}$. This means $\vec{b} \cdot \vec{n} = 0$.

2. Any point on the line lies in the plane. We can check this using the point $\vec{a}$ on the line, i.e., $\vec{a} \cdot \vec{n} = d$.

From the equation of the line, the point on the line is $\vec{a} = 2\hat{i}− 3\hat{j}− \hat{k}$, and the direction vector is $\vec{b} = \hat{i}− \hat{j}+ 2\hat{k}$.

From the equation of the plane, the normal vector is $\vec{n} = 3\hat{i}+ \hat{j}− \hat{k}$, and $d = -2$.

Condition 1: Check if $\vec{b} \cdot \vec{n} = 0$.

Condition 1 is satisfied. The line is parallel to the plane or lies in the plane.

Condition 2: Check if the point $\vec{a}$ lies in the plane, i.e., $\vec{a} \cdot \vec{n} = d$.

The constant $d$ for the plane is -2. We found $\vec{a} \cdot \vec{n} = 4$.

Condition 2 is not satisfied. The point $\vec{a}$ on the line does not lie in the plane.

Since Condition 1 is met ($\vec{b} \cdot \vec{n} = 0$) and Condition 2 is not met ($\vec{a} \cdot \vec{n} \neq d$), the line is parallel to the plane but does not lie in the plane.

Answer:

The statement is False.

Given:

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

The claimed vector equation of the line is $\vec{r}= 5\hat{i}− 4\hat{j}+ 6\hat{k}+ λ (3\hat{i}+ 7\hat{j}+ 2\hat{k})$.

To Check:

Determine if the given statement is True or False.

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:

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

where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

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

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

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

From these, the position vector of a point on the line is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$.

The direction vector parallel to the line is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

Substitute these into the general vector equation form (ii):

Comparing our calculated vector equation (iii) with the claimed vector equation in the question, we see that they are identical.

Answer:

The statement is True.

Given:

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

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

The claimed Cartesian equation of the line is $\frac{x − 5}{2} = \frac{y + 2}{−1} = \frac{z − 4}{3}$.

To Check:

Determine if the given statement is True or False.

Solution:

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

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

The direction ratios of the line are the components of the parallel vector, i.e., $(a, b, c) = (2, 1, 3)$.

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:

Substitute the given values $(x_1, y_1, z_1) = (5, -2, 4)$ and $(a, b, c) = (2, 1, 3)$ into equation (i):

Simplify equation (ii):

The claimed equation in the question is $\frac{x − 5}{2} = \frac{y + 2}{−1} = \frac{z − 4}{3}$.

Comparing our calculated equation (iii) with the claimed equation, we see that the denominator for the y-term is different (1 in our calculation vs -1 in the claim).

This indicates that the direction ratios are different. The claimed direction vector is $2\hat{i}− \hat{j}+ 3\hat{k}$, which is parallel to the claimed line, but not the given vector $2\hat{i}+ \hat{j}+ 3\hat{k}$.

Answer:

The statement is False.

Given:

The foot of the perpendicular from the origin (0, 0, 0) to the plane is the point P(5, -3, -2).

The claimed equation of the plane is $\vec{r} \cdot (5\hat{i}− 3\hat{j}− 2\hat{k}) = 38$.

To Check:

Determine if the given statement is True or False.

Solution:

The vector joining the origin (O) to the foot of the perpendicular (P) is normal to the plane.

The coordinates of the origin are (0, 0, 0).

The coordinates of the foot of the perpendicular are (5, -3, -2).

The vector $\vec{OP}$ is given by $(5-0)\hat{i} + (-3-0)\hat{j} + (-2-0)\hat{k} = 5\hat{i} - 3\hat{j} - 2\hat{k}$.

This vector $\vec{OP}$ is the normal vector $\vec{n}$ to the plane.

The point P(5, -3, -2) lies on the plane. The position vector of this point is $\vec{p} = 5\hat{i} - 3\hat{j} - 2\hat{k}$.

The vector equation of a plane passing through a point with position vector $\vec{p}$ and having a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{p}) \cdot \vec{n} = 0$, which simplifies to $\vec{r} \cdot \vec{n} = \vec{p} \cdot \vec{n}$.

Substitute the values of $\vec{p}$ and $\vec{n}$ into the equation $\vec{r} \cdot \vec{n} = \vec{p} \cdot \vec{n}$:

Calculate the dot product on the right side of equation (i):

Substitute the result back into equation (i):

Equation (iv) is the vector equation of the plane.

Comparing our calculated equation (iv) with the claimed equation $\vec{r}. (5\hat{i}− 3\hat{j}− 2\hat{k}) = 38$, we see that they are identical.

Answer:

The statement is True.