# Ncert Solutions For Class 12 Maths.3 Ex 11.2

## Ncert Solutions For Class 12 Maths Chapter 11.3 Ex 11.2

Q1. Prove 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.

Sol:

Two lines with direction cosines: l1, m1, n1 and l2, m2, n2 are perpendicular to each other, if l1 l2 + m1 m2 + n1 n2 = 0

(i) For the lines with direction cosines, $$\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$$ and $$\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\\$$

Therefore, l1 l2 + m1 m2 + n1 n2 = $$\frac{12}{13} \times \frac{4}{13} + \left ( \frac{-3}{13} \right ) \times \frac{12}{13} + \left ( \frac{-4}{13} \right ) \times \frac{3}{13}\\$$

i.e. l1 l2 + m1 m2 + n1 n2 = $$\frac{48}{169} – \frac{36}{169} – \frac{12}{13}$$

(or) l1 l2 + m1 m2 + n1 n2 = 0

Hence, the lines are perpendicular.

(ii) For the lines with direction cosines, $$\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$$ and $$\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\\$$.

l1 l2 + m1 m2 + n1 n2 = $$\left [ \frac{3}{13} \times \frac{12}{13} \right ] + \left [ \left (\frac{-4}{13} \right ) \times \left (\frac{-3}{13} \right ) \right ] + \left [ \frac{12}{13} \times \left (\frac{-4}{13} \right ) \right ]\\$$

Therefore, l1 l2 + m1 m2 + n1 n2 = $$\frac{12}{169} – \frac{48}{169} + \frac{36}{169}$$

or, l1 l2 + m1 m2 + n1 n2 = 0

Hence, the lines are perpendicular.

(iii) For the lines with direction cosines, $$\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$$ and $$\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\\$$

l1 l2 + m1 m2 + n1 n2 =

Therefore, l1 l2 + m1 m2 + n1 n2 = $$\frac{36}{169} + \frac{12}{169} – \frac{48}{169}$$

i.e. l1 l2 + m1 m2 + n1 n2 = 0

Hence, the lines are perpendicular.

Therefore, all the lines are mutually perpendicular.

Q2. 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).

Sol:

Let PQ be the line joining the points, (1, −1, 2) and (3, 4, − 2), and RS be the line joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios p1 + q1 + r1 of PQ are (3 − 1), [4 − (−1)], and (−2 − 2) that is 2, 5, and −4.

The direction ratios p1 + q1 + r1 of RS are (3 − 0), (5 − 3), and (6 −2) that is 3, 2, and 4.

PQ and RS will be perpendicular to each other, if l1 l2 + m1 m2 + n1 n2 = 0

l1 l2 + m1 m2 + n1 n2 = 2 × 3 + 5 ×2 + (-4) × 4

i.e. l1 l2 + m1 m2 + n1 n2 = 6 + 10 – 16 = 0

Hence, PQ and RS are perpendicular to each other.

Q3. Prove that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and (1, 2, 5).

Sol:

Let PQ be the line through the points, (4, 7, 8) and (2, 3, 4), and RS be the line through the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios p1, q1, r1 of PQ are (2 − 4), (3 − 7), and (4 − 8) that is −2, −4, and -4.

The direction ratios p2, q2, r2 of RS are [1 − (−1)], [2 − (−2)], and (5 − 1) that is 2, 4, and 4.

PQ will be parallel to RS, if $$\frac{p_{1}}{p_{2}} = \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}}$$

$$\frac{p_{1}}{p_{2}} = \frac{-2}{2} = -1\\$$

Also, $$\frac{q_{1}}{q_{2}} = \frac{-4}{4} = -1\\$$

And $$\frac{r_{1}}{r_{2}} = \frac{-4}{4} = -1\\$$

Therefore, $$\frac{p_{1}}{p_{2}} = \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}}$$

Hence, PQ is parallel to RS.

Q4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $$3\hat{i} + 2\hat{j} – 2\hat{k}$$.

Sol:

It is given that the line passes through the point A (1, 2, 3).

Therefore, the position vector through A is,

$$\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}$$ $$\bar{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}$$

It is known that the line which passes through point A and parallel to $$\vec{b}$$ is given by $$\vec{r} = \vec{a} + \lambda \vec{b}$$ where $$\lambda$$ is a constant.

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

This is required equation of the line.

Q5. Find the equation of the line in Cartesian and in vector form that passes through the point with position vector $$2\hat{i} – \hat{j} + 4\hat{k}$$ and is in the direction $$\hat{i} + 2\hat{j} – \hat{k}$$.

Sol:

It is given that line passes through the point with position vector

$$\vec{a} = 2\hat{i} – \hat{j} + 4\hat{k}$$ ——– (1)

$$\vec{b} = \hat{i} + 2\hat{j} – \hat{k}$$ ———- (2)

It is known that a line through a point with position vector $$\vec{a}$$ and parallel to $$\vec{b}$$ is given by the equation,

$$\vec{r} = \vec{a} + \lambda \vec{b}$$ $$\vec{r} = 2\hat{i} – \hat{j} + 4\hat{k} + \lambda \left (\hat{i} + 2\hat{j} – \hat{k} \right )$$

Eliminating $$\lambda$$, we obtain the Cartesian form of equation as,

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

This is the required equation of the given line in Cartesian form.

Q6. Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by $$\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}$$.

Sol:

It is given that line passes through the given point (−2, 4, −5) and it is parallel to $$\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}$$

The direction ratios of the line $$\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}$$ are 3, 5, and 6.

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

Hence, its direction ratios are 3k, 5k, and 6k, where k ≠ 0.

It is known that the equation of the line through the point $$\left ( x_{1}, y_{1}, z_{1} \right )$$ and with direction ratios p, q, r is given by $$\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}$$.

Hence, the equation of the required line is,

$$\frac{x + 2}{3k} = \frac{y – 4}{5k} = \frac{z + 5}{6k}$$ $$\frac{x + 2}{3} = \frac{y – 4}{5} = \frac{z + 5}{6} = k$$

Q7. The Cartesian equation of a line is $$\frac{x – 5}{3} = \frac{y + 4}{7} = \frac{z – 6}{2}$$. Give the vector form of the line.

Sol:

The Cartesian equation of the line is,

$$\frac{x – 5}{3} = \frac{y + 4}{7} = \frac{z – 6}{2}$$ ——— (1)

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

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector, $$\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$$.

It is known that the line through position vector $$\vec{a}$$ and in the direction of the vector $$\vec{b}$$ is given by the equation,

$$\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R$$ $$\vec{r} = \left (5\hat{i} – 4\hat{j} + 6\hat{k} \right ) + \lambda \left ( 3\hat{i} + 7\hat{j} + 2\hat{k} \right )$$

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

Q8. Find the Cartesian and the vector equations of the lines that pass through the origin and (5, −2, 3).

Sol:

The required line passes through the origin. Hence, its position vector is given by,

$$\vec{a} = \vec{0}$$ ————– (1)

The direction ratios of the line through origin and (5, −2, 3) are,

(5 − 0) = 5

(−2 − 0) = −2

(3 − 0) = 3

The line is parallel to the vector given by the equation,

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

The equation of the line in vector form through a point with position vector $$\vec{a}$$ and parallel to $$\vec{b}$$ is,

$$\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R$$ $$\vec{r} = \vec{0} + \lambda \left ( 5\hat{i} – 2\hat{j} + 3\hat{k} \right )$$ $$\vec{r} = \lambda \left ( 5\hat{i} – 2\hat{j} + 3\hat{k} \right )$$

The equation of the line through the point $$\left ( x_{1}, y_{1}, z_{1} \right )$$ and the direction ratios are p, q, r is given as,

$$\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}$$

Hence, the equation of the required line in the Cartesian form is,

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

Q9. Find the Cartesian and the vector equations of the line that passes through the points (3, −2, −5) and (3, −2, 6).

Sol:

Let the line passing through the points, M (3, −2, −5) and N (3, −2, 6), be MN.

Since MN passes through M (3, −2, −5), its position vector is given by, $$\vec{a} = 3\hat{i} – 2\hat{j} – 5\hat{k}$$.

The direction ratios of MN are given by,

(3 − 3) = 0

(−2 + 2) = 0

(6 + 5) = 11

The equation of the vector in the direction of MN is,

$$\vec{b} = 0.\hat{i} – 0.\hat{j} + 11\hat{k}$$ $$\vec{b} = 11\hat{k}$$

The equation of MN in vector form is given by,

$$\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R$$ $$\vec{r} = \left (3\hat{i} – 2\hat{j} – 5\hat{k} \right ) + 11 \lambda \hat{k}$$

The equation of MN in Cartesian form is,

$$\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}$$

i.e. $$\frac{x – 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}$$

Q10. Calculate the angle between the given pairs of lines:

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

(ii) $$\vec{r} = 3\hat{i} + \hat{j} – 2\hat{k} + \lambda \left ( \hat{i} – \hat{j} – 2\hat{k} \right ) \; and \; \vec{r} = 2\hat{i} – \hat{j} – 56\hat{k} + \mu \left ( 3\hat{i} – 5\hat{j} – 4\hat{k} \right )$$

Sol:

(i) Let $$\theta$$ be the angle between the given lines.

The angle between the given pairs of lines is given by,

$$\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right |$$

The given lines are parallel to the vectors, $$\vec{b_{1}} = 3\hat{i} + 2\hat{j} + 6\hat{k}$$ and $$\vec{b_{2}} = \hat{i} + 2\hat{j} + 2\hat{k}$$ respectively.

Therefore,

$$\left |\vec{b_{1}} \right | = \sqrt{3^{2} + 2^{2} + 6^{2}}$$= 7

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{2}} \right | = \sqrt{\left ( 1 \right )^{2} + \left ( 2 \right )^{2} + \left ( 2 \right )^{2}}$$ = 3

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}.\vec{b_{2}} = \left ( 3\hat{i} + 2\hat{j} + 6\hat{k} \right ).\left ( \hat{i} + 2\hat{j} + 2\hat{k} \right )$$

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}.\vec{b_{2}} = 3 \times 1 + 2 \times 2 + 6 \times 2$$ = 3 + 4 + 12 = 19

$$\\\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{19}{7 \times 3}$$

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

(ii) The given lines are parallel to the vectors, $$\vec{b_{1}} = \hat{i} – \hat{j} – 2\hat{k}$$ and $$\vec{b_{2}} = 3\hat{i} – 5\hat{j} – 4\hat{k}$$ respectively.

Therefore,

$$\left |\vec{b_{1}} \right | = \sqrt{\left (-1 \right )^{2} + \left (-1 \right )^{2} + \left (-2 \right )^{2}}= \sqrt{6}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{2}} \right | = \sqrt{\left (3 \right )^{2} + \left (-5 \right )^{2} + \left (-4 \right )^{2}}$$

$$\left |\vec{b_{2}} \right | = \sqrt{50}$$ $$= 5\sqrt{2}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}.\vec{b_{2}} = \left ( \hat{i} – \hat{j} – 2\hat{k} \right ).\left ( 3\hat{i} – 5\hat{j} – 4\hat{k}\right )$$

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}.\vec{b_{2}} = 1 \times 3 – 1 \left ( -5 \right ) – 2\left ( -4 \right )$$ = 3 + 5 + 8 = 16

$$\\\boldsymbol{\Rightarrow }$$ $$\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right | = \frac{16}{\sqrt{6}.5\sqrt{2}}= \frac{16}{\sqrt{2}.\sqrt{3}.5\sqrt{2}} = \frac{16}{10\sqrt{3}}$$

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

$$\\\boldsymbol{\Rightarrow }$$ $$\theta = \cos ^{-1}\left (\frac{8}{5\sqrt{3}} \right )$$

Q11. Calculate the angle between the given pairs of lines:

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

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

Sol:

(i) Let $$\vec{b_{1}}$$ and $$\vec{b_{2}}$$ be the vectors parallel to the pair of lines, $$\frac{x – 2}{2} = \frac{y – 1}{5} = \frac{z + 3}{-3}$$ and $$\frac{x + 2}{-1} = \frac{y – 4}{8} = \frac{z – 5}{4}$$

$$\vec{b_{1}} = 2\hat{i} + 5\hat{j} – 3\hat{k}$$ and

$$\vec{b_{2}} = -\hat{i} + 8\hat{j} + 4\hat{k}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{1}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( 5 \right )^{2} + \left ( -3 \right )^{2}}= \sqrt{38}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{2}} \right | = \sqrt{\left ( -1 \right )^{2} + \left ( 8 \right )^{2} + \left ( 4 \right )^{2}}= \sqrt{81}$$= 9

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}. \vec{b_{2}} = \left (2\hat{i} + 5\hat{j} – 3\hat{k} \right ).\left (-\hat{i} + 8\hat{j} + 4\hat{k} \right )$$

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}. \vec{b_{2}} = 2\left ( -1 \right ) + 5 \times 8 + \left ( -3 \right )4$$= -2 + 40 – 12 = 26

The angle $$\theta$$, between the given pair of lines is given by the relation,

$$\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right | = \frac{26}{9\sqrt{38}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\theta = \cos ^{-1}\left (\frac{26}{9\sqrt{38}} \right )$$

(ii) Let $$\vec{b_{1}}$$ and $$\vec{b_{2}}$$ be the vectors parallel to the given pair of lines, $$\frac{x }{2} = \frac{y}{2} = \frac{z}{1}$$ and $$\frac{x – 5}{4} = \frac{y – 2}{1} = \frac{z – 3}{8}$$ respectively.

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}} = 2\hat{i} + 2\hat{j} + \hat{k}$$ $$\;\;and \;\;\vec{b_{2}} = 4\hat{i} + \hat{j} + 8\hat{k}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{1}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( 2 \right )^{2} + \left ( 1 \right )^{2}}= \sqrt{9}$$= 3

$$\\\boldsymbol{\Rightarrow }$$ $$\left |\vec{b_{2}} \right | = \sqrt{\left ( 4 \right )^{2} + \left ( 1 \right )^{2} + \left ( 8 \right )^{2}} = \sqrt{81}$$ = 9

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}. \vec{b_{2}} = \left ( 2\hat{i} + 2\hat{j} + \hat{k} \right ).\left ( 4\hat{i} + \hat{j} + 8\hat{k} \right )$$

$$\\\boldsymbol{\Rightarrow }$$ $$\vec{b_{1}}. \vec{b_{2}} = 2 \times 4 + 2 \times 1 + 1 \times 8$$ = 8 + 2 + 8 = 18

The angle $$\theta$$, between the given pair of lines is given by the relation,

$$\\\boldsymbol{\Rightarrow }$$ $$\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right |= \frac{18}{3 \times 9}= \frac{2}{3}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\theta = \cos ^{-1}\left (\frac{2}{3} \right )$$

Q12. Find the values of m so the line $$\frac{1 – x}{3} = \frac{7y – 14}{2m} = \frac{z – 3}{2}$$ and $$\frac{7 – 7x}{3m} = \frac{y – 5}{1} = \frac{6 – z}{5}$$ are at right angles.

Sol:

The given equations can be written in the standard form as $$\frac{x – 1}{-3} = \frac{y – 2}{\frac{2m}{7}} = \frac{z – 3}{2}$$ and $$\frac{x – 1}{\frac{-3m}{7}} = \frac{y – 5}{1} = \frac{z – 6}{-5}$$

The direction ratios of the lines are -3, $$\frac{2m}{7}$$, 2 and $$\frac{-3m}{7}$$,1 , -5 respectively.

Two lines with direction ratios, $$a_{1}, b_{1}, c_{1}$$ and $$a_{2}, b_{2}, c_{2}$$ are perpendicular to each other, if $$a_{1}a_{1} + b_{1}b_{2} + c_{1}c_{2} = 0$$

$$\left ( -3 \right ).\left ( \frac{-3m}{7} \right ) + \left ( \frac{2m}{7} \right ). \left ( 1 \right ) + 2.\left ( -5 \right ) = 0$$ $$\frac{9m}{7} + \frac{2m}{7} = 10$$ $$11m = 70$$ $$m = \frac{70}{11}\\$$

Therefore, the value of m is $$\frac{70}{11}$$.

Q13. 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.

Sol:

The equations of the given lines are lines $$\frac{x – 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}$$ and $$\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$$.

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, $$a_{1}, b_{1}, c_{1}$$ and $$a_{2}, b_{2}, c_{2}$$ are perpendicular to each other, if $$a_{1}a_{1} + b_{1}b_{2} + c_{1}c_{2} = 0$$

$$\left (7 \times 1 \right ) + \left (-5 \times 2 \right ) + \left (1 \times 3 \right )\\$$

= 7 – 10 + 3 = 0

Hence, the given lines are perpendicular to each other.

Q14. Find the shortest distance between the given lines,

$$\vec{r} = \left ( \hat{i} + 2\hat{j} + \hat{k} \right ) + \lambda \left ( \hat{i} – \hat{j} + \hat{k} \right )$$ and $$\vec{r} = \left ( 2\hat{i} – \hat{j} – \hat{k} \right ) + \mu \left ( 2\hat{i} + \hat{j} + 2\hat{k} \right )$$

Sol:

The equations of the given lines are,

$$\vec{r} = \left ( \hat{i} + 2\hat{j} + \hat{k} \right ) + \lambda \left ( \hat{i} – \hat{j} + \hat{k} \right )$$ $$\vec{r} = \left ( 2\hat{i} – \hat{j} – \hat{k} \right ) + \mu \left ( 2\hat{i} + \hat{j} + 2\hat{k} \right )$$

It is known that the shortest distance between the lines, $$\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}$$ and $$\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}$$, is given by:

$$\\d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |$$ . . . . . . . . . . . . . (1)

Comparing the given equations, we obtain:

$$\vec{a_{1}} = \hat{i} + 2\hat{j} + \hat{k}$$ , $$\vec{b_{1}} = \hat{i} – \hat{j} + \hat{k}$$

And, $$\vec{a_{2}} = 2\hat{i} – \hat{j} – \hat{k}$$, $$\vec{b_{2}} = 2\hat{i} + \hat{j} + 2\hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{a_{2}} – \vec{a_{1}}= \left (2\hat{i} – \hat{j} – \hat{k} \right ) – \left ( \hat{i} + 2\hat{j} + \hat{k} \right )= \hat{i} – 3\hat{j} – 2\hat{k}\\$$

$$\\\vec{b_{1}} \times \vec{b_{2}}$$ = $$\boldsymbol{\begin{vmatrix} \hat{i}&\hat{j} &\hat{k}\\ 1& -1&1\\ 2& 1& 2\end{vmatrix}}\\$$

$$\vec{b_{1}} \times \vec{b_{2}} = \left ( -2 – 1 \right )\hat{i} – \left ( 2 – 2 \right )\hat{j} + \left ( 1 + 2 \right )\hat{k} = -3\hat{i} + 3\hat{k}$$

And, $$\left |\vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( -3 \right )^{2} + \left ( 3 \right )^{2}} = \sqrt{9 + 9}= 3\sqrt{2}$$

Substituting all the values in equation (1), we obtain:

$$d = \left | \frac{\left ( -3\hat{i} + 3\hat{k}\right ). \left ( \hat{i} – 3\hat{j} – 2\hat{k}\right )}{3\sqrt{2}} \right |= \left | \frac{-3.1 + 3.\left ( -2 \right )}{3\sqrt{2}} \right |\left | =\frac{-9}{3\sqrt{2}} \right |=\frac{3}{\sqrt{2}} =\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3\sqrt{2}}{2}$$

Hence, the shortest distance between the two lines is $$\frac{3\sqrt{2}}{2}$$ units.

Q15. Find the shortest distance between the lines $$\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$$ and $$\frac{x – 3}{1} = \frac{y – 5}{-2} = \frac{z – 7}{1}$$

Sol:

The given lines are as follows,

$$\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$$ and $$\frac{x – 3}{1} = \frac{y – 5}{-2} = \frac{z – 7}{1}$$

It is known that the shortest distance between the two lines, $$\frac{x – x_{1}}{a_{1}} = \frac{y – y_{1}}{b_{1}} = \frac{z – z_{1}}{c_{1}}$$ and $$\frac{x – x_{2}}{a_{2}} = \frac{y – y_{2}}{b_{2}} = \frac{z – z_{2}}{c_{2}}$$ is given by,

$$\boldsymbol{\Rightarrow \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} &c_{1} \\ a_{2}& b_{2} &c_{2} \end{vmatrix} }{\sqrt{(b_{1}c_{2}-b_{1}c_{1})^{2}-(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}}\\$$. . . . . . . . . . . (1)

Comparing the given equations, we obtain:

$$x_{1} = -1, y_{1} = -1, z_{1} = -1$$$$a_{1} = 7, b_{1} = -6, c_{1} = 1$$

And, $$x_{2} = 3, y_{2} = 5, z_{2} = 7$$$$a_{2} = 1, b_{2} = -2, c_{2} = 1$$

Then,

$$\boldsymbol{\Rightarrow \begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} &c_{1} \\ a_{2}& b_{2} &c_{2} \end{vmatrix}\;=\;\begin{vmatrix} 4 & 6 & 8\\ 7 & -6 & 8\\ 1 & -2 & 1 \end{vmatrix}}\\$$

= $$\\4\left ( -6 + 2 \right ) – 6\left ( 7 – 1 \right ) + 8\left ( -14 + 6 \right )$$

= $$– 16 – 36 – 64$$ = – 116

= $$\\\sqrt{\left ( b_{1}c_{2} – b_{2}c_{1} \right )^{2} + \left ( c_{1}a_{2} – c_{2}a_{1} \right )^{2} + \left (a_{1}b_{2} – a_{2}b_{1} \right )^{2} }\\$$

= $$\\\sqrt{\left ( -6 + 2 \right )^{2} + \left ( 1 + 7 \right )^{2} + \left ( – 14 + 6 \right )^{2}}=\sqrt{116}=2\sqrt{29}\\$$

Substituting all the values in equation (1), we obtain:

$$d = \frac{-116}{2\sqrt{29}}= \frac{-58}{\sqrt{29}}= \frac{-2 \times 29}{\sqrt{29}}= -2 \sqrt{29}$$

Since distance is always non-negative, the distance between the given lines is $$2 \sqrt{29}$$ units.

Q16. Find the shortest distance between the lines whose vector equations are $$\vec{r} = \left ( \hat{i} + 2\hat{j} + 3\hat{k} \right ) + \lambda \left ( \hat{i} – 3\hat{j} + 2\hat{k} \right )$$ and $$\vec{r} = \left ( 4\hat{i} + 5\hat{j} + 6\hat{k} \right ) + \mu \left ( 2\hat{i} + 3\hat{j} + \hat{k} \right )$$

Sol:

The given vectors are as follows:

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

It is known that the shortest distance between the lines, $$\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}$$ and $$\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}$$

The lines are given by:

$$d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |$$ . . . . . . . . . . . (1)

Comparing the given equations with $$\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}$$ and $$\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}$$

$$\\\vec{a_{1}} = \hat{i} + 2\hat{j} + 3\hat{k}$$, $$\vec{b_{1}} = \hat{i} – 3\hat{j} + 2\hat{k}$$, $$\vec{a_{2}} = 4\hat{i} + 5\hat{j} + 6\hat{k}$$, $$\vec{b_{2}} = 2\hat{i} + 3\hat{j} + \hat{k}$$

$$\vec{a_{2}} – \vec{a_{1}} = \left ( 4\hat{i} + 5\hat{j} + 6\hat{k} \right ) – \left ( \hat{i} + 2\hat{j} + 3\hat{k} \right )= 3\hat{i} + 3\hat{j} + 3\hat{k}\\$$ $$\\\vec{b_{1}} \times \vec{b_{2}} = \begin{vmatrix} \hat{i} \;\;\;\;\;\; \hat{j} \;\;\;\;\;\; \hat{k}\\ 1 \;\;\; -3 \;\;\;\; 2\\ 2 \;\;\;\;\;\; 3 \;\;\;\;\;\; 1 \end{vmatrix}$$

= $$\\\left ( -3 – 6 \right )\hat{i} – \left ( 1 – 4 \right )\hat{j} + \left ( 3 + 6 \right )\hat{k} = -9 \hat{i} + 3\hat{j} + 9\hat{k}\\$$

$$\\\left |\vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( -9 \right )^{2} + \left ( 3 \right )^{2} + \left ( 9 \right )^{2}}=\sqrt{81 + 9 + 81} =sqrt{171}=3\sqrt{19}\\$$ $$\\\left (\vec{b_{1}} \times \vec{b_{2}} \right ). \left (\vec{a_{2}} – \vec{a_{1}} \right ) = \left (-9 \hat{i} + 3\hat{j} + 9\hat{k} \right ). \left ( 3\hat{i} + 3\hat{j} + 3\hat{k} \right )$$

= $$\\\left (-9 \times 3 \right ) + \left ( 3 \times 3 \right ) + \left ( 9 \times 3 \right )$$ = 9

Substituting all the values in equation (1), we obtain:

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

Hence, the shortest distance between the two given lines is $$\frac{3}{\sqrt{19}}$$ units.

Q17. Find the shortest distance between the lines whose vector equations are $$\vec{r} = \left ( 1 – t \right )\hat{i} + \left ( t – 2 \right )\hat{j} + \left ( 3 – 2t \right )\hat{k}$$ and $$\vec{r} = \left ( s + 1 \right )\hat{i} + \left ( 2s – 1 \right )\hat{j} + \left ( 2s + 1 \right )\hat{k}$$

Sol:

The lines are as follows:

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

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

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

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

It is known that the shortest distance between the lines, $$\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}$$ and $$\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}$$ is given by:

$$\\d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |$$ . . . . . . . . (3)

Comparing the given equations with $$\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}$$ and $$\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}$$,

$$\vec{a_{1}} = \hat{i} – 2\hat{j} + 3\hat{k}$$, $$\vec{b_{1}} = -\hat{i} + \hat{j} – 2\hat{k}$$

And, $$\vec{a_{2}} = \hat{i} – \hat{j} – \hat{k}$$, $$\vec{b_{2}} = \hat{i} + 2\hat{j} – 2\hat{k}$$

$$\vec{a_{2}} – \vec{a_{1}} = \left ( \hat{i} – \hat{j} – \hat{k} \right ) – \left ( \hat{i} – 2\hat{j} + 3\hat{k} \right ) = \hat{j} – 4\hat{k}\\$$ $$\\\vec{b_{1}} \times \vec{b_{2}} = \begin{vmatrix} \hat{i} \;\;\;\;\;\; \hat{j} \;\;\;\;\;\; \hat{k}\\ -1 \;\;\; 1 \;\;\;\; -2\\ \;\;1 \;\;\;\;\;\; 2 \;\;\;\;-2 \end{vmatrix}\\$$

= $$\\\left ( -2 + 4 \right )\hat{i} – \left ( 2 + 2 \right )\hat{j} + \left ( -2 – 1 \right )\hat{k}= 2\hat{i} – 4\hat{j} – 3\hat{k}$$

$$\left | \vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( -4 \right )^{2} + \left ( -3 \right )^{2}}= \sqrt{4 + 16 + 9}= \sqrt{29}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\\\left (\vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}}\right ) = \left (2\hat{i} – 4\hat{j} – 3\hat{k} \right ). \left ( \hat{j} – 4\hat{k} \right )$$ = -4 + 12 = 8

Substituting all the values in equation (3), we obtain:

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

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