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.

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