Rbse Solutions Class 12 Maths Exercise 11.2 | Lines in Space

Get detailed solutions for NCERT Class 12 Maths Exercise 11.2 (Three Dimensional Geometry). Learn to find vector and Cartesian equations of a line passing through a point and parallel to a vector. Master conditions for parallel and perpendicular lines, calculating the angle between two lines, and finding the Shortest Distance (SD) between skew lines using both vector and Cartesian formulas. Essential for Chapter 11.

This exercise covers conditions for perpendicular and parallel lines, finding line equations in vector and Cartesian form, calculating the angle between two lines, and finding the shortest distance between skew lines.

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1. Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

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

Let the lines be $L_1, L_2, L_3$ with direction cosines:

• $L_1: l_1 = \frac{12}{13}, m_1 = \frac{-3}{13}, n_1 = \frac{-4}{13}$
• $L_2: l_2 = \frac{4}{13}, m_2 = \frac{12}{13}, n_2 = \frac{3}{13}$
• $L_3: l_3 = \frac{3}{13}, m_3 = \frac{-4}{13}, n_3 = \frac{12}{13}$

(i) $L_1$ and $L_2$:

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

$$= \frac{48}{169} – \frac{36}{169} – \frac{12}{169} = \frac{48 – 36 – 12}{169} = \frac{0}{169} = 0$$

$L_1 \perp L_2$.

(ii) $L_2$ and $L_3$:

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

$$= \frac{12}{169} – \frac{48}{169} + \frac{36}{169} = \frac{12 – 48 + 36}{169} = \frac{0}{169} = 0$$

$L_2 \perp L_3$.

(iii) $L_1$ and $L_3$:

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

$$= \frac{36}{169} + \frac{12}{169} – \frac{48}{169} = \frac{36 + 12 – 48}{169} = \frac{0}{169} = 0$$

$L_1 \perp L_3$.

Since the dot product of the direction cosines for every pair is zero, the lines are mutually perpendicular.

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

Two lines are perpendicular if the dot product of their direction ratios ($a_1 a_2 + b_1 b_2 + c_1 c_2$) is zero.

(i) Direction Ratios of Line 1 ($L_1$) through $A(1, -1, 2)$ and $B(3, 4, -2)$:

$$a_1 = 3 – 1 = 2$$

$$b_1 = 4 – (-1) = 5$$

$$c_1 = -2 – 2 = -4$$

$$L_1: \langle 2, 5, -4 \rangle$$

(ii) Direction Ratios of Line 2 ($L_2$) through $C(0, 3, 2)$ and $D(3, 5, 6)$:

$$a_2 = 3 – 0 = 3$$

$$b_2 = 5 – 3 = 2$$

$$c_2 = 6 – 2 = 4$$

$$L_2: \langle 3, 2, 4 \rangle$$

(iii) Check for Perpendicularity:

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

$$= 6 + 10 – 16 = 16 – 16 = 0$$

Since the dot product is zero, the two lines are perpendicular.

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

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

(i) Direction Ratios of Line 1 ($L_1$) through $A(4, 7, 8)$ and $B(2, 3, 4)$:

$$a_1 = 2 – 4 = -2$$

$$b_1 = 3 – 7 = -4$$

$$c_1 = 4 – 8 = -4$$

$$L_1: \langle -2, -4, -4 \rangle$$

(ii) Direction Ratios of Line 2 ($L_2$) through $C(-1, -2, 1)$ and $D(1, 2, 5)$:

$$a_2 = 1 – (-1) = 2$$

$$b_2 = 2 – (-2) = 4$$

$$c_2 = 5 – 1 = 4$$

$$L_2: \langle 2, 4, 4 \rangle$$

(iii) Check for Parallelism:

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

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

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

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

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4. Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat{i} + 2\hat{j} – 2\hat{k}$.

A line passing through $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and parallel to $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ has the equation $\vec{r} = \vec{a} + \lambda \vec{b}$.

• Position vector of the point: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$
• Direction vector: $\vec{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}$

(i) Vector Equation:

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

(ii) Cartesian Equation:

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

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

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5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{i} – \hat{j} + 4\hat{k}$ and is in the direction $\hat{i} + 2\hat{j} – \hat{k}$.

• Position vector of the point: $\vec{a} = 2\hat{i} – \hat{j} + 4\hat{k}$
• Direction vector: $\vec{b} = \hat{i} + 2\hat{j} – \hat{k}$

(i) Vector Equation:

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

(ii) Cartesian Equation:

The line passes through $(x_1, y_1, z_1) = (2, -1, 4)$ and has direction ratios $\langle a, b, c \rangle = \langle 1, 2, -1 \rangle$.

$$\mathbf{\frac{x – 2}{1} = \frac{y – (-1)}{2} = \frac{z – 4}{-1} \implies \frac{x – 2}{1} = \frac{y + 1}{2} = \frac{z – 4}{-1}}$$

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6. Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

The required line is parallel to the given line, so it will have the same direction ratios $\langle a, b, c \rangle$.

1. Direction Ratios: The given line $\frac{x – (-3)}{3} = \frac{y – 4}{5} = \frac{z – (-8)}{6}$ has direction ratios $\langle 3, 5, 6 \rangle$.
2. Point: The required line passes through $(x_1, y_1, z_1) = (-2, 4, -5)$.

Cartesian Equation:

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

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

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7. The cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$. Write its vector form.

The Cartesian equation $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ corresponds to the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$.

• Point $\vec{a}$ (Passes through): The line passes through $(x_1, y_1, z_1) = (5, -4, 6)$.$$\vec{a} = 5\hat{i} – 4\hat{j} + 6\hat{k}$$
• Direction $\vec{b}$ (Parallel to): The direction ratios are $\langle a, b, c \rangle = \langle 3, 7, 2 \rangle$.$$\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$$

Vector Equation:

$$\mathbf{\vec{r} = (5\hat{i} – 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k})}$$

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8. Find the angle between the following pairs of lines:

The angle $\theta$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by $\cos \theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| |\vec{b}_2|} \right|$.

(i) $\vec{r} = 2\hat{i} – 5\hat{j} + \hat{k} + \lambda (3\hat{i} + 2\hat{j} + 6\hat{k})$ and $\vec{r} = 7\hat{i} – 6\hat{k} + \mu (\hat{i} + 2\hat{j} + 2\hat{k})$

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

1. Dot Product $\vec{b}_1 \cdot \vec{b}_2$:$$\vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19$$
2. Magnitudes:$$|\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$$$|\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$
3. Angle $\theta$:$$\cos \theta = \left| \frac{19}{7 \cdot 3} \right| = \frac{19}{21}$$$$\mathbf{\theta = \cos^{-1}\left(\frac{19}{21}\right)}$$

(ii) $\vec{r} = 3\hat{i} + \hat{j} – 2\hat{k} + \lambda (\hat{i} – \hat{j} – 2\hat{k})$ and $\vec{r} = 2\hat{i} – \hat{j} – 5\hat{k} + \mu (3\hat{i} – 5\hat{j} – 4\hat{k})$

• $\vec{b}_1 = \hat{i} – \hat{j} – 2\hat{k}$
• $\vec{b}_2 = 3\hat{i} – 5\hat{j} – 4\hat{k}$

1. Dot Product $\vec{b}_1 \cdot \vec{b}_2$:$$\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16$$
2. Magnitudes:$$|\vec{b}_1| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$$$|\vec{b}_2| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$$
3. Angle $\theta$:$$\cos \theta = \left| \frac{16}{\sqrt{6} \cdot 5\sqrt{2}} \right| = \frac{16}{5\sqrt{12}} = \frac{16}{5 \cdot 2\sqrt{3}} = \frac{8}{5\sqrt{3}}$$Rationalizing the denominator: $\cos \theta = \frac{8\sqrt{3}}{15}$.$$\mathbf{\theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)}$$

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9. Find the angle between the following pair of lines:

The angle $\theta$ between two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.

(i) $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ and $\frac{x+2}{1} = \frac{y-4}{8} = \frac{z-5}{4}$

• Line 1 Direction Ratios: $\langle a_1, b_1, c_1 \rangle = \langle 2, 5, -3 \rangle$
• Line 2 Direction Ratios: $\langle a_2, b_2, c_2 \rangle = \langle 1, 8, 4 \rangle$

1. Dot Product:$$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(1) + (5)(8) + (-3)(4) = 2 + 40 – 12 = 30$$
2. Magnitudes:$$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$$$$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$$
3. Angle $\theta$:$$\cos \theta = \left| \frac{30}{9 \cdot \sqrt{38}} \right| = \frac{10}{3\sqrt{38}}$$$$\mathbf{\theta = \cos^{-1}\left(\frac{10}{3\sqrt{38}}\right)}$$

(ii) $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$

• Line 1 Direction Ratios: $\langle a_1, b_1, c_1 \rangle = \langle 2, 2, 1 \rangle$
• Line 2 Direction Ratios: $\langle a_2, b_2, c_2 \rangle = \langle 4, 1, 8 \rangle$

1. Dot Product:$$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$$
2. Magnitudes:$$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$$$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$$
3. Angle $\theta$:$$\cos \theta = \left| \frac{18}{3 \cdot 9} \right| = \frac{18}{27} = \frac{2}{3}$$$$\mathbf{\theta = \cos^{-1}\left(\frac{2}{3}\right)}$$

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10. Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

First, rewrite the line equations in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

(i) Line 1 ($L_1$): $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$

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

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

$$L_1 \text{ Direction Ratios: } \langle a_1, b_1, c_1 \rangle = \langle -3, 2p/7, 2 \rangle$$

(ii) Line 2 ($L_2$): $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$

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

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

$$L_2 \text{ Direction Ratios: } \langle a_2, b_2, c_2 \rangle = \langle -3p/7, 1, -5 \rangle$$

(iii) Condition for Right Angle (Perpendicularity):

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

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

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

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

$$11p = 70 \implies \mathbf{p = 70/11}$$

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11. Show that 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 to each other.

• Line 1 Direction Ratios: $\langle a_1, b_1, c_1 \rangle = \langle 7, -5, 1 \rangle$
• Line 2 Direction Ratios: $\langle a_2, b_2, c_2 \rangle = \langle 1, 2, 3 \rangle$

Check for Perpendicularity:

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

$$= 7 – 10 + 3 = 10 – 10 = 0$$

Since the dot product of the direction ratios is zero, the lines are perpendicular.

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12. Find the shortest distance between the lines $\vec{r} = \hat{i} + 2\hat{j} + \hat{k} + \lambda (\hat{i} – \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} – \hat{j} – \hat{k} + \mu (2\hat{i} + \hat{j} + 2\hat{k})$.

The lines are skew lines. The shortest distance (SD) 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:

$$\text{SD} = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$$

• $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$
• $\vec{b}_1 = \hat{i} – \hat{j} + \hat{k}$
• $\vec{a}_2 = 2\hat{i} – \hat{j} – \hat{k}$
• $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$

1. Calculate $\vec{a}_2 – \vec{a}_1$:$$\vec{a}_2 – \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} – 3\hat{j} – 2\hat{k}$$
2. Calculate $\vec{b}_1 \times \vec{b}_2$:$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2 – 1) – \hat{j}(2 – 2) + \hat{k}(1 – (-2)) = -3\hat{i} + 0\hat{j} + 3\hat{k}$$
3. Calculate $|\vec{b}_1 \times \vec{b}_2|$:$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$
4. Calculate the numerator $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)$:$$(-3\hat{i} + 3\hat{k}) \cdot (\hat{i} – 3\hat{j} – 2\hat{k}) = (-3)(1) + (0)(-3) + (3)(-2) = -3 + 0 – 6 = -9$$
5. Shortest Distance (SD):$$\text{SD} = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}$$Rationalizing: $\text{SD} = \frac{3\sqrt{2}}{2}$ units.$$\mathbf{\text{SD} = \frac{3\sqrt{2}}{2} \text{ units}}$$

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13. Find the shortest distance between the lines $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$.

First, identify the vectors $\vec{a}_1, \vec{b}_1, \vec{a}_2, \vec{b}_2$.

• Line 1: Passes through $(-1, -1, -1)$, Direction $\langle 7, -6, 1 \rangle$.$$\vec{a}_1 = -\hat{i} – \hat{j} – \hat{k}, \quad \vec{b}_1 = 7\hat{i} – 6\hat{j} + \hat{k}$$
• Line 2: Passes through $(3, 5, 7)$, Direction $\langle 1, -2, 1 \rangle$.$$\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}, \quad \vec{b}_2 = \hat{i} – 2\hat{j} + \hat{k}$$

1. Calculate $\vec{a}_2 – \vec{a}_1$:$$\vec{a}_2 – \vec{a}_1 = (3-(-1))\hat{i} + (5-(-1))\hat{j} + (7-(-1))\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}$$
2. Calculate $\vec{b}_1 \times \vec{b}_2$:$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-6 – (-2)) – \hat{j}(7 – 1) + \hat{k}(-14 – (-6))$$$$\vec{b}_1 \times \vec{b}_2 = -4\hat{i} – 6\hat{j} – 8\hat{k}$$
3. Calculate $|\vec{b}_1 \times \vec{b}_2|$:$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$$
4. Calculate the numerator $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)$:$$(-4\hat{i} – 6\hat{j} – 8\hat{k}) \cdot (4\hat{i} + 6\hat{j} + 8\hat{k})$$$$= (-4)(4) + (-6)(6) + (-8)(8) = -16 – 36 – 64 = -116$$
5. Shortest Distance (SD):$$\text{SD} = \left| \frac{-116}{2\sqrt{29}} \right| = \frac{58}{\sqrt{29}}$$Rationalizing: $\text{SD} = \frac{58\sqrt{29}}{29} = 2\sqrt{29}$ units.$$\mathbf{\text{SD} = 2\sqrt{29} \text{ units}}$$

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14. Find the shortest distance between the lines $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda (\hat{i} – 3\hat{j} + 2\hat{k})$ and $\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu (2\hat{i} + 3\hat{j} + \hat{k})$.

• $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$
• $\vec{b}_1 = \hat{i} – 3\hat{j} + 2\hat{k}$
• $\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$
• $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$

1. Calculate $\vec{a}_2 – \vec{a}_1$:$$\vec{a}_2 – \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
2. Calculate $\vec{b}_1 \times \vec{b}_2$:$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 – 6) – \hat{j}(1 – 4) + \hat{k}(3 – (-6))$$$$\vec{b}_1 \times \vec{b}_2 = -9\hat{i} + 3\hat{j} + 9\hat{k}$$
3. Calculate $|\vec{b}_1 \times \vec{b}_2|$:$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$$
4. Calculate the numerator $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)$:$$(-9\hat{i} + 3\hat{j} + 9\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})$$$$= (-9)(3) + (3)(3) + (9)(3) = -27 + 9 + 27 = 9$$
5. Shortest Distance (SD):$$\text{SD} = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}}$$Rationalizing: $\text{SD} = \frac{3\sqrt{19}}{19}$ units.$$\mathbf{\text{SD} = \frac{3\sqrt{19}}{19} \text{ units}}$$

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15. Find the shortest distance between the lines whose vector equations are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} – (2s+1)\hat{k}$.

First, rewrite the lines in the standard form $\vec{r} = \vec{a} + \lambda \vec{b}$.

(i) Line 1 ($\vec{r}_1$) (using parameter $t$):

$$\vec{r}_1 = (\hat{i} – 2\hat{j} + 3\hat{k}) + t (-\hat{i} + \hat{j} – 2\hat{k})$$

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

(ii) Line 2 ($\vec{r}_2$) (using parameter $s$):

$$\vec{r}_2 = (\hat{i} – \hat{j} – \hat{k}) + s (\hat{i} + 2\hat{j} – 2\hat{k})$$

$$\vec{a}_2 = \hat{i} – \hat{j} – \hat{k}, \quad \vec{b}_2 = \hat{i} + 2\hat{j} – 2\hat{k}$$

1. Calculate $\vec{a}_2 – \vec{a}_1$:$$\vec{a}_2 – \vec{a}_1 = (\hat{i} – \hat{j} – \hat{k}) – (\hat{i} – 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} – 4\hat{k}$$
2. Calculate $\vec{b}_1 \times \vec{b}_2$:$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 – (-4)) – \hat{j}(2 – 2) + \hat{k}(-2 – 1)$$$$\vec{b}_1 \times \vec{b}_2 = 2\hat{i} + 0\hat{j} – 3\hat{k}$$
3. Calculate $|\vec{b}_1 \times \vec{b}_2|$:$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$$
4. Calculate the numerator $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)$:$$(2\hat{i} – 3\hat{k}) \cdot (\hat{j} – 4\hat{k}) = (2)(0) + (0)(1) + (-3)(-4) = 0 + 0 + 12 = 12$$
5. Shortest Distance (SD):$$\text{SD} = \left| \frac{12}{\sqrt{13}} \right| = \frac{12}{\sqrt{13}}$$Rationalizing: $\text{SD} = \frac{12\sqrt{13}}{13}$ units.$$\mathbf{\text{SD} = \frac{12\sqrt{13}}{13} \text{ units}}$$