Chapter 11 Three Dimensional Geometry (Additional Questions)

Let the point be $P = (2, 3, -5)$ and the origin be $O = (0, 0, 0)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by the formula:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

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Here, $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (2, 3, -5)$.

Substituting these values into the formula:

Distance $OP = \sqrt{(2 - 0)^2 + (3 - 0)^2 + (-5 - 0)^2}$

Distance $OP = \sqrt{(2)^2 + (3)^2 + (-5)^2}

Distance $OP = \sqrt{4 + 9 + 25}

Distance $OP = \sqrt{38}

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Thus, the distance of the point $(2, 3, -5)$ from the origin is $\sqrt{38}$.

Comparing this with the given options, we find that option (C) is correct.

The final answer is (C) $\sqrt{38}$.

The direction cosines of a line in three-dimensional space are the cosines of the angles that the line makes with the positive x-axis, y-axis, and z-axis, respectively.

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Consider the z-axis.

The z-axis makes an angle of $90^\circ$ with the positive x-axis.

The z-axis makes an angle of $90^\circ$ with the positive y-axis.

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

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Therefore, the direction cosines of the z-axis are:

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

We know that $\cos 90^\circ = 0$ and $\cos 0^\circ = 1$.

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

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Comparing this with the given options, we find that option (C) is correct.

The final answer is (C) $(0, 0, 1)$.

Let the direction cosines of the line be $l, m, n$.

The direction cosines are defined as the cosines of the angles the line makes with the positive x, y, and z axes, respectively.

So, $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.

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A fundamental property of direction cosines is that the sum of the squares of the direction cosines of any line in three-dimensional space is always equal to 1.

Mathematically, this is expressed as:

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

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Substituting the definitions of $l, m, n$ in terms of $\alpha, \beta, \gamma$, we get:

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

This simplifies to:

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

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Therefore, the value of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$ is 1.

Comparing this with the given options, we find that option (B) is correct.

The final answer is (B) $1$.

Let the two given points be $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.

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The direction ratios of a line segment joining two points in three-dimensional space are proportional to the differences in their corresponding coordinates.

For the points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the differences in coordinates are:

Difference in x-coordinates: $x_2 - x_1$

Difference in y-coordinates: $y_2 - y_1$

Difference in z-coordinates: $z_2 - z_1$

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Therefore, the direction ratios of the line segment joining $A$ and $B$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.

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Comparing this with the given options, we find that option (B) is correct.

The final answer is (B) $(x_2-x_1, y_2-y_1, z_2-z_1)$.

Let $\vec{r}$ be the position vector of an arbitrary point $P$ on the line.

The line passes through a point $A$ with position vector $\vec{a}$.

The vector joining point $A$ to point $P$ is $\vec{AP} = \vec{r} - \vec{a}$.

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The line is parallel to the vector $\vec{b}$.

Since the vector $\vec{AP}$ lies on the line and the line is parallel to $\vec{b}$, the vector $\vec{AP}$ must be parallel to $\vec{b}$.

If two vectors are parallel, one is a scalar multiple of the other.

Thus, we can write:

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

where $\lambda$ is a scalar parameter.

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Rearranging the equation to solve for $\vec{r}$, we get the vector equation of the line:

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

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Comparing this equation with the given options, we find that option (A) is correct.

The final answer is (A) $\vec{r} = \vec{a} + \lambda \vec{b}$.

Let the line pass through the point $A(x_1, y_1, z_1)$.

Let $(a, b, c)$ be the direction ratios of the line.

Let $P(x, y, z)$ be an arbitrary point on the line.

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The direction ratios of the line segment joining $A(x_1, y_1, z_1)$ and $P(x, y, z)$ are $(x - x_1, y - y_1, z - z_1)$.

Since the points $A$ and $P$ lie on the same line, the direction ratios of the segment $AP$ must be proportional to the direction ratios of the line itself.

Therefore, the differences in coordinates $(x - x_1, y - y_1, z - z_1)$ are proportional to the direction ratios $(a, b, c)$.

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This proportionality can be written as the Cartesian equation of the line:

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

This form is valid provided $a, b, c$ are not zero. If any of the direction ratios is zero, the corresponding numerator must also be zero.

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Comparing this equation with the given options, we find that option (A) is correct.

The final answer is (A) $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

Let the two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.

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

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The formula for the cosine of the angle $\theta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is a standard result in three-dimensional geometry.

It is given by the dot product of the unit vectors along the directions of the lines, which in terms of direction cosines is:

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

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If we are looking for the acute angle between the lines, we take the absolute value of the expression:

$\cos \theta_{\text{acute}} = |l_1 l_2 + m_1 m_2 + n_1 n_2|$

However, the question asks for $\cos \theta$, which refers to the angle itself, which can be acute or obtuse depending on the line orientations chosen. The fundamental relationship between the angle and direction cosines is given by the first formula.

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Comparing the standard formula with the given options, we find that option (A) matches the formula for $\cos \theta$.

Option (B) gives the cosine of the acute angle.

Option (C) is equivalent to option (A) because for direction cosines, $l_1^2+m_1^2+n_1^2 = 1$ and $l_2^2+m_2^2+n_2^2 = 1$, simplifying the denominator to 1. However, option (A) is the more direct and simplified form when using direction cosines.

Option (D) is incorrect as it does not represent the cosine of the angle between lines.

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The final answer is (A) $l_1 l_2 + m_1 m_2 + n_1 n_2$.

Let the direction ratios of the two lines be $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$.

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Two lines in three-dimensional space are perpendicular if and only if the sum of the products of their corresponding direction ratios is zero.

This condition arises from the formula for the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:

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

For the lines to be perpendicular, the angle between them must be $90^\circ$ ($\frac{\pi}{2}$ radians).

If $\theta = 90^\circ$, then $\cos \theta = \cos 90^\circ = 0$.

Setting the formula for $\cos \theta$ to 0, we get:

$\frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} = 0$

Assuming the lines are not degenerate (i.e., $\sqrt{a_1^2+b_1^2+c_1^2} \neq 0$ and $\sqrt{a_2^2+b_2^2+c_2^2} \neq 0$), the denominator is non-zero.

Therefore, the condition for perpendicularity is that the numerator must be zero:

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

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Comparing this condition with the given options, we find that option (A) is correct.

Option (C) represents the condition for the lines to be parallel.

The final answer is (A) $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Let the two skew lines be $L_1$ and $L_2$, given by the vector equations:

$L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1$

$L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2$

where $\vec{a}_1$ and $\vec{a}_2$ are the position vectors of points on lines $L_1$ and $L_2$ respectively, and $\vec{b}_1$ and $\vec{b}_2$ are vectors parallel to lines $L_1$ and $L_2$ respectively. Since the lines are skew, $\vec{b}_1$ and $\vec{b}_2$ are not parallel, and the lines do not intersect.

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The shortest distance between two skew lines is the length of the common perpendicular to both lines.

The direction vector of the common perpendicular is perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. This direction is given by the cross product $\vec{b}_1 \times \vec{b}_2$.

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Consider a vector joining any point on $L_1$ (say with position vector $\vec{a}_1$) to any point on $L_2$ (say with position vector $\vec{a}_2$). This vector is $\vec{a}_2 - \vec{a}_1$.

The shortest distance between the lines is the magnitude of the projection of the vector $(\vec{a}_2 - \vec{a}_1)$ onto the direction of the common perpendicular, which is $(\vec{b}_1 \times \vec{b}_2)$.

The projection of vector $\vec{u}$ onto vector $\vec{v}$ is given by $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$.

Here, $\vec{u} = \vec{a}_2 - \vec{a}_1$ and $\vec{v} = \vec{b}_1 \times \vec{b}_2$.

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So, the scalar projection of $(\vec{a}_2 - \vec{a}_1)$ onto $(\vec{b}_1 \times \vec{b}_2)$ is $\frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|}$.

Since distance is always non-negative, we take the absolute value of this scalar projection.

Therefore, the shortest distance ($d$) between the two skew lines is:

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

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Comparing this formula with the given options, we find that option (B) is correct.

The term $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ is the scalar triple product $[\vec{a}_2 - \vec{a}_1, \vec{b}_1, \vec{b}_2]$.

The final answer is (B) $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

The Cartesian equation of a plane is given as:

$ax+by+cz+d'=0$

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The standard form of the Cartesian equation of a plane is $Ax + By + Cz + D = 0$.

In this standard form, the coefficients of $x, y, z$ (i.e., $A, B, C$) are the direction ratios of a vector normal to the plane.

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Comparing the given equation $ax+by+cz+d'=0$ with the standard form $Ax+By+Cz+D=0$, we can see that $A=a$, $B=b$, and $C=c$. The constant term is $D=d'$.

Therefore, the direction ratios of the normal to the plane $ax+by+cz+d'=0$ are $(a, b, c)$.

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Comparing this result with the given options, we find that option (A) is correct.

Option (B) includes the constant term $d'$, which is not part of the direction ratios of the normal vector.

Options (C) and (D) do not represent the direction ratios of the normal vector derived directly from the coefficients of the Cartesian equation.

The final answer is (A) $(a, b, c)$.

Let's analyze each statement regarding the angle between two planes.

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(A) The angle between two planes is the angle between their normal vectors.

This is the fundamental definition of the angle between two intersecting planes. The angle between the planes is defined as the angle between the lines perpendicular to them at the point of intersection. These perpendicular lines are parallel to the normal vectors of the planes. Thus, the angle between the planes is equal to the angle between their normal vectors.

This statement is correct.

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(B) If the planes are parallel, the angle between them is $0^\circ$ or $180^\circ$.

If two planes are parallel, their normal vectors are also parallel. The angle between parallel vectors can be $0^\circ$ (if they point in the same direction) or $180^\circ$ (if they point in opposite directions). Since the angle between planes is defined as the angle between their normals, the angle between parallel planes is $0^\circ$ or $180^\circ$. By convention, the angle between parallel planes is usually taken as $0^\circ$. However, stating $0^\circ$ or $180^\circ$ based on the normal vector direction is also correct.

This statement is correct.

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(C) If the planes are perpendicular, the angle between their normals is $90^\circ$.

Two planes are perpendicular if the angle between them is $90^\circ$. Based on the definition in statement (A), the angle between the planes is the angle between their normal vectors. Therefore, if the planes are perpendicular, their normal vectors must be perpendicular, which means the angle between their normals is $90^\circ$.

This statement is correct.

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(D) The angle between planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ is $\cos^{-1} \left| \frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}} \right|$.

The normal vector to the plane $A_1x+B_1y+C_1z+D_1=0$ has direction ratios $(A_1, B_1, C_1)$. Let $\vec{n}_1 = A_1 \hat{i} + B_1 \hat{j} + C_1 \hat{k}$.

The normal vector to the plane $A_2x+B_2y+C_2z+D_2=0$ has direction ratios $(A_2, B_2, C_2)$. Let $\vec{n}_2 = A_2 \hat{i} + B_2 \hat{j} + C_2 \hat{k}$.

The cosine of the angle $\theta$ between the two normal vectors $\vec{n}_1$ and $\vec{n}_2$ is given by the dot product formula:

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

$\cos \theta = \frac{(A_1 \hat{i} + B_1 \hat{j} + C_1 \hat{k}) \cdot (A_2 \hat{i} + B_2 \hat{j} + C_2 \hat{k})}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$

$\cos \theta = \frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$

The angle between the planes is the angle between the normal vectors. To ensure we get the acute angle between the planes (which is typically considered), we take the absolute value of $\cos \theta$.

Let $\phi$ be the angle between the planes. Then $\cos \phi = |\cos \theta|$, so:

$\cos \phi = \left| \frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}} \right|$

Thus, the angle between the planes is $\phi = \cos^{-1} \left| \frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}} \right|$.

This statement is correct.

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Since the question asks to select all that apply and all four statements are correct descriptions or properties related to the angle between two planes, all options (A), (B), (C), and (D) are correct.

The final answer is (A), (B), (C), and (D).

Let's examine the given Assertion (A) and Reason (R).

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Assertion (A): The lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-4}{4} = \frac{y-5}{6} = \frac{z-6}{8}$ are parallel.

The direction ratios of the first line are $(a_1, b_1, c_1) = (2, 3, 4)$.

The direction ratios of the second line are $(a_2, b_2, c_2) = (4, 6, 8)$.

For two lines to be parallel, their direction ratios must be proportional.

Let's check the ratios of corresponding direction ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{4}{8} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, the direction ratios are proportional. Therefore, the lines are parallel.

Assertion (A) is True.

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Reason (R): Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Here $\frac{2}{4} = \frac{3}{6} = \frac{4}{8} = \frac{1}{2}$.

The first part of Reason (R) states the condition for the parallelism of two lines in terms of their direction ratios. This condition is a standard result in three-dimensional geometry and is correct.

The second part of Reason (R) applies this condition to the given lines and correctly calculates the ratios of their direction ratios as $\frac{2}{4} = \frac{3}{6} = \frac{4}{8} = \frac{1}{2}$.

Reason (R) is True.

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Now, we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) makes a specific claim about the parallelism of two given lines.

Reason (R) provides the general principle for checking parallelism using direction ratios and demonstrates that the given lines satisfy this principle, thus supporting the claim made in Assertion (A).

Therefore, Reason (R) is the correct explanation for Assertion (A).

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Based on this analysis, the correct option is (A).

The final answer is (A) Both A and R are true and R is the correct explanation of A.

Let the point be $P(x_1, y_1, z_1)$ and the equation of the plane be $Ax+By+Cz+D=0$.

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The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax+By+Cz+D=0$ is given by a standard formula.

The formula is derived by considering the projection of the vector from a point on the plane to the given point onto the normal vector of the plane.

The magnitude of the normal vector to the plane $Ax+By+Cz+D=0$ is $|\vec{n}| = \sqrt{A^2+B^2+C^2}$.

The numerator involves substituting the coordinates of the point $(x_1, y_1, z_1)$ into the equation of the plane, i.e., $Ax_1+By_1+Cz_1+D$. Since distance must be non-negative, the absolute value of this expression is taken.

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The formula for the distance ($d$) is:

$d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$

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Comparing this formula with the given options, we find that option (C) is correct.

Option (A) is missing the denominator, which normalizes the expression by the magnitude of the normal vector.

Option (B) is missing the absolute value in the numerator, which is required to ensure the distance is non-negative.

Option (D) is the distance formula between two points, which is not applicable here.

The final answer is (C) $\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.

Let the given point be $P(1, 2, -1)$. The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$.

The plane is perpendicular to the normal vector $\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}$. The direction ratios of the normal are $(A, B, C) = (2, 3, -1)$.

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The equation of a plane passing through a point $(x_1, y_1, z_1)$ and having direction ratios of the normal as $(A, B, C)$ is given by the Cartesian form:

$A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$

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Substituting the given values $(x_1, y_1, z_1) = (1, 2, -1)$ and $(A, B, C) = (2, 3, -1)$ into the formula, we get:

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

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

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Comparing this equation with the given options, we see that it matches option (A).

We can also expand option (A) to get the general form of the Cartesian equation:

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

$2x + 3y - z - 9 = 0$

This is the equation of the plane.

Option (D) is the vector equation $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ written in component form, which is also a correct equation for the plane, but option (A) is the Cartesian form derived directly from the standard formula $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. Since option (A) is one of the choices, it is the intended answer in Cartesian form.

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The final answer is (A) $2(x-1) + 3(y-2) - 1(z+1) = 0$.

The equation of a plane passing through three non-collinear points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by $(\vec{r}-\vec{a}) \cdot ((\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})) = 0$.

Here, $\vec{r}$ is the position vector of an arbitrary point $P$ on the plane.

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The vectors involved in the scalar triple product are $\vec{u} = \vec{r}-\vec{a}$, $\vec{v} = \vec{b}-\vec{a}$, and $\vec{w} = \vec{c}-\vec{a}$.

The vector $\vec{r}-\vec{a}$ is the vector $\vec{AP}$ from point $A$ to point $P$.

The vector $\vec{b}-\vec{a}$ is the vector $\vec{AB}$ from point $A$ to point $B$.

The vector $\vec{c}-\vec{a}$ is the vector $\vec{AC}$ from point $A$ to point $C$.

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The expression $(\vec{r}-\vec{a}) \cdot ((\vec{b}-\vec{a}) \times (\vec{c}-\vec{a}))$ is the scalar triple product of the vectors $\vec{r}-\vec{a}$, $\vec{b}-\vec{a}$, and $\vec{c}-\vec{a}$.

The scalar triple product of three vectors is zero if and only if the three vectors are coplanar.

The geometric interpretation of the scalar triple product is the volume of the parallelepiped formed by the three vectors. If this volume is zero, the vectors must lie in the same plane.

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Since the equation of the plane is given by the condition that the scalar triple product of $\vec{r}-\vec{a}$, $\vec{b}-\vec{a}$, and $\vec{c}-\vec{a}$ is zero, it means that these three vectors are coplanar.

The vector $\vec{r}-\vec{a}$ represents the vector from $A$ to any point $P$ on the plane. The vectors $\vec{b}-\vec{a}$ and $\vec{c}-\vec{a}$ lie in the plane containing points $A, B, C$. If $\vec{r}-\vec{a}$ is coplanar with $\vec{b}-\vec{a}$ and $\vec{c}-\vec{a}$, then the point $P$ lies in the plane defined by $A, B, C$.

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Thus, the scalar triple product being zero indicates that the vectors $\vec{r}-\vec{a}$, $\vec{b}-\vec{a}$, and $\vec{c}-\vec{a}$ are coplanar.

Comparing this with the given options, we find that option (C) is correct.

The final answer is (C) Coplanar.

The equation of the plane is given by $2x - 3y + 4z = 12$.

To find the intercepts on the coordinate axes, we set two variables to zero and solve for the third.

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To find the x-intercept, set $y=0$ and $z=0$:

$2x - 3(0) + 4(0) = 12$

$2x = 12$

$x = \frac{12}{2} = 6$

The x-intercept is 6. The point where the plane crosses the x-axis is $(6, 0, 0)$.

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To find the y-intercept, set $x=0$ and $z=0$:

$2(0) - 3y + 4(0) = 12$

$-3y = 12$

$y = \frac{12}{-3} = -4$

The y-intercept is -4. The point where the plane crosses the y-axis is $(0, -4, 0)$.

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To find the z-intercept, set $x=0$ and $y=0$:

$2(0) - 3(0) + 4z = 12$

$4z = 12$

$z = \frac{12}{4} = 3$

The z-intercept is 3. The point where the plane crosses the z-axis is $(0, 0, 3)$.

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The intercepts made by the plane on the x, y, and z axes are 6, -4, and 3, respectively.

The standard way to represent the intercepts is the ordered triplet of these values, which is $(6, -4, 3)$.

Let's examine the options:

(A) $(6, -4, 3)$ - Lists the correct intercepts.

(B) $(12, -12/3, 12/4) = (6, -4, 3)$ - Shows the calculation of intercepts by dividing the constant term (12) by the coefficients of $x, y, z$ (2, -3, 4) respectively, which is a correct method, and also shows the resulting values $(6, -4, 3)$.

(C) $(6, -4, 3)$ (Intercepts are $x, y, z$ values when other variables are zero) - States the correct intercepts and provides the definition of intercepts.

(D) All of the above are related to finding the intercepts. - Since (A), (B), and (C) all present or explain how to find the correct intercepts, this statement is also true.

However, in multiple choice questions, we look for the option that most directly and completely answers the question or provides the best representation of the answer. Options (A), (B), and (C) all identify the correct intercepts. Option (B) specifically shows the division $D/A, D/B, D/C$ which leads to the intercepts from the equation $Ax+By+Cz=D$. This method is directly applicable here ($A=2, B=-3, C=4, D=12$). Option (C) provides the values and the definition, which is also quite complete.

Among the options presenting the values, (B) gives the method used to find them from the equation form $Ax+By+Cz=D$, making it a strong candidate. Option (A) just lists the numbers. Option (C) lists the numbers and the definition. Option (B) shows the calculation. Let's consider option (B) as it explicitly shows the calculation leading to the intercepts from the given equation form.

The final answer is (B) $(12, -12/3, 12/4) = (6, -4, 3)$.

Let the two given points be $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.

Let $P(x, y, z)$ be an arbitrary point on the line passing through $A$ and $B$.

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The vector $\vec{AP}$ from point $A$ to point $P$ is given by:

$\vec{AP} = (x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}$

The vector $\vec{AB}$ from point $A$ to point $B$ is given by:

$\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

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Since points $A, B,$ and $P$ are collinear (they lie on the same line), the vector $\vec{AP}$ must be parallel to the vector $\vec{AB}$.

If two vectors are parallel, one is a scalar multiple of the other. Thus, $\vec{AP} = \lambda \vec{AB}$ for some scalar $\lambda$.

$(x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k} = \lambda [(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}]$

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Equating the corresponding components, we get:

$x - x_1 = \lambda (x_2 - x_1)$

$y - y_1 = \lambda (y_2 - y_1)$

$z - z_1 = \lambda (z_2 - z_1)$

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Assuming $x_2 - x_1 \neq 0$, $y_2 - y_1 \neq 0$, and $z_2 - z_1 \neq 0$, we can express $\lambda$ from each equation:

$\frac{x - x_1}{x_2 - x_1} = \lambda$

$\frac{y - y_1}{y_2 - y_1} = \lambda$

$\frac{z - z_1}{z_2 - z_1} = \lambda$

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Since all expressions are equal to the same parameter $\lambda$, we can write the Cartesian equation of the line as:

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

This is the standard Cartesian form of a line passing through two points.

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Comparing this equation with the given options, we find that option (B) matches this form.

Note that option (C) is also a valid form using $(x_2, y_2, z_2)$ as the reference point: $\frac{x-x_2}{x_1-x_2} = \frac{y-y_2}{y_1-y_2} = \frac{z-z_2}{z_1-z_2}$, which is equivalent to option (C) as $\frac{x-x_2}{-(x_2-x_1)} = \frac{y-y_2}{-(y_2-y_1)} = \frac{z-z_2}{-(z_2-z_1)}$. Option (D) is also an equivalent rearrangement. However, option (B) is the most commonly presented standard form using the first point $(x_1, y_1, z_1)$ as the base.

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The final answer is (B) $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.

The vector equations of the two lines are given by:

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

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

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The direction vector of Line 1 is $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$. Its components (direction ratios) are $(a_1, b_1, c_1) = (2, -1, 1)$.

The direction vector of Line 2 is $\vec{b}_2 = \hat{i} + \hat{j} + 2\hat{k}$. Its components (direction ratios) are $(a_2, b_2, c_2) = (1, 1, 2)$.

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

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

Alternatively, using direction ratios:

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

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Substituting the components of the direction vectors into the formula:

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

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

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

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Now, calculate $\cos \theta$:

$\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}

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The angle $\theta$ is then found by taking the inverse cosine:

$\theta = \cos^{-1}(\frac{1}{2})$

$\theta = 60^\circ$ or $\frac{\pi}{3}$ radians

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Let's examine the given options:

(A) $\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}$ - This is the correct formula for $\cos \theta$ with the numbers substituted.

(B) $\frac{2-1+2}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ - This shows the calculation results for the dot product and magnitudes, and the final value of $\cos \theta$.

(C) $\cos^{-1}(\frac{1}{2}) = 60^\circ$ or $\pi/3$ radians - This is the value of the angle $\theta$ itself, derived from $\cos \theta = 1/2$.

(D) All of the above steps/values lead to the angle. - Options (A), (B), and (C) are indeed consecutive steps or results in finding the angle between the lines.

Since the question asks for $\cos \theta = \dots$, options (A) and (B) provide expressions for or the value of $\cos \theta$. Option (C) gives the angle $\theta$. Option (D) states that (A), (B), and (C) are all related to finding the angle. Given the options, the most appropriate answer is that all options are relevant steps or results in the process.

The final answer is (D) All of the above steps/values lead to the angle.

We are given two parallel lines with vector equations:

Line 1: $\vec{r} = \vec{a}_1 + \lambda \vec{b}$

Line 2: $\vec{r} = \vec{a}_2 + \mu \vec{b}$

Here, $\vec{a}_1$ is the position vector of a point on Line 1, $\vec{a}_2$ is the position vector of a point on Line 2, and $\vec{b}$ is the common direction vector for both lines (since they are parallel).

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The shortest distance between two parallel lines is the perpendicular distance from any point on one line to the other line.

Consider a point $A_1$ on Line 1 with position vector $\vec{a}_1$, and a point $A_2$ on Line 2 with position vector $\vec{a}_2$.

The vector connecting these two points is $\vec{A_1A_2} = \vec{a}_2 - \vec{a}_1$.

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The shortest distance is the magnitude of the projection of the vector $\vec{A_1A_2}$ onto a vector that is perpendicular to the direction vector $\vec{b}$.

Alternatively, consider the parallelogram formed by the vectors $(\vec{a}_2 - \vec{a}_1)$ and $\vec{b}$. The area of this parallelogram is given by the magnitude of their cross product:

Area $= |(\vec{a}_2 - \vec{a}_1) \times \vec{b}|$.

The area of a parallelogram is also equal to the product of its base and height. We can take the base to be the magnitude of the direction vector $\vec{b}$, i.e., $|\vec{b}|$. The corresponding height is the shortest distance ($d$) between the two parallel lines.

Area $= |\vec{b}| \cdot d$

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Equating the two expressions for the area:

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

Solving for the shortest distance $d$:

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

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Comparing this formula with the given options, we find that option (B) matches the derived formula.

The final answer is (B) $\frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.

Let the equations of the two planes be:

$P_1: \vec{r} \cdot \vec{n}_1 = d_1$

$P_2: \vec{r} \cdot \vec{n}_2 = d_2$

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These equations can be rewritten in the form $P=0$ as:

Plane 1: $\vec{r} \cdot \vec{n}_1 - d_1 = 0$

Plane 2: $\vec{r} \cdot \vec{n}_2 - d_2 = 0$

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The equation of a plane passing through the intersection of two planes $P_1=0$ and $P_2=0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is an arbitrary scalar constant.

Using this principle, the equation of the plane passing through the intersection of the given planes is:

$(\vec{r} \cdot \vec{n}_1 - d_1) + \lambda (\vec{r} \cdot \vec{n}_2 - d_2) = 0$

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Let's examine the given options:

(C) $(\vec{r} \cdot \vec{n}_1 - d_1) + \lambda (\vec{r} \cdot \vec{n}_2 - d_2) = 0$

This is the standard form $P_1 + \lambda P_2 = 0$, where $P_1 = \vec{r} \cdot \vec{n}_1 - d_1$ and $P_2 = \vec{r} \cdot \vec{n}_2 - d_2$. This equation represents the family of planes passing through the intersection of the two planes (except possibly the plane $P_2$ itself, which corresponds to $\lambda \to \infty$, or can be obtained by writing the equation as $\mu P_1 + P_2 = 0$). This form is correct.

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(A) $\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$

Let's rearrange option (C):

$\vec{r} \cdot \vec{n}_1 - d_1 + \lambda (\vec{r} \cdot \vec{n}_2) - \lambda d_2 = 0$

$\vec{r} \cdot \vec{n}_1 + \lambda (\vec{r} \cdot \vec{n}_2) = d_1 + \lambda d_2$

Using the property of dot product $\vec{r} \cdot \vec{n}_1 + \lambda (\vec{r} \cdot \vec{n}_2) = \vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2)$, we get:

$\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$

This matches option (A). So, option (A) is an equivalent form of option (C).

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(B) $\vec{r} \cdot (\vec{n}_1 - \lambda \vec{n}_2) = d_1 - \lambda d_2$

This equation is of the form $\vec{r} \cdot (\vec{n}_1 + k \vec{n}_2) = d_1 + k d_2$ where $k = -\lambda$. Since $\lambda$ can be any real number, $k = -\lambda$ can also be any real number. Therefore, this equation represents the same family of planes as (A) and (C), just with a different parameterization (letting the parameter be $-\lambda$ instead of $\lambda$).

So, option (B) is also an equivalent form.

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(D) All of the above are equivalent forms.

Since options (A), (B), and (C) are all equivalent representations of the equation of a plane passing through the intersection of the given planes (differing only in algebraic manipulation and possibly the sign of the parameter), statement (D) is correct.

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The final answer is (D) All of the above are equivalent forms.

Let's analyze each statement to determine which condition(s) imply that two planes are parallel.

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(A) Their normal vectors are parallel.

A normal vector to a plane is perpendicular to every line and vector lying in or parallel to the plane. If two planes are parallel, they have the same orientation in space. This means that a line perpendicular to one plane is also perpendicular to the other. Since the normal vectors are along these perpendicular lines, the normal vectors of two parallel planes must be parallel.

Conversely, if the normal vectors of two planes are parallel, then both planes are perpendicular to the same direction. This implies the planes are parallel to each other (they either coincide or are distinct but parallel).

This statement is correct.

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(B) Their equations are $A_1x+B_1y+C_1z=D_1$ and $A_2x+B_2y+C_2z=D_2$ such that $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.

The normal vector to the plane $A_1x+B_1y+C_1z=D_1$ is $\vec{n}_1 = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$.

The normal vector to the plane $A_2x+B_2y+C_2z=D_2$ is $\vec{n}_2 = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$.

Two vectors $\vec{n}_1$ and $\vec{n}_2$ are parallel if their corresponding components are proportional, i.e., if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.

From statement (A), if the normal vectors are parallel, the planes are parallel. Therefore, this condition on the coefficients implies the planes are parallel.

This statement is correct.

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(C) The angle between them is $90^\circ$.

The angle between two planes is defined as the acute angle between their normal vectors. If the angle between the planes is $90^\circ$, this means the angle between their normal vectors is $90^\circ$, indicating the normal vectors are perpendicular. If the normal vectors are perpendicular, the planes are perpendicular, not parallel.

This statement is incorrect.

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(D) A vector parallel to one plane is also parallel to the other.

If two planes are parallel, any vector lying in or parallel to one plane must also lie in or be parallel to the other plane. This is because parallel planes share the same set of directions that are parallel to them.

Conversely, if any vector parallel to one plane is also parallel to the other, it means that the direction space of both planes is the same. This can only happen if the planes are parallel.

This statement is correct.

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Based on the analysis, statements (A), (B), and (D) are correct conditions that imply two planes are parallel.

The final answer is (A), (B), and (D).

Let's analyze the given Assertion (A) and Reason (R).

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Assertion (A): The lines $\frac{x-1}{0} = \frac{y-2}{1} = \frac{z-3}{0}$ and $\frac{x-1}{1} = \frac{y-2}{0} = \frac{z-3}{0}$ are perpendicular.

For the first line, the equation is $\frac{x-1}{0} = \frac{y-2}{1} = \frac{z-3}{0}$. The direction ratios of this line are $(a_1, b_1, c_1) = (0, 1, 0)$. This implies $x-1=0$ and $z-3=0$, so $x=1$ and $z=3$. The line is parallel to the y-axis and passes through the point $(1, 2, 3)$.

For the second line, the equation is $\frac{x-1}{1} = \frac{y-2}{0} = \frac{z-3}{0}$. The direction ratios of this line are $(a_2, b_2, c_2) = (1, 0, 0)$. This implies $y-2=0$ and $z-3=0$, so $y=2$ and $z=3$. The line is parallel to the x-axis and passes through the point $(1, 2, 3)$.

The first line is parallel to the y-axis (direction $(0, 1, 0)$) and the second line is parallel to the x-axis (direction $(1, 0, 0)$). The x-axis and y-axis are perpendicular. Therefore, lines parallel to them are perpendicular.

Alternatively, we can use the condition for perpendicularity based on direction ratios.

Assertion (A) is True.

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Reason (R): Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Here $\frac{2}{4} = \frac{3}{6} = \frac{4}{8} = \frac{1}{2}$.

The first part of the reason states the condition for parallel lines, not perpendicular lines. The condition for perpendicular lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

The second part of the reason, "Here $\frac{2}{4} = \frac{3}{6} = \frac{4}{8} = \frac{1}{2}$", appears to be copied from a previous question (specifically Question 12) discussing parallel lines, and is irrelevant to the lines mentioned in Assertion (A) of this question.

Reason (R) makes an incorrect statement about the condition for perpendicularity (it gives the condition for parallelism) and uses irrelevant values in its application.

Reason (R) is False.

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Since Assertion (A) is true and Reason (R) is false, we look for the option that matches this combination.

Comparing with the given options:

(A) Both A and R are true and R is the correct explanation of A. (Incorrect, R is false)

(B) Both A and R are true but R is not the correct explanation of A. (Incorrect, R is false)

(C) A is true but R is false. (Correct)

(D) A is false but R is true. (Incorrect, A is true and R is false)

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The final answer is (C) A is true but R is false.

Let's match the geometric entities with their corresponding representations or properties:

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(i) Line in space

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is typically given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{r}$ is the position vector of an arbitrary point on the line and $\lambda$ is a scalar parameter.

This matches option (b).

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(ii) Plane in space

The vector equation of a plane in normal form is typically given by $\vec{r} \cdot \vec{n} = d$, where $\vec{r}$ is the position vector of an arbitrary point on the plane, $\vec{n}$ is a normal vector to the plane, and $d$ is a scalar constant related to the distance of the plane from the origin.

This matches option (a).

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(iii) Direction cosines

For any line in three-dimensional space, the sum of the squares of its direction cosines $(l, m, n)$ is always equal to 1, i.e., $l^2 + m^2 + n^2 = 1$.

This matches option (c).

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(iv) Shortest distance between skew lines

The formula for the shortest distance 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 $\frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$. The expression in the numerator, $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$, is a scalar triple product of the vectors $\vec{a}_2 - \vec{a}_1$, $\vec{b}_1$, and $\vec{b}_2$.

This matches option (d).

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Summary of matches:

(i) - (b)

(ii) - (a)

(iii) - (c)

(iv) - (d)

Let the line be parallel to the x-axis.

The direction of the positive x-axis is given by the unit vector $\hat{i}$.

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

A line parallel to the x-axis makes the following angles with the coordinate axes:

Angle with the positive x-axis ($\alpha$) = $0^\circ$

Angle with the positive y-axis ($\beta$) = $90^\circ$

Angle with the positive z-axis ($\gamma$) = $90^\circ$

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The direction cosines are $(\cos \alpha, \cos \beta, \cos \gamma)$.

Calculating the cosines of these angles:

$\cos \alpha = \cos 0^\circ = 1$

$\cos \beta = \cos 90^\circ = 0$

$\cos \gamma = \cos 90^\circ = 0$

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Thus, the direction cosines of a line parallel to the x-axis are $(1, 0, 0)$.

Note that $(-1, 0, 0)$ are also direction cosines for a line parallel to the x-axis (pointing in the negative direction).

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Comparing this with the given options, option (A) represents the direction cosines of the positive x-axis, which is a line parallel to the x-axis.

The final answer is (A) $(1, 0, 0)$.

The general vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by:

$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$

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Expanding this equation, we get:

$\vec{r} \cdot \vec{n} - \vec{a} \cdot \vec{n} = 0$

or

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

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In this question, the plane passes through the origin $(0, 0, 0)$.

The position vector of the origin is $\vec{a} = \vec{0}$.

The normal vector to the plane is given as $\vec{n}$.

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Substituting $\vec{a} = \vec{0}$ into the equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, we get:

$\vec{r} \cdot \vec{n} = \vec{0} \cdot \vec{n}

Since the dot product of any vector with the zero vector is zero, $\vec{0} \cdot \vec{n} = 0$.

So, the equation of the plane becomes:

$\vec{r} \cdot \vec{n} = 0$

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This equation represents the set of all points $\vec{r}$ such that the vector $\vec{r}$ is perpendicular to the normal vector $\vec{n}$. Since the plane passes through the origin, any vector $\vec{r}$ from the origin to a point on the plane lies entirely within the plane. If $\vec{r}$ is perpendicular to the normal $\vec{n}$, the point lies on the plane. This confirms the equation is correct.

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Comparing our derived equation with the given options, we find that option (A) matches.

The final answer is (A) $\vec{r} \cdot \vec{n} = 0$.

The vector equations of the two lines are given by:

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

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

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The direction vector of Line 1 is $\vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$. Its components (direction ratios) are $(a_1, b_1, c_1) = (2, -1, 1)$.

The direction vector of Line 2 is $\vec{b}_2 = \hat{i} + \hat{j} + 2\hat{k}$. Its components (direction ratios) are $(a_2, b_2, c_2) = (1, 1, 2)$.

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

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

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Let's calculate the dot product and magnitudes:

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

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

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

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Now, calculate $\cos \theta$:

$\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}

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The angle $\theta$ is then found by taking the inverse cosine:

$\theta = \cos^{-1}(\frac{1}{2})$

$\theta = 60^\circ$ or $\frac{\pi}{3}$ radians

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Let's examine the given options in the question text:

(A) $\cos^{-1}(\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}})$ - This shows the structure of the formula for $\cos \theta$ applied to the given vectors, and then takes the inverse cosine to find the angle.

(B) $\cos^{-1}(\frac{2-1+2}{\sqrt{6}\sqrt{6}}) = \cos^{-1}(\frac{3}{6}) = \cos^{-1}(\frac{1}{2})$ - This shows the evaluation of the dot product and magnitudes in the numerator and denominator, leading to the value of $\cos \theta = 1/2$, and then takes the inverse cosine to find the angle. This matches our calculation steps and result.

(C) $\cos^{-1}(\frac{1}{2}) = 60^\circ$ or $\pi/3$ radians - This states the final value of the angle in degrees and radians, based on the calculated value of $\cos \theta = 1/2$. This also matches our result.

(D) All of the above steps/values lead to the angle. - Since options (A), (B), and (C) all represent valid steps in the calculation of the angle or the final value of the angle, this statement is correct.

Although the question asks for "$\cos \theta = \dots$", and options (A), (B), (C) are expressions for $\theta$ (using $\cos^{-1}$), the calculation within options (A) and (B) correctly shows the value of $\cos \theta = 1/2$. Option (B) provides the direct value of $\cos \theta = 1/2$ inside the $\cos^{-1}$. Option (D) confirms that all these relate to finding the angle.

Considering the options provided, option (D) is the most appropriate answer, as it acknowledges that (A), (B), and (C) are all relevant to finding the angle.

The final answer is (D) All of the above steps/values lead to the angle.

Let the given point be $(x_1, y_1, z_1) = (1, -1, 2)$.

Let the equation of the plane be $Ax+By+Cz+D=0$, which is $3x+4y-12z+13=0$.

Here, $A=3$, $B=4$, $C=-12$, and $D=13$.

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The distance $d$ of the point $(x_1, y_1, z_1)$ from the plane $Ax+By+Cz+D=0$ is given by the formula:

$d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$

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Substituting the given values into the formula:

$d = \frac{|3(1) + 4(-1) + (-12)(2) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}}$

This expression matches option (A).

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Now, let's evaluate the expression:

Numerator: $|3(1) + 4(-1) + (-12)(2) + 13| = |3 - 4 - 24 + 13| = |-1 - 24 + 13| = |-25 + 13| = |-12| = 12$

Denominator: $\sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{25 + 144} = \sqrt{169} = 13$

So, $d = \frac{12}{13}$.

The calculation steps and the result $\frac{12}{13}$ match option (B).

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The distance is $\frac{12}{13}$ units.

This value with units matches option (C).

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Since options (A), (B), and (C) are all valid steps or results in the calculation of the distance, option (D) which states that all of the above are related to the calculation is correct.

The final answer is (D) All of the above are steps/values to calculate the distance.

Let the position vector of point A be $\vec{a} = 10\hat{i} + 20\hat{j} + 50\hat{k}$.

Let the position vector of point B be $\vec{b} = 50\hat{i} + 60\hat{j} + 80\hat{k}$.

The vector representing the direction of the line from A to B is $\vec{b} - \vec{a}$.

$\vec{b} - \vec{a} = (50-10)\hat{i} + (60-20)\hat{j} + (80-50)\hat{k} = 40\hat{i} + 40\hat{j} + 30\hat{k}$

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The vector equation of the line passing through A and B is $\vec{r} = \vec{a} + \lambda (\vec{b}-\vec{a})$, where $\lambda$ is a scalar parameter.

$\vec{r} = (10\hat{i} + 20\hat{j} + 50\hat{k}) + \lambda (40\hat{i} + 40\hat{j} + 30\hat{k})$

$\vec{r} = (10 + 40\lambda)\hat{i} + (20 + 40\lambda)\hat{j} + (50 + 30\lambda)\hat{k}$

The Cartesian coordinates of a point on the line are $(x, y, z)$, where $x = 10 + 40\lambda$, $y = 20 + 40\lambda$, $z = 50 + 30\lambda$.

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The equation of the restricted airspace plane is $x+y+z=150$.

To find if the line intersects the plane, we substitute the expressions for $x, y, z$ from the line equation into the plane equation:

$(10 + 40\lambda) + (20 + 40\lambda) + (50 + 30\lambda) = 150$

Combine the terms:

$(40\lambda + 40\lambda + 30\lambda) + (10 + 20 + 50) = 150$

$110\lambda + 80 = 150$

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Solve for $\lambda$:

$110\lambda = 150 - 80$

$110\lambda = 70$

$\lambda = \frac{70}{110} = \frac{7}{11}$

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The value of $\lambda = \frac{7}{11}$ is a real number, which means the line (extending infinitely) intersects the plane. The drone's path is the line segment from point A to point B.

Point A corresponds to $\lambda = 0$ (since $\vec{a} + 0(\vec{b}-\vec{a}) = \vec{a}$).

Point B corresponds to $\lambda = 1$ (since $\vec{a} + 1(\vec{b}-\vec{a}) = \vec{b}$).

The line segment from A to B corresponds to the range $0 \leq \lambda \leq 1$.

The calculated value $\lambda = \frac{7}{11}$ lies within this range ($0 \leq \frac{7}{11} \leq 1$).

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Since the intersection point corresponds to a value of $\lambda$ between 0 and 1, the drone's path (the line segment AB) intersects the restricted airspace plane.

Therefore, the answer is Yes.

Comparing with the options, option (A) correctly states that a value of $\lambda$ exists for intersection, implying the line intersects the plane, and based on the context of the problem involving the path from A to B, it implies the segment intersects.

The final answer is (A) Yes (A value of $\lambda$ exists for intersection).

Let the equations of the two parallel planes be $P_1: Ax+By+Cz=D_1$ and $P_2: Ax+By+Cz=D_2$.

The planes are parallel because their normal vectors are the same (or proportional). The direction ratios of the normal vector are $(A, B, C)$.

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The distance between two parallel planes can be found by taking any point on one plane and calculating its perpendicular distance to the other plane.

Let's rewrite the second plane's equation in the form $Ax+By+Cz+D=0$: $Ax+By+Cz-D_2=0$.

The distance of a point $(x_0, y_0, z_0)$ from the plane $Ax+By+Cz+D=0$ is given by $\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$.

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Consider a point on the first plane $Ax+By+Cz=D_1$. For simplicity, let's consider the points where the plane intersects the coordinate axes (assuming $A, B, C \neq 0$). The point on the x-axis is $(\frac{D_1}{A}, 0, 0)$.

Now, we find the distance of this point $(\frac{D_1}{A}, 0, 0)$ from the second plane $Ax+By+Cz-D_2=0$.

Distance $d = \frac{|A(\frac{D_1}{A}) + B(0) + C(0) - D_2|}{\sqrt{A^2+B^2+C^2}}$

$d = \frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}}$

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This formula holds true regardless of the point chosen on the first plane and is the standard formula for the distance between two parallel planes $Ax+By+Cz=D_1$ and $Ax+By+Cz=D_2$. The absolute value in the numerator ensures that the distance is non-negative.

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Comparing our derived formula with the given options, we find that option (B) matches.

The final answer is (B) $\frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}}$.

Let's consider the possible relationships between a line and a plane in three-dimensional space.

There are three fundamental possibilities:

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1. The line and the plane are parallel. In this case, they do not intersect at any point.

2. The line intersects the plane at exactly one point. This occurs when the line is not parallel to the plane.

3. The line lies entirely within the plane. In this case, every point on the line is also a point on the plane, meaning they intersect at infinitely many points.

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The given statement lists two of these possibilities: "parallel" and "intersecting at a single point".

The missing third possibility is that the line lies entirely within the plane.

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Let's check the options:

(A) Be perpendicular to: Perpendicularity describes the angle of intersection, not the number of intersection points. A line perpendicular to a plane intersects it at a single point (unless the line is contained within the plane, which is covered by option B).

(B) Lie entirely within: This is the third fundamental relationship where the line has infinite points in common with the plane.

(C) Be tangent to: Tangency is a concept usually associated with curves and surfaces. While a line lying in a plane could be considered "tangent" in a loose sense, it's not one of the standard distinct classifications of the relationship between a line and a plane in this context.

(D) Intersect at two points: A straight line can intersect a flat plane at most at one point, unless the entire line is contained within the plane (in which case it intersects at infinitely many points). It cannot intersect at exactly two points.

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Therefore, the statement is correctly completed by the phrase "Lie entirely within".

The final answer is (B) Lie entirely within.

A plane that cuts off intercepts $a, b, c$ on the x, y, and z axes respectively passes through the points $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$.

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The general equation of a plane is $Ax + By + Cz + D = 0$.

If the plane passes through $(a, 0, 0)$, substituting these coordinates into the equation gives:

$A(a) + B(0) + C(0) + D = 0$

$Aa + D = 0 \implies A = -\frac{D}{a}$

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If the plane passes through $(0, b, 0)$, substituting these coordinates gives:

$A(0) + B(b) + C(0) + D = 0$

$Bb + D = 0 \implies B = -\frac{D}{b}$

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If the plane passes through $(0, 0, c)$, substituting these coordinates gives:

$A(0) + B(0) + C(c) + D = 0$

$Cc + D = 0 \implies C = -\frac{D}{c}$

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Substitute these values of $A, B, C$ back into the general equation of the plane:

$(-\frac{D}{a})x + (-\frac{D}{b})y + (-\frac{D}{c})z + D = 0

Assuming $D \neq 0$ (otherwise the plane passes through the origin and intercepts are all zero or undefined), we can divide the entire equation by $-D$:

$\frac{1}{a}x + \frac{1}{b}y + \frac{1}{c}z - 1 = 0$

Rearranging the terms, we get the intercept form of the plane equation:

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

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This is the standard equation of a plane that cuts off intercepts $a, b, c$ on the x, y, and z axes respectively.

Comparing this derived equation with the given options, we find that option (B) matches.

The final answer is (B) $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.

The given equation of the plane is $3x-4y+12z=5$.

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The direction ratios of the normal vector to a plane with the Cartesian equation $Ax+By+Cz+D=0$ are $(A, B, C)$.

For the given plane $3x-4y+12z-5=0$, the direction ratios of the normal are $(3, -4, 12)$.

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A line perpendicular to the plane is parallel to the normal vector of the plane.

Therefore, the direction ratios of a line perpendicular to the plane are proportional to the direction ratios of the normal vector.

The direction ratios of the line are proportional to $(3, -4, 12)$.

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Let the direction ratios be $a=3$, $b=-4$, $c=12$.

The magnitude of the vector with these direction ratios is $\sqrt{a^2+b^2+c^2} = \sqrt{3^2+(-4)^2+12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$.

The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}$, $\frac{b}{\sqrt{a^2+b^2+c^2}}$, $\frac{c}{\sqrt{a^2+b^2+c^2}}$.

So, the direction cosines are $(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13})$.

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Now let's check the options:

(A) $(3, -4, 12)$: These are the direction ratios of the normal. Direction cosines are proportional to direction ratios. So, $(3, -4, 12)$ is proportional to $(\frac{3}{13}, -\frac{4}{13}, \frac{12}{13})$ (the proportionality constant is 13). This statement is correct.

(B) $(-3, 4, -12)$: These are also direction ratios proportional to $(3, -4, 12)$ (the proportionality constant is -1). So, $(-3, 4, -12)$ is also proportional to the direction cosines $(\frac{3}{13}, -\frac{4}{13}, \frac{12}{13})$. This statement is correct.

(C) $(3/13, -4/13, 12/13)$: These are the direction cosines themselves. Direction cosines are always proportional to themselves (with proportionality constant 1). This statement is correct.

(D) All of the above are proportional to the direction cosines.: Since (A), (B), and (C) are all proportional to the direction cosines, this statement is correct.

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The question asks what the direction cosines are proportional to. All the given options (A), (B), and (C) provide sets of numbers that are proportional to the direction cosines. Therefore, the most appropriate answer is (D).

The final answer is (D) All of the above are proportional to the direction cosines.

The equation of the given plane is $x-y+z=5$.

The general form of a linear equation in three variables representing a plane is $Ax+By+Cz+D=0$. In this case, the equation can be written as $x-y+z-5=0$.

The coefficients of $x, y, z$ in the Cartesian equation of a plane represent the direction ratios of the normal vector to the plane. For the plane $x-y+z=5$, the direction ratios of the normal vector are $(1, -1, 1)$.

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Two planes are parallel if and only if their normal vectors are parallel.

This means that the normal vector of the new plane must be parallel to the normal vector of the given plane $x-y+z=5$.

So, the normal vector of the new plane can also be taken as proportional to $(1, -1, 1)$. For simplicity, we can use the same direction ratios $(1, -1, 1)$ for the normal of the new plane.

The equation of a plane with normal direction ratios $(A, B, C) = (1, -1, 1)$ has the form $1x - 1y + 1z = k$, or $x-y+z = k$, where $k$ is a constant.

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The plane passes through the point $(1, 2, 3)$.

To find the value of $k$, we substitute the coordinates of this point $(x=1, y=2, z=3)$ into the equation of the plane $x-y+z = k$:

$1 - 2 + 3 = k$

$-1 + 3 = k$

$2 = k$

So, the value of the constant $k$ is 2.

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The equation of the plane passing through $(1, 2, 3)$ and parallel to $x-y+z=5$ is:

$x-y+z = 2$

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Let's look at the given options:

(A) $x-y+z = 5$: This is the equation of the original plane, not a parallel plane passing through $(1, 2, 3)$. The point $(1, 2, 3)$ does not satisfy this equation ($1-2+3 = 2 \neq 5$).

(B) $x-y+z = 1-2+3 = 2$: This correctly substitutes the point $(1, 2, 3)$ into the left side of the equation $x-y+z = k$ to find $k=2$, and presents the resulting equation $x-y+z=2$.

(C) $x-y+z - 2 = 0$: This is a rearrangement of the equation $x-y+z=2$ (by subtracting 2 from both sides). This is an equivalent form of the equation of the plane.

(D) Both (B) and (C) are equivalent: Since option (C) is just the rearrangement of option (B), they represent the same plane. This statement is correct.

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The question asks for "The equation of the plane". Options (B) and (C) both provide the correct equation, just in slightly different forms. Option (D) explicitly states that (B) and (C) are equivalent, which is true. In a multiple-choice question where multiple options are correct forms, and one option states that these forms are equivalent, option (D) is typically the most comprehensive and intended answer.

The final answer is (D) Both (B) and (C) are equivalent.

Two lines in three-dimensional space are defined as skew lines if they satisfy two conditions simultaneously:

1. They are not parallel.

2. They do not intersect.

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When two lines are not parallel and do not intersect, they cannot lie in the same plane. Lines that lie in the same plane are called coplanar lines, and coplanar lines are either parallel or intersecting.

Skew lines are therefore non-coplanar lines that are not parallel.

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Let's evaluate the given options based on this definition:

(A) They are parallel and distinct: This describes two parallel lines that do not coincide. Parallel lines are coplanar, so they are not skew lines.

(B) They intersect at a single point: Intersecting lines are coplanar, so they are not skew lines.

(C) They are not parallel and do not intersect: This is the definition of skew lines.

(D) They lie in the same plane: Lines that lie in the same plane are coplanar. As discussed, skew lines are non-coplanar.

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Thus, the necessary condition for two lines in space to be skew lines is that they are not parallel and do not intersect.

The final answer is (C) They are not parallel and do not intersect.

Let the equation of the line be $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{b}$ is the direction vector of the line.

Let the equation of the plane be $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector to the plane.

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The angle $\phi$ between the line and the plane is defined as the acute angle between the line and its projection onto the plane.

The angle $\theta$ between the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ is given by $\cos \theta = \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}$. The acute angle between the vectors is found using $|\cos \theta| = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.

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Consider the relationship between the line, the plane, and the normal vector. The line makes an angle $\phi$ with the plane and an angle $\theta$ with the normal vector $\vec{n}$. The normal vector $\vec{n}$ is perpendicular to the plane.

If the line is perpendicular to the plane, then $\phi = 90^\circ$. In this case, the line is parallel to the normal vector, so $\theta = 0^\circ$ or $180^\circ$. The relationship would not be $\phi = 90^\circ - \theta$.

If the line is parallel to the plane, then $\phi = 0^\circ$. In this case, the line's direction vector $\vec{b}$ is perpendicular to the normal vector $\vec{n}$, so $\theta = 90^\circ$. Here, $0^\circ = 90^\circ - 90^\circ$, so the relationship $\phi = 90^\circ - \theta$ holds.

In general, the angle $\phi$ between the line and the plane and the angle $\theta$ between the line and the normal to the plane are complementary (assuming $\phi$ and $\theta$ are acute angles that sum to $90^\circ$ or are such that one is $90^\circ$ minus the other).

The relationship is $\phi = 90^\circ - \theta$.

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From this complementary relationship, we can also express a trigonometric relationship:

$\sin \phi = \sin (90^\circ - \theta)

Using the trigonometric identity $\sin(90^\circ - \theta) = \cos \theta$, we get:

$\sin \phi = \cos \theta$

Since $\phi$ is typically taken as the acute angle between the line and the plane ($0^\circ \leq \phi \leq 90^\circ$), $\sin \phi \geq 0$. The angle $\theta$ between the direction vector $\vec{b}$ and the normal vector $\vec{n}$ can range from $0^\circ$ to $180^\circ$. To ensure $\sin \phi$ is non-negative, we take the absolute value of $\cos \theta$:

$\sin \phi = |\cos \theta|$

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Let's check the given options:

(A) $\phi = \theta$: This is generally incorrect.

(B) $\phi = 90^\circ - \theta$: This is a correct relationship between the angles.

(C) $\sin \phi = |\cos \theta|$: This is also a correct relationship between the angles, derived from (B).

(D) Both (B) and (C) are related to the angle calculation: Since both (B) and (C) provide valid and related formulas for the angle calculation, this statement is correct.

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In the context of multiple-choice questions where multiple related correct statements or steps are given, the option indicating that all relevant statements are correct is often the intended answer.

The final answer is (D) Both (B) and (C) are related to the angle calculation.

Let's examine the given Assertion (A) and Reason (R).

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Assertion (A): The equation $ax+by+cz+d=0$ represents a plane in space.

The general form of a linear equation in three variables $x, y, z$ is $Ax+By+Cz+D=0$. Provided that at least one of the coefficients $A, B, C$ is non-zero, this equation always represents a plane in three-dimensional space.

Here, the equation is $ax+by+cz+d=0$. Assuming $a, b,$ and $c$ are not all zero, this equation is indeed the Cartesian equation of a plane.

Assertion (A) is True (under the standard assumption that $a, b, c$ are not all zero).

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Reason (R): This is the general linear equation in three variables $x, y, z$, which defines a plane.

The equation $ax+by+cz+d=0$ is indeed the general form of a linear equation in three variables $x, y, z$. A fundamental result in three-dimensional geometry is that any linear equation in $x, y, z$ (where the coefficients of $x, y, z$ are not all zero) represents a plane, and conversely, every plane can be represented by such a linear equation.

Reason (R) is True.

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Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the given equation represents a plane. Reason (R) provides the justification for this statement by defining the equation as the general linear equation in three variables and stating the geometric fact that such equations define planes.

Thus, Reason (R) correctly explains why Assertion (A) is true.

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Based on this analysis, the correct option is (A).

The final answer is (A) Both A and R are true and R is the correct explanation of A.

The equation of the xy-plane is $z=0$.

The normal vector to the xy-plane is perpendicular to it and parallel to the z-axis. The direction ratios of the z-axis (and hence the normal to the xy-plane) are $(0, 0, 1)$.

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A plane parallel to the xy-plane must have its normal vector parallel to the normal vector of the xy-plane.

Thus, the normal vector of the new plane is also parallel to the z-axis, and its direction ratios can be taken as $(0, 0, 1)$ (or any non-zero multiple thereof).

The general equation of a plane with normal direction ratios $(A, B, C)$ is $Ax+By+Cz = k$, where $k$ is a constant.

Substituting the direction ratios $(A, B, C) = (0, 0, 1)$, the equation of the plane parallel to the xy-plane takes the form:

$0x + 0y + 1z = k$

This simplifies to:

$z = k$

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The plane passes through the point $(1, 2, 3)$.

To find the value of $k$, we substitute the coordinates of this point into the equation $z = k$.

Since the z-coordinate of the point is 3, we have:

$3 = k$

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Therefore, the equation of the plane parallel to the xy-plane and passing through the point $(1, 2, 3)$ is:

$z = 3$

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Comparing this equation with the given options, we find that option (C) is correct.

The final answer is (C) $z = 3$.

Given:

Equation of Line 1 ($L_1$): $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$

Equation of Line 2 ($L_2$, the z-axis): $\frac{x-0}{0} = \frac{y-0}{0} = \frac{z-0}{1}$

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To Find:

The shortest distance between line $L_1$ and the z-axis.

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Solution:

The given line $L_1$ passes through the point with position vector $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and is parallel to the vector $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

The z-axis ($L_2$) passes through the origin (point with position vector $\vec{a}_2 = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$) and is parallel to the vector $\vec{b}_2 = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$.

The shortest distance 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:

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

First, we calculate the vector connecting the points $\vec{a}_1$ and $\vec{a}_2$:

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

Next, we calculate the cross product of the direction vectors $\vec{b}_1$ and $\vec{b}_2$:

$$\vec{b}_1 \times \vec{b}_2 = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times (0\hat{i} + 0\hat{j} + 1\hat{k})$$ $$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & 0 & 1 \end{vmatrix}$$ $$= \hat{i}((3)(1) - (4)(0)) - \hat{j}((2)(1) - (4)(0)) + \hat{k}((2)(0) - (3)(0))$$ $$= \hat{i}(3 - 0) - \hat{j}(2 - 0) + \hat{k}(0 - 0)$$ $$\vec{b}_1 \times \vec{b}_2 = 3\hat{i} - 2\hat{j} + 0\hat{k} = 3\hat{i} - 2\hat{j}$$

Now, we calculate the dot product of $(\vec{a}_2 - \vec{a}_1)$ and $(\vec{b}_1 \times \vec{b}_2)$, which forms the numerator of the formula:

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + 0\hat{k})$$ $$= (-1)(3) + (-2)(-2) + (-3)(0) = -3 + 4 + 0 = 1$$

Then, we calculate the magnitude of the cross product vector $(\vec{b}_1 \times \vec{b}_2)$, which forms the denominator of the formula:

$$|\vec{b}_1 \times \vec{b}_2| = |3\hat{i} - 2\hat{j}| = \sqrt{(3)^2 + (-2)^2 + (0)^2} = \sqrt{9 + 4 + 0} = \sqrt{13}$$

Finally, we substitute these values into the shortest distance formula:

$$d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{1}{\sqrt{13}} \right|$$ $$d = \frac{1}{\sqrt{13}}$$

The calculated shortest distance between the given line and the z-axis is $\frac{1}{\sqrt{13}}$ units.

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Comparison with Options:

The calculated shortest distance is $\frac{1}{\sqrt{13}}$ units. Let's compare this result with the given options:

• (A) $5/\sqrt{13}$ units
• (B) $13/\sqrt{5}$ units
• (C) $0$ units
• (D) $5$ units

The calculated distance $\frac{1}{\sqrt{13}}$ does not match any of the provided options.

Based on the standard formula and calculations for the shortest distance between skew lines using the provided information, the result is consistently $\frac{1}{\sqrt{13}}$. There appears to be a discrepancy in the options provided for this question.

If forced to select an option despite the mismatch, option (A) $\mathbf{5/\sqrt{13}}$ is the only option that shares the same denominator ($\sqrt{13}$) as our calculated result. This might suggest a potential error in the numerator value within the intended problem or options.

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Alternate Solution (Using Perpendicular Vector Between General Points):

Let a general point P on the line $L_1$ be parameterized by $\lambda$:

$$P = (1 + 2\lambda, 2 + 3\lambda, 3 + 4\lambda)$$

Let a general point Q on the z-axis ($L_2$) be parameterized by $\mu$:

$$Q = (0, 0, \mu)$$

The vector connecting points P and Q is $\vec{PQ}$:

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

The shortest distance between the lines is the magnitude of $\vec{PQ}$ when $\vec{PQ}$ is perpendicular to both direction vectors $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b}_2 = \hat{k}$.

$\vec{PQ} \cdot \vec{b}_1 = 0$:

$$-(1 + 2\lambda)(2) - (2 + 3\lambda)(3) + (\mu - 3 - 4\lambda)(4) = 0$$ $$-2 - 4\lambda - 6 - 9\lambda + 4\mu - 12 - 16\lambda = 0$$ $$-20 - 29\lambda + 4\mu = 0$$

$\vec{PQ} \cdot \vec{b}_2 = 0$:

$$-(1 + 2\lambda)(0) - (2 + 3\lambda)(0) + (\mu - 3 - 4\lambda)(1) = 0$$ $$\mu - 3 - 4\lambda = 0$$

Substitute equation (ii) into equation (i):

$$29\lambda - 4(4\lambda + 3) = -20$$ $$29\lambda - 16\lambda - 12 = -20$$ $$13\lambda = -8 \implies \lambda = -\frac{8}{13}$$

Substitute the value of $\lambda$ back into equation (ii) to find $\mu$:

$$\mu = 4\left(-\frac{8}{13}\right) + 3 = -\frac{32}{13} + \frac{39}{13} = \frac{7}{13}$$

Now, substitute the values of $\lambda$ and $\mu$ back into the expression for $\vec{PQ}$ to find the shortest distance vector:

$$\vec{PQ} = -\left(1 + 2\left(-\frac{8}{13}\right)\right)\hat{i} - \left(2 + 3\left(-\frac{8}{13}\right)\right)\hat{j} + \left(\frac{7}{13} - 3 - 4\left(-\frac{8}{13}\right)\right)\hat{k}$$ $$\vec{PQ} = -\left(1 - \frac{16}{13}\right)\hat{i} - \left(2 - \frac{24}{13}\right)\hat{j} + \left(\frac{7}{13} - \frac{39}{13} + \frac{32}{13}\right)\hat{k}$$ $$\vec{PQ} = -\left(\frac{13 - 16}{13}\right)\hat{i} - \left(\frac{26 - 24}{13}\right)\hat{j} + \left(\frac{7 - 39 + 32}{13}\right)\hat{k}$$ $$\vec{PQ} = -\left(-\frac{3}{13}\right)\hat{i} - \left(\frac{2}{13}\right)\hat{j} + \left(\frac{0}{13}\right)\hat{k}$$ $$\vec{PQ} = \frac{3}{13}\hat{i} - \frac{2}{13}\hat{j} + 0\hat{k}$$

The shortest distance is the magnitude of this vector:

$$d = |\vec{PQ}| = \left| \frac{3}{13}\hat{i} - \frac{2}{13}\hat{j} \right| = \sqrt{\left(\frac{3}{13}\right)^2 + \left(-\frac{2}{13}\right)^2 + 0^2}$$ $$d = \sqrt{\frac{9}{169} + \frac{4}{169} + 0} = \sqrt{\frac{13}{169}} = \sqrt{\frac{1}{13}} = \frac{1}{\sqrt{13}}$$

This alternate method confirms the shortest distance is $\frac{1}{\sqrt{13}}$ units, reinforcing the earlier conclusion that the provided options do not include the correct calculated distance.

Given:

Points $P(x_1, y_1, z_1) = (2, 4, 5)$ and $Q(x_2, y_2, z_2) = (3, 5, -4)$.

The line segment PQ is divided by the xy-plane.

Let the ratio of division be $\lambda:1$.

---

To Find:

The ratio $\lambda:1$ in which the xy-plane divides the line segment PQ.

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Solution:

Let the point that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in the ratio $m:n$ be $R(x, y, z)$. The coordinates of R are given by the section formula:

$$R(x, y, z) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$$

In this problem, the points are $P(2, 4, 5)$ and $Q(3, 5, -4)$, and the ratio is $m:n = \lambda:1$. So $m = \lambda$ and $n = 1$. The coordinates of the dividing point are:

$$R(x, y, z) = \left( \frac{\lambda(3) + 1(2)}{\lambda+1}, \frac{\lambda(5) + 1(4)}{\lambda+1}, \frac{\lambda(-4) + 1(5)}{\lambda+1} \right)$$ $$R(x, y, z) = \left( \frac{3\lambda + 2}{\lambda+1}, \frac{5\lambda + 4}{\lambda+1}, \frac{-4\lambda + 5}{\lambda+1} \right)$$

The dividing point lies on the xy-plane. The equation of the xy-plane is $z=0$. Therefore, the z-coordinate of the dividing point must be $0$.

For a fraction to be equal to zero, the numerator must be zero, provided the denominator is not zero. So, we have:

$$-4\lambda + 5 = 0$$

Solving for $\lambda$:

$$\lambda = \frac{-5}{-4} = \frac{5}{4}$$

The value of $\lambda$ is $\frac{5}{4}$. Since the ratio is $\lambda:1$, the ratio is $\frac{5}{4}:1$. Multiplying both sides by 4 gives the ratio $5:4$.

The sign of $\lambda$ determines whether the division is internal or external. If $\lambda$ is positive, the division is internal. If $\lambda$ is negative, the division is external.

Since $\lambda = \frac{5}{4}$, which is a positive value, the division is internal.

Thus, the xy-plane divides the line segment joining $(2, 4, 5)$ and $(3, 5, -4)$ in the ratio $5:4$ internally.

---

Let's check the given options based on our finding $\lambda = 5/4$ and internal division:

• (A) $\frac{-4\lambda+5}{\lambda+1} = 0 \implies -4\lambda+5=0 \implies \lambda = 5/4$. This part correctly derives the value of $\lambda$. However, it does not explicitly state the type of division in the answer option text itself, though the equation leading to $\lambda=5/4$ is correct.
• (B) $\lambda = 5/4$ (Ratio is $5:4$ externally). This correctly finds the value of $\lambda$ and the ratio but incorrectly states that the division is external.
• (C) $\lambda = 5/4$ (Ratio is $5:4$ internally). This correctly finds the value of $\lambda$, the ratio, and correctly states that the division is internal.
• (D) $\lambda = -5/4$ (Ratio is $5:4$ externally). This incorrectly calculates the value of $\lambda$ and incorrectly states the type of division (as $\lambda$ would need to be negative for external division, but the correct value $5/4$ is positive).

Option (C) matches our calculation and interpretation of the positive ratio value.

The final answer is $\lambda = 5/4$, which means the ratio is $5:4$ internally.

---

The correct option is (C) $\lambda = 5/4$ (Ratio is $5:4$ internally).

Given:

The point $(x_1, y_1, z_1) = (1, 0, 0)$.

The equation of the plane is $x+y+z=1$, which can be written as $x+y+z-1=0$.

Comparing the plane equation with the general form $Ax+By+Cz+D=0$, we have $A=1$, $B=1$, $C=1$, and $D=-1$.

---

To Find:

The value of $\lambda$ used in finding the foot of the perpendicular.

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Solution:

The foot of the perpendicular from the point $(x_1, y_1, z_1)$ to the plane $Ax+By+Cz+D=0$ is a point that lies on the plane. This point is also the intersection of the plane and the line passing through $(x_1, y_1, z_1)$ and perpendicular to the plane.

The direction ratios of the normal to the plane $Ax+By+Cz+D=0$ are $(A, B, C)$. Since the line is perpendicular to the plane, its direction ratios are also proportional to $(A, B, C)$.

The equation of the line passing through $(x_1, y_1, z_1) = (1, 0, 0)$ and having direction ratios $(A, B, C) = (1, 1, 1)$ is given by:

$$\frac{x-x_1}{A} = \frac{y-y_1}{B} = \frac{z-z_1}{C} = \lambda$$

Substituting the given values:

$$\frac{x-1}{1} = \frac{y-0}{1} = \frac{z-0}{1} = \lambda$$ $$\frac{x-1}{1} = \frac{y}{1} = \frac{z}{1} = \lambda$$

Any point on this line can be expressed in terms of $\lambda$ by setting each part equal to $\lambda$:

$$x-1 = \lambda \implies x = 1 + \lambda$$ $$y = \lambda$$ $$z = \lambda$$

So, a general point on the line is $(1+\lambda, \lambda, \lambda)$.

The foot of the perpendicular is this point that also lies on the plane $x+y+z=1$. Therefore, the coordinates of this point must satisfy the equation of the plane.

Substitute the coordinates $(1+\lambda, \lambda, \lambda)$ into the plane equation $x+y+z=1$:

$$(1+\lambda) + (\lambda) + (\lambda) = 1$$

Now, we solve this linear equation for $\lambda$:

$$1 + \lambda + \lambda + \lambda = 1$$ $$1 + 3\lambda = 1$$

Subtract 1 from both sides:

$$3\lambda = 1 - 1$$ $$3\lambda = 0$$

Divide by 3:

$$\lambda = \frac{0}{3}$$ $$\lambda = 0$$

The value of $\lambda$ is $0$.

---

We compare this result with the given options:

• (A) $-1/3$
• (B) $1/3$
• (C) $0$
• (D) $-1$

Our calculated value $\lambda=0$ matches option (C).

---

The final answer is $\lambda = 0$.

The correct option is (C) $0$.

Let the two points be $P_1 = (1, 2, -3)$ and $P_2 = (-1, -2, 1)$.

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

For the given points $P_1(1, 2, -3)$ and $P_2(-1, -2, 1)$:

$x_2 - x_1 = -1 - 1 = -2$

$y_2 - y_1 = -2 - 2 = -4$

$z_2 - z_1 = 1 - (-3) = 1 + 3 = 4$

So, the direction ratios of the line $P_1P_2$ are $(-2, -4, 4)$. Let these be $a = -2$, $b = -4$, and $c = 4$.

---

The magnitude of the direction ratios vector $(a, b, c)$ is given by $\sqrt{a^2 + b^2 + c^2}$.

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

---

The direction cosines of the line with direction ratios $(a, b, c)$ and magnitude $r = \sqrt{a^2 + b^2 + c^2}$ are given by $(\frac{a}{r}, \frac{b}{r}, \frac{c}{r})$.

Direction cosine $l = \frac{-2}{6} = -\frac{1}{3}$

Direction cosine $m = \frac{-4}{6} = -\frac{2}{3}$

Direction cosine $n = \frac{4}{6} = \frac{2}{3}$

---

Therefore, the direction cosines of the line passing through the points (1, 2, -3) and (-1, -2, 1) are $(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.

We are asked to find the vector equation of a line passing through a given point and parallel to a given vector.

---

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

---

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

---

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.

---

Substituting the given values of $\vec{a}$ and $\vec{b}$ into the equation:

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

---

This is the required vector equation of the line.

Let the two given points be $P_1 = (3, -2, -5)$ and $P_2 = (3, -2, 6)$.

---

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

For the given points $P_1(3, -2, -5)$ and $P_2(3, -2, 6)$:

$a = x_2 - x_1 = 3 - 3 = 0$

$b = y_2 - y_1 = -2 - (-2) = -2 + 2 = 0$

$c = z_2 - z_1 = 6 - (-5) = 6 + 5 = 11$

The direction ratios of the line $P_1P_2$ are $(0, 0, 11)$.

---

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

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

When a direction ratio is zero, the corresponding numerator must also be zero. Using point $P_1(3, -2, -5)$, we have $(x_1, y_1, z_1) = (3, -2, -5)$ and $(a, b, c) = (0, 0, 11)$.

Since $a = 0$, we must have $x - x_1 = 0$, which means $x - 3 = 0$, or $x = 3$.

Since $b = 0$, we must have $y - y_1 = 0$, which means $y - (-2) = 0$, or $y = -2$.

Since $c = 11 \neq 0$, the third part of the equation is $\frac{z - z_1}{c} = \frac{z - (-5)}{11} = \frac{z + 5}{11}$.

---

Thus, the Cartesian equations of the line are given by:

and $\frac{z + 5}{11}$ is proportional to the variation along the line.

---

The equation of the line can be written as the system of equations:

$x = 3$, $y = -2$.

This represents a line parallel to the z-axis, passing through the point (3, -2, -5) (and (3, -2, 6)).

The given equations of the lines are in Cartesian form:

Line 1: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{6}$

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

---

The direction ratios of a line given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ are $(a, b, c)$.

For Line 1, the direction ratios are $(a_1, b_1, c_1) = (2, 3, 6)$.

For Line 2, the direction ratios are $(a_2, b_2, c_2) = (1, 2, -2)$.

---

Let $\theta$ be the angle between the two lines. The cosine of the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:

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

---

Calculate the numerator:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(1) + (3)(2) + (6)(-2) = 2 + 6 - 12 = 8 - 12 = -4$.

The absolute value of the numerator is $|-4| = 4$.

---

Calculate the magnitude of the direction ratios vector for Line 1:

$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

---

Calculate the magnitude of the direction ratios vector for Line 2:

$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

---

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

$\cos \theta = \frac{|-4|}{(7)(3)} = \frac{4}{21}$

---

The angle $\theta$ between the lines is the inverse cosine of this value:

$\theta = \cos^{-1}\left(\frac{4}{21}\right)$

Given:

Point through which the plane passes: $(1, 0, -2)$.

Vector normal to the plane: $\vec{n} = \hat{i} + \hat{j} - \hat{k}$.

---

To Find:

Vector equation of the plane.

---

Solution:

Let the given point be $A$. The position vector of point $A$ is $\vec{a} = 1\hat{i} + 0\hat{j} - 2\hat{k} = \hat{i} - 2\hat{k}$.

---

The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

where $\vec{r}$ is the position vector of any point on the plane, given by $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

---

Substitute the values of $\vec{a}$ and $\vec{n}$ into the equation:

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

---

Calculate the dot product on the left side:

$(x)(1) + (y)(1) + (z)(-1) = x + y - z$

---

Calculate the dot product on the right side ($\vec{a} \cdot \vec{n}$):

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

---

Equating the two dot products, we get the vector equation of the plane:

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

---

Alternatively, using the form $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$:

$((x\hat{i} + y\hat{j} + z\hat{k}) - (\hat{i} - 2\hat{k})) \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$

$((x-1)\hat{i} + (y-0)\hat{j} + (z-(-2))\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$

$((x-1)\hat{i} + y\hat{j} + (z+2)\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$

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

$x - 1 + y - z - 2 = 0$

$x + y - z - 3 = 0$

This is the Cartesian form of the equation of the plane, which corresponds to the vector equation $\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 3$ since $x\hat{i} + y\hat{j} + z\hat{k} = \vec{r}$.

---

The required vector equation is $\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 3$.

The given vector equation of the plane is $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 5$.

---

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

---

The vector normal to the plane is given by $\vec{n} = 2\hat{i} - 3\hat{j} + 4\hat{k}$.

---

Substitute the expression for $\vec{r}$ into the vector equation:

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

---

Perform the dot product:

$(x)(2) + (y)(-3) + (z)(4) = 5$

$2x - 3y + 4z = 5$

---

This equation is the Cartesian equation of the plane.

We are asked to find the distance of a given point from a given plane.

---

The given point is $P(2, 5, -3)$. The position vector of this point is $\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}$.

---

The given vector equation of the plane is $\vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4$.

This equation is in the form $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = 6\hat{i} - 3\hat{j} + 2\hat{k}$ is the normal vector to the plane, and $d = 4$.

---

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

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

---

Calculate the dot product $\vec{a} \cdot \vec{n}$:

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

$\vec{a} \cdot \vec{n} = (2)(6) + (5)(-3) + (-3)(2) = 12 - 15 - 6 = -3 - 6 = -9$.

---

Calculate the magnitude of the normal vector $|\vec{n}|$.

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

---

Substitute these values into the distance formula:

Distance $= \frac{|-9 - 4|}{7} = \frac{|-13|}{7} = \frac{13}{7}$.

---

Alternatively, convert the plane equation to Cartesian form.

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

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

$6x - 3y + 2z = 4$, or $6x - 3y + 2z - 4 = 0$.

This is in the form $Ax + By + Cz + D = 0$, where $A=6$, $B=-3$, $C=2$, and $D=-4$.

---

The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

---

Here, $(x_1, y_1, z_1) = (2, 5, -3)$.

Numerator $= |(6)(2) + (-3)(5) + (2)(-3) + (-4)| = |12 - 15 - 6 - 4| = |-3 - 6 - 4| = |-9 - 4| = |-13| = 13$.

Denominator $= \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.

---

Distance $= \frac{13}{7}$.

---

The distance of the point (2, 5, -3) from the given plane is $\frac{13}{7}$ units.

The given equation of the line is in Cartesian form:

$\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z-5}{4}$

---

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

Comparing the given equation with the standard form, we can identify the direction ratios of the line from the denominators.

The direction ratios are $(a, b, c) = (2, -1, 4)$.

---

To find the direction cosines $(l, m, n)$, we use the direction ratios $(a, b, c)$ and the magnitude $r = \sqrt{a^2 + b^2 + c^2}$. The direction cosines are given by $l = \frac{a}{r}$, $m = \frac{b}{r}$, $n = \frac{c}{r}$.

---

First, calculate the magnitude $r$:

$r = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$.

---

Now, calculate the direction cosines:

$l = \frac{2}{\sqrt{21}}$

$m = \frac{-1}{\sqrt{21}}$

$n = \frac{4}{\sqrt{21}}$

---

Thus, the direction ratios are $(2, -1, 4)$ and the direction cosines are $(\frac{2}{\sqrt{21}}, \frac{-1}{\sqrt{21}}, \frac{4}{\sqrt{21}})$.

Given:

The plane passes through the origin $(0, 0, 0)$.

The direction ratios of the normal to the plane are $(2, -3, 1)$. These are the components of the normal vector $\vec{n}$.

---

To Find:

The equation of the plane.

---

Solution:

Let the point on the plane be $O(0, 0, 0)$. The position vector of the origin is $\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$.

---

The normal vector to the plane is $\vec{n} = 2\hat{i} - 3\hat{j} + 1\hat{k}$.

---

The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

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

---

Calculate the right-hand side, $\vec{a} \cdot \vec{n}$:

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

$\vec{a} \cdot \vec{n} = (0)(2) + (0)(-3) + (0)(1) = 0 + 0 + 0 = 0$.

---

Substitute the values into the vector equation:

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

This is the vector equation of the plane.

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To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = 0$

Perform the dot product:

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

$2x - 3y + z = 0$

This is the Cartesian equation of the plane.

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Alternatively, using the Cartesian form directly: The equation of a plane passing through $(x_1, y_1, z_1)$ with normal direction ratios $(A, B, C)$ is $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.

Here, $(x_1, y_1, z_1) = (0, 0, 0)$ and $(A, B, C) = (2, -3, 1)$.

Substituting these values:

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

$2x - 3y + z = 0$

This is the Cartesian equation, which is consistent with the vector form.

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The equation of the plane is $\vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = 0$ in vector form, or $2x - 3y + z = 0$ in Cartesian form.

Given:

The equations of the two planes are:

Plane 1: $2x + y - 2z = 5$

Plane 2: $3x - 6y - 2z = 7$

---

To Find:

The angle between the two planes.

---

Solution:

The equation of a plane in Cartesian form is $Ax + By + Cz = D$. The coefficients $(A, B, C)$ are the direction ratios of the normal vector to the plane.

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For Plane 1 ($2x + y - 2z = 5$), the direction ratios of the normal vector $\vec{n_1}$ are $(A_1, B_1, C_1) = (2, 1, -2)$.

So, the normal vector is $\vec{n_1} = 2\hat{i} + 1\hat{j} - 2\hat{k}$.

---

For Plane 2 ($3x - 6y - 2z = 7$), the direction ratios of the normal vector $\vec{n_2}$ are $(A_2, B_2, C_2) = (3, -6, -2)$.

So, the normal vector is $\vec{n_2} = 3\hat{i} - 6\hat{j} - 2\hat{k}$.

---

The angle $\theta$ between two planes is defined as the angle between their normal vectors. The cosine of the angle between two vectors $\vec{n_1}$ and $\vec{n_2}$ is given by:

$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$

---

Calculate the dot product $\vec{n_1} \cdot \vec{n_2}$:

$\vec{n_1} \cdot \vec{n_2} = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (3\hat{i} - 6\hat{j} - 2\hat{k})$

$\vec{n_1} \cdot \vec{n_2} = (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4$.

---

Calculate the magnitude of $\vec{n_1}$:

$|\vec{n_1}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.

---

Calculate the magnitude of $\vec{n_2}$:

$|\vec{n_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.

---

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

$\cos \theta = \frac{|4|}{(3)(7)} = \frac{4}{21}$

---

The angle $\theta$ between the planes is the inverse cosine of this value:

$\theta = \cos^{-1}\left(\frac{4}{21}\right)$

Given:

The point is $(x_1, y_1, z_1) = (3, -2, 1)$.

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

---

To Find:

The distance of the point from the plane.

---

Solution:

The equation of the plane is in the Cartesian form $Ax + By + Cz + D = 0$, where $A=2$, $B=-1$, $C=2$, and $D=3$.

---

The formula for the distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

---

Substitute the coordinates of the point $(x_1, y_1, z_1) = (3, -2, 1)$ and the coefficients of the plane equation $(A=2, B=-1, C=2, D=3)$ into the formula.

---

Calculate the numerator:

$Ax_1 + By_1 + Cz_1 + D = (2)(3) + (-1)(-2) + (2)(1) + 3$

$= 6 + 2 + 2 + 3 = 8 + 2 + 3 = 10 + 3 = 13$.

The absolute value of the numerator is $|13| = 13$.

---

Calculate the denominator:

$\sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.

---

Substitute the numerator and denominator values into the distance formula:

Distance $= \frac{13}{3}$.

---

The distance of the point (3, -2, 1) from the plane $2x - y + 2z + 3 = 0$ is $\frac{13}{3}$ units.

Given:

The intercepts of the plane on the x, y, and z axes are $a=2$, $b=3$, and $c=4$ respectively.

---

To Find:

The equation of the plane.

---

Solution:

The intercept form of the equation of a plane with intercepts $a$, $b$, and $c$ on the x, y, and z axes is given by:

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

---

Substitute the given values of the intercepts $a=2$, $b=3$, and $c=4$ into the intercept form equation:

$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$

---

To convert this to the standard Cartesian form ($Ax + By + Cz = D$), we can find a common denominator for the fractions, which is the LCM of 2, 3, and 4. The LCM(2, 3, 4) is 12.

---

Multiply the entire equation by 12:

$12 \left(\frac{x}{2} + \frac{y}{3} + \frac{z}{4}\right) = 12(1)$

$12 \times \frac{x}{2} + 12 \times \frac{y}{3} + 12 \times \frac{z}{4} = 12$

$6x + 4y + 3z = 12$

---

The equation of the plane can be written in intercept form as $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$, or in Cartesian form as $6x + 4y + 3z = 12$.

Given:

Equation of the line: $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} + 2\hat{k})$

Equation of the plane: $\vec{r} \cdot (3\hat{i} - 6\hat{j} - 2\hat{k}) = 12$

---

To Find:

The angle between the line and the plane.

---

Solution:

The equation of the line is in the form $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{b}$ is the direction vector of the line.

From the given line equation, the direction vector of the line is $\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}$.

---

The equation of the plane is in the form $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector to the plane.

From the given plane equation, the normal vector to the plane is $\vec{n} = 3\hat{i} - 6\hat{j} - 2\hat{k}$.

---

Let $\phi$ be the angle between the line and the plane. The angle $\phi$ between the line with direction vector $\vec{b}$ and the plane with normal vector $\vec{n}$ is given by the formula:

$\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

---

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

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

$\vec{b} \cdot \vec{n} = (2)(3) + (-1)(-6) + (2)(-2) = 6 + 6 - 4 = 12 - 4 = 8$.

---

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

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

---

Calculate the magnitude of the normal vector $|\vec{n}|$:

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

---

Substitute these values into the formula for $\sin \phi$:

$\sin \phi = \frac{|8|}{(3)(7)} = \frac{8}{21}$

---

The angle $\phi$ between the line and the plane is the inverse sine of this value:

$\phi = \sin^{-1}\left(\frac{8}{21}\right)$

Given:

A line makes equal angles with the coordinate axes (x, y, and z axes).

---

To Find:

The direction cosines of the line.

---

Solution:

Let the angles made by the line with the positive 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 $n = \cos \gamma$.

---

According to the problem statement, the line makes equal angles with the coordinate axes.

---

A fundamental property of direction cosines is that the sum of the squares of the direction cosines of any line is equal to 1.

---

Since $\alpha = \beta = \gamma$, we can substitute this into equation (i):

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

$3 \cos^2 \alpha = 1$

---

Solve for $\cos^2 \alpha$:

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

---

Take the square root to find $\cos \alpha$:

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

---

Since $\alpha = \beta = \gamma$, the direction cosines are:

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

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

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

---

The direction cosines can be $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ or $(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$ or other combinations of signs. However, the question asks for the direction cosines of "a line". By convention, when a line makes equal angles, the direction cosines are usually taken with the same sign.

---

Therefore, the direction cosines of a line that makes equal angles with the coordinate axes are $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ or $(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$.

Given:

The plane passes through the point $(1, 2, 3)$.

The plane is parallel to the plane $3x + 4y - 5z = 7$.

---

To Find:

The equation of the new plane.

---

Solution:

Two parallel planes have the same normal vector (or proportional normal vectors).

---

The equation of the given plane is $3x + 4y - 5z = 7$. The coefficients of $x$, $y$, and $z$ are the components of the normal vector to this plane.

So, the normal vector to the given plane is $\vec{n} = 3\hat{i} + 4\hat{j} - 5\hat{k}$.

---

Since the required plane is parallel to the given plane, it has the same normal vector $\vec{n} = 3\hat{i} + 4\hat{j} - 5\hat{k}$.

---

The required plane passes through the point $(1, 2, 3)$. Let this point be $A$. The position vector of point $A$ is $\vec{a} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

---

The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

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

---

Calculate the dot product $\vec{a} \cdot \vec{n}$:

$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + 4\hat{j} - 5\hat{k})$

$\vec{a} \cdot \vec{n} = (1)(3) + (2)(4) + (3)(-5) = 3 + 8 - 15 = 11 - 15 = -4$.

---

Substitute the values into the vector equation:

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

This is the vector equation of the plane.

---

To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

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

Perform the dot product:

$3x + 4y - 5z = -4$

Or, $3x + 4y - 5z + 4 = 0$.

---

Alternatively, using the Cartesian form directly: The equation of a plane parallel to $Ax + By + Cz = D$ is $Ax + By + Cz = k$.

The required plane is parallel to $3x + 4y - 5z = 7$, so its equation is of the form $3x + 4y - 5z = k$.

---

Since the plane passes through the point $(1, 2, 3)$, these coordinates must satisfy the equation:

$3(1) + 4(2) - 5(3) = k$

$3 + 8 - 15 = k$

$11 - 15 = k$

$k = -4$

---

Substitute the value of $k$ back into the equation:

$3x + 4y - 5z = -4$

Or, $3x + 4y - 5z + 4 = 0$.

This is the Cartesian equation, which is consistent with the vector form.

---

The equation of the plane is $\vec{r} \cdot (3\hat{i} + 4\hat{j} - 5

Given:

The equations of the two lines in Cartesian form:

Line 1 ($L_1$): $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

Line 2 ($L_2$): $\frac{x-1}{-1} = \frac{y-2}{0} = \frac{z-3}{2}$

---

To Find:

The shortest distance between the two lines.

---

Solution:

First, we convert the Cartesian equations to vector form $\vec{r} = \vec{a} + \lambda\vec{b}$ or $\vec{r} = \vec{c} + \mu\vec{d}$.

---

For Line 1: $\frac{x-0}{1} = \frac{y-0}{2} = \frac{z-0}{3}$.

This line passes through the point $(0, 0, 0)$. The position vector of this point is $\vec{a_1} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$.

The direction vector of this line is $\vec{b_1} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

The vector equation of Line 1 is $\vec{r} = \vec{0} + \lambda(1\hat{i} + 2\hat{j} + 3\hat{k})$.

---

For Line 2: $\frac{x-1}{-1} = \frac{y-2}{0} = \frac{z-3}{2}$.

This line passes through the point $(1, 2, 3)$. The position vector of this point is $\vec{a_2} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

The direction vector of this line is $\vec{b_2} = -1\hat{i} + 0\hat{j} + 2\hat{k}$. Note that the denominator 0 for the y-term indicates that the y-component of the direction vector is 0.

The vector equation of Line 2 is $\vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(-1\hat{i} + 0\hat{j} + 2\hat{k})$.

---

We need to determine if the lines are parallel or skew. The lines are parallel if their direction vectors are proportional. $\vec{b_1} = (1, 2, 3)$ and $\vec{b_2} = (-1, 0, 2)$. These vectors are not proportional since $\frac{1}{-1} \neq \frac{2}{0}$ (undefined) $\neq \frac{3}{2}$. Thus, the lines are not parallel. They are either intersecting or skew.

---

The shortest distance 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:

Distance $= \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$

If this distance is 0, the lines intersect (which is a special case of coplanar lines).

---

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (1\hat{i} + 2\hat{j} + 3\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k}) = \hat{i} + 2\hat{j} + 3\hat{k}$.

---

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 0 & 2 \end{vmatrix}$

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

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

$= 4\hat{i} - 5\hat{j} + 2\hat{k}$.

---

Now, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

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

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

$= 4 - 10 + 6$

$= -6 + 6 = 0$.

---

Since the numerator $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is 0, the shortest distance between the lines is 0. This means the lines are coplanar and, since they are not parallel, they must intersect.

---

The shortest distance between the given lines is 0.

Given:

Equation of Plane 1: $x + y + z = 6 \implies x + y + z - 6 = 0$

Equation of Plane 2: $2x + 3y + 4z = -5 \implies 2x + 3y + 4z + 5 = 0$

Point through which the plane passes: $(1, 1, 1)$

---

To Find:

The equation of the plane passing through the intersection of the two given planes and the point (1, 1, 1).

---

Solution:

The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a constant.

---

Using the given planes, the equation of the required plane is:

---

Since the plane (i) passes through the point (1, 1, 1), the coordinates of this point must satisfy the equation. Substitute $x=1$, $y=1$, $z=1$ into equation (i):

$(1 + 1 + 1 - 6) + \lambda(2(1) + 3(1) + 4(1) + 5) = 0$

$(3 - 6) + \lambda(2 + 3 + 4 + 5) = 0$

$-3 + \lambda(14) = 0$

$14\lambda = 3$

$\lambda = \frac{3}{14}$

---

Now, substitute the value of $\lambda = \frac{3}{14}$ back into equation (i):

$(x + y + z - 6) + \frac{3}{14}(2x + 3y + 4z + 5) = 0$

---

Multiply the entire equation by 14 to eliminate the fraction:

$14(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0$

$14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0$

---

Combine the like terms:

$(14x + 6x) + (14y + 9y) + (14z + 12z) + (-84 + 15) = 0$

$20x + 23y + 26z - 69 = 0$

---

This is the required equation of the plane.

Given:

The point from which the perpendicular is drawn is the origin $O(0, 0, 0)$.

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

---

To Find:

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

---

Solution:

Let $P$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane $2x - 3y + 4z - 6 = 0$. The line segment $OP$ is perpendicular to the plane.

---

The direction ratios of the normal to the plane $Ax + By + Cz + D = 0$ are $(A, B, C)$.

For the plane $2x - 3y + 4z - 6 = 0$, the direction ratios of the normal are $(2, -3, 4)$.

---

The line $OP$ is parallel to the normal vector of the plane. Therefore, the direction ratios of the line $OP$ are also $(2, -3, 4)$.

---

The equation of the line $OP$ passing through the origin $(0, 0, 0)$ and having direction ratios $(2, -3, 4)$ is given in Cartesian form as:

---

Any point on the line $OP$ can be represented by the coordinates $(2\lambda, -3\lambda, 4\lambda)$ for some scalar $\lambda$.

---

The foot of the perpendicular $P$ lies on both the line $OP$ and the plane $2x - 3y + 4z - 6 = 0$. Therefore, the coordinates of $P$ must satisfy the equation of the plane.

---

Substitute the coordinates $(2\lambda, -3\lambda, 4\lambda)$ into the plane equation:

$2(2\lambda) - 3(-3\lambda) + 4(4\lambda) - 6 = 0$

$4\lambda + 9\lambda + 16\lambda - 6 = 0$

$29\lambda - 6 = 0$

$29\lambda = 6$

$\lambda = \frac{6}{29}$

---

Now, substitute the value of $\lambda$ back into the coordinates of the general point on the line to find the coordinates of the foot of the perpendicular $P$:

$x = 2\lambda = 2 \left(\frac{6}{29}\right) = \frac{12}{29}$

$y = -3\lambda = -3 \left(\frac{6}{29}\right) = -\frac{18}{29}$

$z = 4\lambda = 4 \left(\frac{6}{29}\right) = \frac{24}{29}$

---

The coordinates of the foot of the perpendicular are $\left(\frac{12}{29}, -\frac{18}{29}, \frac{24}{29}\right)$.

Given:

Point A: $(3, -2, -5)$

Point B: $(3, -2, 6)$

---

To Find:

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

---

Solution:

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

---

The direction vector of the line passing through A and B is given by $\vec{b} - \vec{a}$.

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

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

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

---

Let the direction vector be $\vec{d} = 11\hat{k}$. Since any scalar multiple of the direction vector represents the same direction, we can also use $\vec{d}' = \hat{k}$.

---

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

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

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

---

Using point A (position vector $\vec{a}$) and the direction vector $\hat{k}$:

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

---

This is the required vector equation of the line.

Given:

Equation of Line 1 ($L_1$): $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$

Equation of Line 2 ($L_2$): $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

---

To Show:

The two lines are perpendicular to each other.

---

Solution:

The direction ratios of a line given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ are $(a, b, c)$. These are the components of the direction vector of the line.

---

For Line 1, $\frac{x-5}{7} = \frac{y-(-2)}{-5} = \frac{z-0}{1}$. The direction ratios are $(a_1, b_1, c_1) = (7, -5, 1)$.

The direction vector of Line 1 is $\vec{b_1} = 7\hat{i} - 5\hat{j} + 1\hat{k}$.

---

For Line 2, $\frac{x-0}{1} = \frac{y-0}{2} = \frac{z-0}{3}$. The direction ratios are $(a_2, b_2, c_2) = (1, 2, 3)$.

The direction vector of Line 2 is $\vec{b_2} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

---

Two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ are perpendicular if and only if their dot product is zero, i.e., $\vec{b_1} \cdot \vec{b_2} = 0$, which is equivalent to $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

---

Calculate the dot product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$:

$\vec{b_1} \cdot \vec{b_2} = (7\hat{i} - 5\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})$

$= (7)(1) + (-5)(2) + (1)(3)$

$= 7 - 10 + 3$

$= -3 + 3 = 0$.

---

Since the dot product of their direction vectors is 0, the lines are perpendicular to each other.

---

Therefore, the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular.

Given:

The plane passes through the point $(a, b, c)$.

The plane is parallel to the coordinate plane XOY.

---

To Find:

The equation of the plane.

---

Solution:

The coordinate plane XOY is the plane where the z-coordinate is always 0. Its equation is $z = 0$.

---

The normal vector to the plane $Ax + By + Cz + D = 0$ is $(A, B, C)$. For the plane $z=0$ (or $0x + 0y + 1z + 0 = 0$), the normal vector is $(0, 0, 1)$. This vector is along the z-axis, which is perpendicular to the XOY plane.

---

A plane parallel to the XOY plane will also have its normal vector parallel to the z-axis. The direction ratios of the normal vector can be taken as $(0, 0, 1)$ (or any non-zero multiple of $(0, 0, 1)$).

So, the normal vector to the required plane is $\vec{n} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$.

---

The required plane passes through the point $(a, b, c)$. The position vector of this point is $\vec{a_0} = a\hat{i} + b\hat{j} + c\hat{k}$.

---

The vector equation of a plane passing through a point with position vector $\vec{a_0}$ and normal to vector $\vec{n}$ is given by:

$\vec{r} \cdot \vec{n} = \vec{a_0} \cdot \vec{n}$

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

---

Calculate the dot product $\vec{a_0} \cdot \vec{n}$:

$\vec{a_0} \cdot \vec{n} = (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (0\hat{i} + 0\hat{j} + 1\hat{k})$

$\vec{a_0} \cdot \vec{n} = (a)(0) + (b)(0) + (c)(1) = 0 + 0 + c = c$.

---

Substitute the values into the vector equation:

$\vec{r} \cdot (\hat{k}) = c$

This is the vector equation of the plane.

---

To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (0\hat{i} + 0\hat{j} + 1\hat{k}) = c$

Perform the dot product:

$x(0) + y(0) + z(1) = c$

$z = c$

---

Alternatively, using the Cartesian form directly: A plane parallel to the XOY plane has an equation of the form $z = k$.

Since the plane passes through the point $(a, b, c)$, these coordinates must satisfy the equation:

$c = k$

---

Substitute the value of $k$ back into the equation:

$z = c$

This is the Cartesian equation, which is consistent with the vector form.

---

The equation of the plane passing through $(a, b, c)$ and parallel to the XOY plane is $z = c$.

Given:

Equation of the line: $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$

Equation of the plane: $x - 2y + 2z = 5$

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To Find:

The angle between the line and the plane.

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Solution:

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

From the given line equation, the direction ratios are $(3, 4, 12)$.

The direction vector of the line is $\vec{b} = 3\hat{i} + 4\hat{j} + 12\hat{k}$.

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The equation of the plane is in the Cartesian form $Ax + By + Cz = D$. The coefficients $(A, B, C)$ are the components of the normal vector to the plane.

From the given plane equation, the components of the normal vector are $(1, -2, 2)$.

The normal vector to the plane is $\vec{n} = 1\hat{i} - 2\hat{j} + 2\hat{k}$.

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Let $\phi$ be the angle between the line and the plane. The angle $\phi$ is related to the angle $\theta$ between the direction vector of the line $\vec{b}$ and the normal vector of the plane $\vec{n}$ by $\phi = 90^\circ - \theta$.

The cosine of the angle $\theta$ between $\vec{b}$ and $\vec{n}$ is given by:

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

The sine of the angle $\phi$ between the line and the plane is given by:

$\sin \phi = |\cos \theta| = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$

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

$\vec{b} \cdot \vec{n} = (3\hat{i} + 4\hat{j} + 12\hat{k}) \cdot (1\hat{i} - 2\hat{j} + 2\hat{k})$

$= (3)(1) + (4)(-2) + (12)(2)$

$= 3 - 8 + 24 = 19$.

The absolute value is $|\vec{b} \cdot \vec{n}| = |19| = 19$.

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

$|\vec{b}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

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Calculate the magnitude of the normal vector $|\vec{n}|$:

$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

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Substitute these values into the formula for $\sin \phi$:

$\sin \phi = \frac{19}{(13)(3)} = \frac{19}{39}$

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The angle $\phi$ between the line and the plane is the inverse sine of this value:

$\phi = \sin^{-1}\left(\frac{19}{39}\right)$

Given:

Equation of Line 1 ($L_1$): $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$

Equation of Line 2 ($L_2$): $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$

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To Find:

The shortest distance between the two lines.

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Solution:

The given lines are in the form $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$.

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From the equation of Line 1:

The position vector of a point on the line is $\vec{a_1} = \hat{i} + 2\hat{j} + \hat{k}$.

The direction vector of the line is $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$.

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From the equation of Line 2:

The position vector of a point on the line is $\vec{a_2} = 2\hat{i} - \hat{j} - \hat{k}$.

The direction vector of the line is $\vec{b_2} = 2\hat{i} + \hat{j} + 2\hat{k}$.

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We first check if the lines are parallel by checking if $\vec{b_1}$ and $\vec{b_2}$ are proportional. Since $\frac{1}{2} \neq \frac{-1}{1}$, the direction vectors are not proportional, so the lines are not parallel. The lines are either intersecting or skew.

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The shortest distance 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:

$D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$

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Calculate the vector connecting points on the two lines, $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k})$

$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k}$

$\vec{a_2} - \vec{a_1} = \hat{i} - 3\hat{j} - 2\hat{k}$

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Calculate the cross product of the direction vectors $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$

$= \hat{i}[(-1)(2) - (1)(1)] - \hat{j}[(1)(2) - (1)(2)] + \hat{k}[(1)(1) - (-1)(2)]$

$= \hat{i}[-2 - 1] - \hat{j}[2 - 2] + \hat{k}[1 + 2]$

$= -3\hat{i} - 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k}$

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Calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ (the numerator of the distance formula):

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k})$

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

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

The absolute value of the numerator is $|-9| = 9$.

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Calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$ (the denominator of the distance formula):

$|\vec{b_1} \times \vec{b_2}| = |-3\hat{i} + 3\hat{k}| = \sqrt{(-3)^2 + 0^2 + 3^2}$

$= \sqrt{9 + 0 + 9} = \sqrt{18}$

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

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Substitute the calculated values into the distance formula:

$D = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}$

Rationalize the denominator:

$D = \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.

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

Given:

Equation of the line: $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

Equation of the plane perpendicular to the required plane: $x - y + 2z = 3$

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To Find:

The equation of the plane containing the given line and perpendicular to the given plane.

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Solution:

The equation of the line is in the form $\vec{r} = \vec{a_1} + \lambda\vec{b}$.

The line passes through the point with position vector $\vec{a_1} = \hat{i} + 2\hat{j} - 4\hat{k}$. This means the point $(1, 2, -4)$ lies on the required plane.

The direction vector of the line is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. Since the line lies in the required plane, its direction vector $\vec{b}$ is parallel to the required plane.

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The equation of the given plane is $x - y + 2z = 3$. The normal vector to this plane is $\vec{n_2} = \hat{i} - \hat{j} + 2\hat{k}$.

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Let the normal vector of the required plane be $\vec{n_1}$.

Since the required plane contains the line, its normal vector $\vec{n_1}$ must be perpendicular to the direction vector of the line $\vec{b}$.

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Since the required plane is perpendicular to the plane $x - y + 2z = 3$, their normal vectors are perpendicular.

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Thus, the normal vector $\vec{n_1}$ is perpendicular to both $\vec{b}$ and $\vec{n_2}$. Therefore, $\vec{n_1}$ is parallel to the cross product of $\vec{b}$ and $\vec{n_2}$.

$\vec{n_1} \propto \vec{b} \times \vec{n_2}$

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Calculate the cross product $\vec{b} \times \vec{n_2}$:

$\vec{b} \times \vec{n_2} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \times (\hat{i} - \hat{j} + 2\hat{k})$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & -1 & 2 \end{vmatrix}$

$= \hat{i}[(3)(2) - (6)(-1)] - \hat{j}[(2)(2) - (6)(1)] + \hat{k}[(2)(-1) - (3)(1)]$

$= \hat{i}[6 + 6] - \hat{j}[4 - 6] + \hat{k}[-2 - 3]$

$= 12\hat{i} - (-2)\hat{j} - 5\hat{k} = 12\hat{i} + 2\hat{j} - 5\hat{k}$.

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We can take the normal vector of the required plane as $\vec{n_1} = 12\hat{i} + 2\hat{j} - 5\hat{k}$.

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The required plane passes through the point $(1, 2, -4)$ with position vector $\vec{a_1} = \hat{i} + 2\hat{j} - 4\hat{k}$ and has normal vector $\vec{n_1} = 12\hat{i} + 2\hat{j} - 5\hat{k}$.

The vector equation of the plane is $\vec{r} \cdot \vec{n_1} = \vec{a_1} \cdot \vec{n_1}$.

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Calculate $\vec{a_1} \cdot \vec{n_1}$:

$\vec{a_1} \cdot \vec{n_1} = (\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (12\hat{i} + 2\hat{j} - 5\hat{k})$

$= (1)(12) + (2)(2) + (-4)(-5)$

$= 12 + 4 + 20 = 36$.

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The vector equation of the plane is $\vec{r} \cdot (12\hat{i} + 2\hat{j} - 5\hat{k}) = 36$.

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To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (12\hat{i} + 2\hat{j} - 5\hat{k}) = 36$

Performing the dot product gives:

$12x + 2y - 5z = 36$

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This is the required equation of the plane in Cartesian form.

Given:

Three non-collinear points on the plane: $A(3, -1, 2)$, $B(5, 2, 4)$, and $C(-1, -1, 6)$.

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To Find:

The equation of the plane passing through these three points.

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Solution:

Let the equation of the plane be $Ax + By + Cz + D = 0$. Since the points A, B, and C lie on the plane, their coordinates must satisfy this equation.

For point A(3, -1, 2): $3A - B + 2C + D = 0$ ... (1)

For point B(5, 2, 4): $5A + 2B + 4C + D = 0$ ... (2)

For point C(-1, -1, 6): $-A - B + 6C + D = 0$ ... (3)

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We have a system of three linear equations in four variables A, B, C, and D. We can eliminate D by subtracting equations.

Subtract (1) from (2):

$(5A + 2B + 4C + D) - (3A - B + 2C + D) = 0$

$5A + 2B + 4C + D - 3A + B - 2C - D = 0$

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Subtract (3) from (2):

$(5A + 2B + 4C + D) - (-A - B + 6C + D) = 0$

$5A + 2B + 4C + D + A + B - 6C - D = 0$

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Now we have a system of two equations (4) and (5) in three variables A, B, and C:

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

$6A + 3B - 2C = 0$

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Subtract equation (4) from equation (5):

$(6A + 3B - 2C) - (2A + 3B + 2C) = 0$

$6A + 3B - 2C - 2A - 3B - 2C = 0$

$4A - 4C = 0$

$4A = 4C \implies A = C$

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Substitute $A = C$ into equation (4):

$2A + 3B + 2A = 0$

$4A + 3B = 0$

$3B = -4A \implies B = -\frac{4}{3}A$

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So we have $C = A$ and $B = -\frac{4}{3}A$. Let's choose a value for A, for example, $A = 3$ (to avoid fractions). If $A = 3$, then $C = 3$ and $B = -\frac{4}{3}(3) = -4$.

So, $(A, B, C) = (3, -4, 3)$.

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Now substitute these values of A, B, and C back into equation (1) to find D:

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

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

$9 + 4 + 6 + D = 0$

$19 + D = 0 \implies D = -19$.

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The equation of the plane is $Ax + By + Cz + D = 0$. Substitute the values of A, B, C, and D:

$3x - 4y + 3z - 19 = 0$

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Alternate Method (using vectors):

Let the points be $A(3, -1, 2)$, $B(5, 2, 4)$, and $C(-1, -1, 6)$.

Form two vectors in the plane using these points, for example, $\vec{AB}$ and $\vec{AC}$.

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

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

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The normal vector to the plane is perpendicular to both $\vec{AB}$ and $\vec{AC}$. So, the normal vector $\vec{n}$ is parallel to $\vec{AB} \times \vec{AC}$.

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Calculate the cross product $\vec{AB} \times \vec{AC}$:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix}$

$= \hat{i}[(3)(4) - (2)(0)] - \hat{j}[(2)(4) - (2)(-4)] + \hat{k}[(2)(0) - (3)(-4)]$

$= \hat{i}[12 - 0] - \hat{j}[8 + 8] + \hat{k}[0 + 12]$

$= 12\hat{i} - 16\hat{j} + 12\hat{k}$.

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We can take the normal vector $\vec{n}$ as a simpler proportional vector by dividing by the common factor 4: $\vec{n} = 3\hat{i} - 4\hat{j} + 3\hat{k}$.

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The required plane passes through point A (position vector $\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$) and has normal vector $\vec{n} = 3\hat{i} - 4\hat{j} + 3\hat{k}$.

The vector equation of the plane is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.

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Calculate $\vec{a} \cdot \vec{n}$:

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

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

$= 9 + 4 + 6 = 19$.

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The vector equation is $\vec{r} \cdot (3\hat{i} - 4\hat{j} + 3\hat{k}) = 19$.

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To get the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} - 4\hat{j} + 3\hat{k}) = 19$

$3x - 4y + 3z = 19$

Or, $3x - 4y + 3z - 19 = 0$.

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Both methods yield the same equation. The equation of the plane is $3x - 4y + 3z - 19 = 0$.

Given:

The point $P(1, 2, 3)$.

The line $L$: $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$.

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To Find:

The coordinates of the foot of the perpendicular from P to L.

The length of the perpendicular from P to L.

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Solution:

The equation of the line $L$ is given in Cartesian form. Let's set it equal to a parameter, say $k$:

$\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = k$

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Any point on the line $L$ can be represented by its coordinates in terms of $k$:

$x - 6 = 3k \implies x = 6 + 3k$

$y - 7 = 2k \implies y = 7 + 2k$

$z - 7 = -2k \implies z = 7 - 2k$

Let the foot of the perpendicular from point $P(1, 2, 3)$ to the line $L$ be $Q$. The coordinates of $Q$ can be written as $(6 + 3k, 7 + 2k, 7 - 2k)$ for some value of $k$.

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The direction vector of the line $L$ is $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.

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The vector $\vec{PQ}$ connects the point $P(1, 2, 3)$ to the point $Q(6 + 3k, 7 + 2k, 7 - 2k)$ on the line.

$\vec{PQ} = (6 + 3k - 1)\hat{i} + (7 + 2k - 2)\hat{j} + (7 - 2k - 3)\hat{k}$

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

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Since $PQ$ is the perpendicular from $P$ to the line $L$, the vector $\vec{PQ}$ must be perpendicular to the direction vector of the line $\vec{b}$. The dot product of perpendicular vectors is zero.

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$(5 + 3k)(3) + (5 + 2k)(2) + (4 - 2k)(-2) = 0$

$15 + 9k + 10 + 4k - 8 + 4k = 0$

Combine the $k$ terms and constant terms:

$(9k + 4k + 4k) + (15 + 10 - 8) = 0$

$17k + 17 = 0$

$17k = -17$

$k = -1$

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Now, substitute the value of $k = -1$ back into the coordinates of point $Q$ to find the coordinates of the foot of the perpendicular:

$x = 6 + 3(-1) = 6 - 3 = 3$

$y = 7 + 2(-1) = 7 - 2 = 5$

$z = 7 - 2(-1) = 7 + 2 = 9$

The coordinates of the foot of the perpendicular $Q$ are $(3, 5, 9)$.

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To find the length of the perpendicular, we need to find the magnitude of the vector $\vec{PQ}$ with the calculated value of $k = -1$.

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

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

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

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The length of the perpendicular is the magnitude of $\vec{PQ}$:

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

$= \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

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The coordinates of the foot of the perpendicular are $(3, 5, 9)$, and the length of the perpendicular is 7 units.

Given:

Equation of Plane 1 ($P_1$): $x + 2y + 3z - 4 = 0$

Equation of Plane 2 ($P_2$): $2x + y - z + 5 = 0$}

Equation of the line ($L$): $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4}$.

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To Find:

The equation of the plane passing through the line of intersection of $P_1$ and $P_2$, and parallel to the line $L$.

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Solution:

The equation of a plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a constant.

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The equation of the required plane is:

Rearranging the terms, we get:

$(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$

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The normal vector to this plane is $\vec{n} = (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k}$.

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The given line $L$ is $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4}$. The direction vector of this line is $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

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Since the required plane is parallel to the line $L$, the normal vector of the plane $\vec{n}$ must be perpendicular to the direction vector of the line $\vec{b}$. The dot product of perpendicular vectors is zero.

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

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

$2 + 4\lambda + 6 + 3\lambda + 12 - 4\lambda = 0$

Combine the $\lambda$ terms and constant terms:

$(4\lambda + 3\lambda - 4\lambda) + (2 + 6 + 12) = 0$

$3\lambda + 20 = 0$

$3\lambda = -20$

$\lambda = -\frac{20}{3}$

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Substitute the value of $\lambda = -\frac{20}{3}$ back into the equation of the plane (i):

$(x + 2y + 3z - 4) + \left(-\frac{20}{3}\right)(2x + y - z + 5) = 0$

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Multiply the entire equation by 3 to eliminate the fraction:

$3(x + 2y + 3z - 4) - 20(2x + y - z + 5) = 0$

$3x + 6y + 9z - 12 - 40x - 20y + 20z - 100 = 0$

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Combine the like terms:

$(3x - 40x) + (6y - 20y) + (9z + 20z) + (-12 - 100) = 0$

$-37x - 14y + 29z - 112 = 0$

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Multiply by -1 to make the coefficient of x positive:

$37x + 14y - 29z + 112 = 0$

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This is the required equation of the plane.

Given:

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

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

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To Find:

The coordinates of the image of point P in the line L.

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Solution:

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

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The equation of the line $L$ in Cartesian form is $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.

Let the common ratio be $k$. Any point on the line $L$ can be written in terms of $k$ as:

So, the coordinates of a general point $Q$ on the line are $(k, 1 + 2k, 2 + 3k)$.

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The direction vector of the line $L$ is $\vec{b} = 1\hat{i} + 2\hat{j} + 3\hat{k}$.

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The vector $\vec{PQ}$ connecting the point $P(1, 6, 3)$ to the point $Q(k, 1 + 2k, 2 + 3k)$ on the line is:

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

$\vec{PQ} = (k - 1)\hat{i} + (2k - 5)\hat{j} + (3k - 1)\hat{k}$.

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Since $Q$ is the foot of the perpendicular from $P$ to the line, the vector $\vec{PQ}$ is perpendicular to the direction vector of the line $\vec{b}$. Their dot product must be zero.

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$(k - 1)(1) + (2k - 5)(2) + (3k - 1)(3) = 0$

$k - 1 + 4k - 10 + 9k - 3 = 0$

Combine like terms:

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

$14k - 14 = 0$

$14k = 14$

$k = 1$

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Now, substitute the value of $k=1$ back into the coordinates of point $Q$ to find the coordinates of the foot of the perpendicular:

$x$-coordinate of $Q = k = 1$

$y$-coordinate of $Q = 1 + 2k = 1 + 2(1) = 3$

$z$-coordinate of $Q = 2 + 3k = 2 + 3(1) = 5$

So, the foot of the perpendicular is $Q(1, 3, 5)$.

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Since $Q(1, 3, 5)$ is the midpoint of $P(1, 6, 3)$ and its image $P'(x', y', z')$, we can use the midpoint formula:

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Solve for $x'$, $y'$, and $z'$:

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

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

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

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

Given:

Equation of Line 1 ($L_1$): $\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$

Equation of Line 2 ($L_2$): $\frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$

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To Show:

The lines $L_1$ and $L_2$ are coplanar.

To Find:

The equation of the plane containing these lines.

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Solution:

First, we identify a point on each line and their direction vectors from the Cartesian equations $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

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For Line 1: $\frac{x-(-3)}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$.

A point on $L_1$ is $A_1(-3, 1, 5)$. The position vector is $\vec{a_1} = -3\hat{i} + \hat{j} + 5\hat{k}$.

The direction vector of $L_1$ is $\vec{b_1} = -3\hat{i} + \hat{j} + 5\hat{k}$.

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For Line 2: $\frac{x-(-1)}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$.

A point on $L_2$ is $A_2(-1, 2, 5)$. The position vector is $\vec{a_2} = -\hat{i} + 2\hat{j} + 5\hat{k}$.

The direction vector of $L_2$ is $\vec{b_2} = -\hat{i} + 2\hat{j} + 5\hat{k}$.

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Two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ are coplanar if and only if the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ is equal to zero.

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Calculate the vector connecting points $A_1$ and $A_2$: $\vec{a_2} - \vec{a_1}$

$\vec{a_2} - \vec{a_1} = (-\hat{i} + 2\hat{j} + 5\hat{k}) - (-3\hat{i} + \hat{j} + 5\hat{k})$

$= (-1 - (-3))\hat{i} + (2 - 1)\hat{j} + (5 - 5)\hat{k} = 2\hat{i} + \hat{j} + 0\hat{k}$.

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Calculate the cross product of the direction vectors: $\vec{b_1} \times \vec{b_2}$

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$

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

$= \hat{i}(5 - 10) - \hat{j}(-15 + 5) + \hat{k}(-6 + 1)$

$= -5\hat{i} - (-10)\hat{j} - 5\hat{k} = -5\hat{i} + 10\hat{j} - 5\hat{k}$.

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Calculate the scalar triple product: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$

$= (2\hat{i} + \hat{j} + 0\hat{k}) \cdot (-5\hat{i} + 10\hat{j} - 5\hat{k})$

$= (2)(-5) + (1)(10) + (0)(-5)$

$= -10 + 10 + 0 = 0$.

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Since $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$, the lines are coplanar.

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To find the equation of the plane containing these lines, we use a point on one of the lines (e.g., $A_1$) and the normal vector to the plane. The normal vector $\vec{n}$ is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$, so it is parallel to their cross product $\vec{b_1} \times \vec{b_2}$.

We calculated $\vec{b_1} \times \vec{b_2} = -5\hat{i} + 10\hat{j} - 5\hat{k}$. We can use a simpler proportional vector for the normal: $\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$ (by dividing by -5).

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Using point $A_1(-3, 1, 5)$ with position vector $\vec{a_1} = -3\hat{i} + \hat{j} + 5\hat{k}$ and normal vector $\vec{n} = \hat{i} - 2\hat{j} + \hat{k}$, the vector equation of the plane is $\vec{r} \cdot \vec{n} = \vec{a_1} \cdot \vec{n}$.

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Calculate $\vec{a_1} \cdot \vec{n}$:

$\vec{a_1} \cdot \vec{n} = (-3\hat{i} + \hat{j} + 5\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})$

$= (-3)(1) + (1)(-2) + (5)(1) = -3 - 2 + 5 = 0$.

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

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To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0$

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

$x - 2y + z = 0$.

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The equation of the plane containing the two lines is $x - 2y + z = 0$.

Given:

Equation of Line 1 ($L_1$): $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$

Equation of Line 2 ($L_2$): $\frac{x-4}{5} = \frac{y-1}{2} = \frac{z-0}{1}$

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To Find:

The shortest distance between the two lines.

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Solution:

The given lines are in the Cartesian form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. We can convert them to vector form $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$.

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For Line 1:

A point on the line is $A_1(1, 2, 3)$. The position vector is $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The direction vector of the line is $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

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For Line 2:

A point on the line is $A_2(4, 1, 0)$. The position vector is $\vec{a_2} = 4\hat{i} + \hat{j} + 0\hat{k}$.

The direction vector of the line is $\vec{b_2} = 5\hat{i} + 2\hat{j} + 1\hat{k}$.

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Check if the lines are parallel: The direction vectors $\vec{b_1} = (2, 3, 4)$ and $\vec{b_2} = (5, 2, 1)$ are not proportional ($\frac{2}{5} \neq \frac{3}{2}$), so the lines are not parallel. They are either intersecting or skew.

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The shortest distance 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:

$D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$

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Calculate the vector connecting points on the two lines, $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (4\hat{i} + \hat{j}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

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

$= 3\hat{i} - \hat{j} - 3\hat{k}$.

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Calculate the cross product of the direction vectors $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times (5\hat{i} + 2\hat{j} + \hat{k})$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}$

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

$= \hat{i}(3 - 8) - \hat{j}(2 - 20) + \hat{k}(4 - 15)$

$= -5\hat{i} - (-18)\hat{j} - 11\hat{k} = -5\hat{i} + 18\hat{j} - 11\hat{k}$.

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Calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ (the numerator of the distance formula):

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3\hat{i} - \hat{j} - 3\hat{k}) \cdot (-5\hat{i} + 18\hat{j} - 11\hat{k})$

$= (3)(-5) + (-1)(18) + (-3)(-11)$

$= -15 - 18 + 33$

$= -33 + 33 = 0$.

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Since the numerator is 0, the shortest distance between the lines is 0. This means the lines are coplanar and intersect.

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The shortest distance between the given lines is 0.

Given:

The plane passes through two points: $A(1, 2, 3)$ and $B(3, -1, 2)$.

The plane is parallel to the line $L$: $\frac{x-1}{1} = \frac{y-2}{0} = \frac{z-3}{2}$.

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To Find:

The equation of the plane.

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Solution:

To find the equation of a plane, we need a point on the plane and a vector normal to the plane.

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We are given two points on the plane: $A(1, 2, 3)$ and $B(3, -1, 2)$. We can use point $A$ as a point on the plane. The position vector of point $A$ is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

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The vector joining the two points $A$ and $B$ lies in the plane. Let's find the vector $\vec{AB}$:

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

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

Since $\vec{AB}$ lies in the plane, the normal vector to the plane must be perpendicular to $\vec{AB}$.

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The given line $L$ is $\frac{x-1}{1} = \frac{y-2}{0} = \frac{z-3}{2}$. The direction vector of this line is $\vec{b} = 1\hat{i} + 0\hat{j} + 2\hat{k}$.

Since the plane is parallel to the line $L$, the direction vector $\vec{b}$ is parallel to the plane. This means the normal vector to the plane must also be perpendicular to the direction vector $\vec{b}$.

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The normal vector $\vec{n}$ to the required plane is perpendicular to both $\vec{AB}$ and $\vec{b}$. Therefore, $\vec{n}$ is parallel to the cross product $\vec{AB} \times \vec{b}$.

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Calculate the cross product $\vec{AB} \times \vec{b}$:

$\vec{AB} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 0 & 2 \end{vmatrix}$

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

$= \hat{i}[-6 - 0] - \hat{j}[4 + 1] + \hat{k}[0 + 3]$

$= -6\hat{i} - 5\hat{j} + 3\hat{k}$.

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We can take the normal vector of the required plane as $\vec{n} = -6\hat{i} - 5\hat{j} + 3\hat{k}$.

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The required plane passes through point A with position vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and has normal vector $\vec{n} = -6\hat{i} - 5\hat{j} + 3\hat{k}$.

The vector equation of a plane is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$, where $\vec{r}$ is the position vector of any point on the plane ($\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$).

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Calculate the dot product $\vec{a} \cdot \vec{n}$:

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

$= (1)(-6) + (2)(-5) + (3)(3)$

$= -6 - 10 + 9 = -16 + 9 = -7$.

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Substitute the values into the vector equation:

$\vec{r} \cdot (-6\hat{i} - 5\hat{j} + 3\hat{k}) = -7$

Multiplying by -1, we get:

$\vec{r} \cdot (6\hat{i} + 5\hat{j} - 3\hat{k}) = 7$

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To find the Cartesian equation, substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (6\hat{i} + 5\hat{j} - 3\hat{k}) = 7$

$6x + 5y - 3z = 7$

Or, $6x + 5y - 3z - 7 = 0$.

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The equation of the plane is $6x + 5y - 3z = 7$.

To Find:

The equation of the sphere which passes through the point (1, -2, 3) and touches the coordinate planes.

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Solution:

The general equation of a sphere with center $(a, b, c)$ and radius $R$ is given by:

$(x-a)^2 + (y-b)^2 + (z-c)^2 = R^2$

It is given that the sphere touches the coordinate planes (xy-plane, yz-plane, zx-plane).

The distance from the center $(a, b, c)$ to the yz-plane ($x=0$) is $|a|$.

The distance from the center $(a, b, c)$ to the zx-plane ($y=0$) is $|b|$.

The distance from the center $(a, b, c)$ to the xy-plane ($z=0$) is $|c|$.

Since the sphere touches these planes, these distances must be equal to the radius $R$.

Therefore, $|a| = |b| = |c| = R$.

Let the radius be $r$. So, $R=r$.

The sphere passes through the point $(1, -2, 3)$. This point lies in the octant where $x$ is positive, $y$ is negative, and $z$ is positive.

For a sphere to touch all three coordinate planes and contain a point in a specific octant, its center must also lie in that same octant.

Thus, the coordinates of the center $(a, b, c)$ must have signs $(+, -, +)$.

So, $a = r$, $b = -r$, and $c = r$.

The center of the sphere is $(r, -r, r)$ and the radius is $r$.

Substituting these into the general equation of the sphere, we get:

$(x-r)^2 + (y-(-r))^2 + (z-r)^2 = r^2$

$(x-r)^2 + (y+r)^2 + (z-r)^2 = r^2$

Since the sphere passes through the point $(1, -2, 3)$, these coordinates must satisfy the equation:

$(1-r)^2 + (-2+r)^2 + (3-r)^2 = r^2$

Expanding the terms:

$(1 - 2r + r^2) + (4 - 4r + r^2) + (9 - 6r + r^2) = r^2$

Combining like terms:

$ (r^2 + r^2 + r^2) + (-2r - 4r - 6r) + (1 + 4 + 9) = r^2$

$3r^2 - 12r + 14 = r^2$

$2r^2 - 12r + 14 = 0$

Dividing the equation by 2:

This is a quadratic equation in $r$. We can solve it using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a=1$, $b=-6$, $c=7$.

$r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)}}{2(1)}$

$r = \frac{6 \pm \sqrt{36 - 28}}{2}$

$r = \frac{6 \pm \sqrt{8}}{2}$

$r = \frac{6 \pm 2\sqrt{2}}{2}$

$r = 3 \pm \sqrt{2}$

So, we have two possible values for the radius, which means there are two spheres that satisfy the given conditions.

The general equation of our sphere can be written as $x^2 + y^2 + z^2 - 2rx + 2ry - 2rz + 2r^2 = 0$.

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Case 1: When $r = 3 + \sqrt{2}$

The equation of the sphere is:

$x^2 + y^2 + z^2 - 2(3+\sqrt{2})x + 2(3+\sqrt{2})y - 2(3+\sqrt{2})z + 2(3+\sqrt{2})^2 = 0$

First, we calculate $2(3+\sqrt{2})^2$:

$2(3^2 + 2(3)(\sqrt{2}) + (\sqrt{2})^2) = 2(9 + 6\sqrt{2} + 2) = 2(11 + 6\sqrt{2}) = 22 + 12\sqrt{2}$.

So, the equation is:

$x^2 + y^2 + z^2 - (6+2\sqrt{2})x + (6+2\sqrt{2})y - (6+2\sqrt{2})z + (22 + 12\sqrt{2}) = 0$

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Case 2: When $r = 3 - \sqrt{2}$

The equation of the sphere is:

$x^2 + y^2 + z^2 - 2(3-\sqrt{2})x + 2(3-\sqrt{2})y - 2(3-\sqrt{2})z + 2(3-\sqrt{2})^2 = 0$

First, we calculate $2(3-\sqrt{2})^2$:

$2(3^2 - 2(3)(\sqrt{2}) + (\sqrt{2})^2) = 2(9 - 6\sqrt{2} + 2) = 2(11 - 6\sqrt{2}) = 22 - 12\sqrt{2}$.

So, the equation is:

$x^2 + y^2 + z^2 - (6-2\sqrt{2})x + (6-2\sqrt{2})y - (6-2\sqrt{2})z + (22 - 12\sqrt{2}) = 0$

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Therefore, the two possible equations for the sphere are:

1. $x^2 + y^2 + z^2 - (6+2\sqrt{2})x + (6+2\sqrt{2})y - (6+2\sqrt{2})z + 22 + 12\sqrt{2} = 0$

2. $x^2 + y^2 + z^2 - (6-2\sqrt{2})x + (6-2\sqrt{2})y - (6-2\sqrt{2})z + 22 - 12\sqrt{2} = 0$

Given:

The two lines are given by the vector equations:

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To Find:

The equation of the plane that contains these two lines.

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Solution:

Let the given lines be $L_1$ and $L_2$.

From equation (i), the line $L_1$ passes through the point with position vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and is parallel to the vector $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$.

From equation (ii), the line $L_2$ is parallel to the vector $\vec{b}_2 = 2\hat{i} + \hat{j} + 4\hat{k}$.

The plane that contains both lines $L_1$ and $L_2$ will pass through the point $\vec{a}$ and will be parallel to both vectors $\vec{b}_1$ and $\vec{b}_2$.

The normal vector $\vec{n}$ to the plane is perpendicular to both direction vectors $\vec{b}_1$ and $\vec{b}_2$. We can find $\vec{n}$ by calculating their cross product.

$\vec{n} = \vec{b}_1 \times \vec{b}_2$

$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 4 \end{vmatrix}$

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

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

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

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

Let's calculate the scalar product $\vec{a} \cdot \vec{n}$:

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

$\vec{a} \cdot \vec{n} = (1)(-5) + (2)(-2) + (3)(3)$

$\vec{a} \cdot \vec{n} = -5 - 4 + 9 = 0$

So, the vector equation of the required plane is:

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

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To find the Cartesian equation of the plane, we substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ into the vector equation.

$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-5\hat{i} - 2\hat{j} + 3\hat{k}) = 0$

$(x)(-5) + (y)(-2) + (z)(3) = 0$

$-5x - 2y + 3z = 0$

Multiplying the entire equation by -1 to make the leading coefficient positive, we get:

$5x + 2y - 3z = 0$

This is the required equation of the plane.

Given:

A point P(1, 8, 4).

A line passing through the points A(0, -1, 3) and B(2, -3, -1).

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To Find:

The coordinates of the foot of the perpendicular drawn from point P to the line joining A and B.

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Solution:

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

The direction ratios of the line are given by the difference in the coordinates of A and B:

$a = 2 - 0 = 2$

$b = -3 - (-1) = -2$

$c = -1 - 3 = -4$

The direction ratios are (2, -2, -4). We can simplify these by dividing by 2 to get (1, -1, -2).

The equation of the line passing through point A(0, -1, 3) with these direction ratios 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's represent this common ratio by a parameter $\lambda$.

Let M be the coordinates of the foot of the perpendicular from point P to the line. Since M lies on the line, its coordinates can be expressed in terms of $\lambda$ as:

$x = \lambda$

$y = -\lambda - 1$

$z = -2\lambda + 3$

So, the coordinates of M are $(\lambda, -\lambda - 1, -2\lambda + 3)$.

Now, we find the direction ratios of the line segment PM, where P is (1, 8, 4).

$a' = x_M - x_P = \lambda - 1$

$b' = y_M - y_P = (-\lambda - 1) - 8 = -\lambda - 9$

$c' = z_M - z_P = (-2\lambda + 3) - 4 = -2\lambda - 1$

The direction ratios of PM are $(\lambda - 1, -\lambda - 9, -2\lambda - 1)$.

Since PM is perpendicular to the line AB, the dot product of their direction ratios must be zero.

The direction ratios of line AB are $(1, -1, -2)$.

The condition for perpendicularity is $aa' + bb' + cc' = 0$.

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

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

Combine the terms:

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

$6\lambda + 10 = 0$

$6\lambda = -10$

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

Now we substitute this value of $\lambda$ back into the coordinates of M to find the foot of the perpendicular.

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

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

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

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