Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.5

Read

Discuss

Question 1. Find the shortest distance between the pair of lines whose vector equation is:

(i) $(\vec{r}=3\hat{i}+8\hat{j}+3\hat{k}+λ(3\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=-3\hat{i}-7\hat{j}+6\hat{k}+μ(-3\hat{i}+2\hat{j}+4\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=-3\hat{i}-7\hat{j}+6\hat{k}-(3\hat{i}+8\hat{j}+3\hat{k})$

= $-6\hat{i}-15\hat{j}+3\hat{k}$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\3&-1&1\\-3&2&4\\\end{array}\right|$

= $\hat{i}(-4-2)-\hat{j}(12+3)+\hat{k}(6-3)$

= $-6\hat{i}-15\hat{j}+3\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-6\hat{i}-15\hat{j}+3\hat{k}).(-6\hat{i}-15\hat{j}+3\hat{k})$

= 36 + 225 + 9

= 270

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-6)^2+(-15)^2+(3)^2}$

= $\sqrt{36+225+9}$

= √270

On substituting the values in the formula, we have

SD = 270/√270

= √270

Shortest distance between the given pair of lines is 3√30 units.

(ii) $\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+λ(\hat{i}-2\hat{j}+7\hat{k})$ and $\vec{r}=-\hat{i}-\hat{j}-\hat{k}+μ(7\hat{i}-6\hat{j}+\hat{k}).$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=-\hat{i}-\hat{j}-\hat{k}-(3\hat{i}+5\hat{j}+7\hat{k})$

= $-4\hat{i}-6\hat{j}-8\hat{k}$

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

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&7\\7&-6&1\\\end{array}\right|$

= $8(5\hat{i}+6\hat{j}+\hat{k})$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-2(2\hat{i}+3\hat{j}+4\hat{k})).(8(5\hat{i}+6\hat{j}+\hat{k}))$

= – 16 × 32

= – 512

$|\vec{b_1}×\vec{b_2}|=8\sqrt{(5)^2+(6)^2+(1)^2}$

= $8\sqrt{25+36+1}$

= $8\sqrt{62}$

On substituting the values in the formula, we have

SD = $|\frac{-512}{8\sqrt{62}}|$

Shortest distance between the given pair of lines is $|\frac{512}{\sqrt{3968}}|$ units.

(iii) $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(2\hat{i}+3\hat{j}+4\hat{k})$ and $\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+μ(3\hat{i}+4\hat{j}+5\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})$

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

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\\\end{array}\right|$

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

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+2\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

= 1

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(2)^2+(-1)^2}$

= $\sqrt{6}$

On substituting the values in the formula, we have

SD = $|\frac{1}{\sqrt6}|$

Shortest distance between the given pair of lines is 1/√6 units.

(iv) $\vec{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-t)\hat{k}$ and $\vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$

Solution:

Above equations can be re-written as:

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

and, $\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$

and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D = $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

= 9/3√2

Shortest distance is 3/√2 units.

(v) $\vec{r}=(λ-1)\hat{i}+(λ+1)\hat{j}-(1+λ)\hat{k}$ and $\vec{r}=(1-μ)\hat{i}+(2μ-1)\hat{j}+(μ+2)\hat{k}$

Solution:

The given equations can be written as:

\ $\vec{r}=(-\hat{i}+\hat{j}-\hat{k})+λ(\hat{i}+\hat{j}-\hat{k})$ and $\vec{r}=\hat{i}-\hat{j}+2\hat{k}+µ(-\hat{i}+2\hat{j}+\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(2\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}+3\hat{k})$

= 15

$|\vec{b_1}×\vec{b_2}|=\sqrt{(3)^2+(3)^2}$

= 3√2

Thus, distance between the lines is $|\frac{15}{3\sqrt2}| = \frac{5}{\sqrt2}$ units.

(vi) $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+λ(2\hat{i}-5\hat{j}+2\hat{k})$ and $\vec{r}=(\hat{i}+2\hat{j}+\hat{k})+μ(\hat{i}-\hat{j}+\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D = $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})$

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(3)^2}$

= 3√2

Substituting the values in the formula, we have

The distance between the lines is $|\frac{9}{3\sqrt2}| = \frac{3}{\sqrt2}$ units.

(vii) $\vec{r}=\hat{i}+\hat{j}+λ(2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2\hat{i}+\hat{j}-\hat{k}+µ(3\hat{i}-5\hat{j}+2\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})$

= 10

Substituting the values in the formula, we have:

The distance between the lines is 10/√59 units.

(viii) $\vec{r}=(8+3λ)\hat{i}-(9+16λ)\hat{j}+(10+7λ)\hat{k}$ and $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+µ(3\hat{i}+8\hat{j}-5\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})$

= 1176

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(3)^2}$

= 84

Substituting the values in the formula, we have:

The distance between the lines is 1176/84 = 14 units.

Question 2. Find the shortest distance between the pair of lines whose cartesian equation is:

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

Solution:

The given lines can be written as:

$\vec{r}=\hat{i}+2\hat{j}+\hat{k}+λ(2\hat{i}+4\hat{j}+3\hat{k})$ and $\vec{r}=2\hat{i}+3\hat{j}+5\hat{k}+µ(3\hat{i}+4\hat{j} +5\hat{k})$

$\vec{a_2}-\vec{a_1}=2\hat{i}+3\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})$

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

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\\\end{array}\right|$

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

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

= –1

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(2)^2+(-1)^2}$

= √6

On substituting the values in the formula, we have:

SD = 1/√6 units.

(ii) $\frac{x-1}{2}=\frac{y+1}{3}=z$ and $\frac{x+1}{3}=\frac{y-2}{1};z=2$

Solution:

The given equations can also be written as:

$\vec{r}=\hat{i}-\hat{j}+λ(2\hat{i}+3\hat{j}+\hat{k})$ and \ $\vec{r}=-\hat{i}+2\hat{j}+2\hat{k}+μ(3\hat{i}+\hat{j})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&1\\3&1&0\\\end{array}\right|$

= $-\hat{i}+3\hat{j}-7\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

= 3

SD = 3/√59 units.

(iii) $\frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}$ and $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}$

Solution:

The given equations can be re-written as:

$\vec{r}=\hat{i}-2\hat{j}+3\hat{k}+ λ(-\hat{i}+\hat{j}+2\hat{k})$ and $\vec{r}=\hat{i}-\hat{j}-\hat{k}+µ(\hat{i}+2\hat{j}-2\hat{k})$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\-1&1&2\\1&2&-2\\\end{array}\right|$

= √29

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

= 8

SD = 8/√29 units.

(iv) $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{-1}$ and $\vec{r}=\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$

Solution:

The given equations can be re-written as:

$\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+λ(\hat{i}-2\hat{j}+\hat{k})$ and $\vec{r}=(-\hat{i}-\hat{j}-\hat{k})+µ(7\hat{i}-6\hat{j}+\hat{k})$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&1\\7&-6&1\\\end{array}\right|$

= $4\hat{i}+6\hat{j}+8\hat{k}$

SD = 58/√29 units.

Question 3. By computing the shortest distance determine whether the pairs of lines intersect or not:

(i) $\vec{r}=(\hat{i}-\hat{j})+λ(2\hat{i}+\hat{k})$ and $\vec{r}=2\hat{i}-\hat{j}+µ(\hat{i}+\hat{j}-\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&0&1\\1&1&-1\\\end{array}\right|$

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

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

= –1

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(3)^2+(2)^2}$

= √14

⇒ SD = 1/√14 units ≠ 0

Hence the given pair of lines does not intersect.

(ii) $\vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})$ and $\vec{r}=(4\hat{i}-\hat{k})+µ(2\hat{i}+3\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\3&-1&0\\2&0&3\\\end{array}\right|$

= $-3\hat{i}-9\hat{j}+2\hat{k}\vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j}) \vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(3\hat{i}-\hat{j}).(-3\hat{i}-9\hat{j}+2\hat{k})$

= 0

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(-9)^2+(2)^2}$

= √94

⇒ SD = 0/√94 units = 0

Hence the given pair of lines are intersecting.

(iii) $\frac{x-1}{2}=\frac{y+1}{3}=z$ and $\frac{x+1}{5}=\frac{y-2}{1};z=2$

Solution:

Given lines can be re-written as:

$\vec{r}=(\hat{i}-\hat{j})+λ(2\hat{i}+3\hat{j}+\hat{k})$ and $\vec{r}=(-\hat{i}+2\hat{j}+2\hat{k})+μ(5\hat{i}+\hat{j})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&1\\5&1&0\\\end{array}\right|$

= $-\hat{i}+5\hat{j}-13\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-2\hat{i}+3\hat{j}+2\hat{k}).(-\hat{i}+5\hat{j}-13\hat{k})$

= −9

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(5)^2+(-13)^2}$

= √195

⇒ SD = 9/√195 units ≠ 0

Hence the given pair of lines does not intersect.

(iv) $\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$

Solution:

Given lines can be re-written as:

$\vec{r}=(5\hat{i}+7\hat{j}-3\hat{k})+λ(4\hat{i}-5\hat{j}-5\hat{k})$ and $\vec{r}=(8\hat{i}+7\hat{j}+5\hat{k})+μ(7\hat{i}+\hat{j}+3\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\4&5&-5\\7&1&3\\\end{array}\right|$

= $-10\hat{i}-47\hat{j}+39\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(3\hat{i}+8\hat{k}).(-10\hat{i}-47\hat{j}+39\hat{k})$

= 282

⇒ SD = 282/√3 units ≠ 0

Hence the given pair of lines does not intersect.

Question 4. Find the shortest distance between the following:

(i) $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(-\hat{i}+\hat{j}-\hat{k})$

Solution:

The second given line can be re-written as: $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(\hat{i}-\hat{j}+\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|$

$(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-3&-4\\1&-1&1\\\end{array}\right|$

= $-7\hat{i}-5\hat{j}+2\hat{k}$

$|(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(-7)^2+(-5)^2+2^2}$

= $\sqrt{78}$

$\vec{|b|}=\sqrt3$

⇒ SD = $\frac{\sqrt{78}}{\sqrt{3}} = \sqrt{26}$ units.

(ii) $\vec{r}=(\hat{i}+\hat{j})+λ(2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+µ(4\hat{i}-2\hat{j}+2\hat{k})$

Solution:

The second given line can be re-written as: $\vec{r}=(2\hat{i}+\hat{j}-\hat{k})+µ'(2\hat{i}-\hat{j}+\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|$

$(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&0&-1\\2&-1&1\\\end{array}\right|$

= $-\hat{i}-3\hat{j}-\hat{k}$

⇒ $|(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(-1)^2+(-3)^2+(-1)^2}$

= √11

$\vec{|b|}=\sqrt6$

⇒ SD = √11/√6 units.

Question 5. Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines:

(i) (0, 0, 0) and (1, 0, 2) (ii) (1, 3, 0) and (0, 3, 0)

Solution:

Equation of the line passing through the vertices (0, 0, 0) and (1, 0, 2) is given by:

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

Similarly, the equation of the line passing through the vertices (1, 3, 0) and (0, 3, 0):

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

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&0&2\\-1&0&0\\\end{array}\right|$

= $-2\hat{j}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+3\hat{j}).(-2\hat{j})$

= −6

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-2)^2}$

= 2

⇒ SD = |-6/2| = 3 units.

Question 6. Write the vector equations of the following lines and hence find the shortest distance between them:

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6};\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}$

Solution:

The given equations can be written as:

$\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r}=(3\hat{i}+3\hat{j}-5\hat{k})+µ2\hat{i}+3\hat{j}+6\hat{k})$

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|$

$(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\\\end{array}\right|$

= $9\hat{i}-14\hat{j}+4\hat{k}$

⇒ $|(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(9)^2+(-14)^2+(4)^2}$

= $\sqrt{293}$

\vec{|b|}= 7

⇒ SD = √293/7 units.

Question 7. Find the shortest distance between the following:

(i) $\vec{r}=\hat{i}+2\hat{j}+\hat{k}+λ(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(2\hat{i}+\hat{j}+2\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=2\hat{i}-\hat{j}-\hat{k}-\hat{i}-2\hat{j}-\hat{k}$

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

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-1&1\\2&1&2\\\end{array}\right|$

= $-3\hat{i}+3\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=9$

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(-3)^2}$

= 3√2

⇒ SD = 3/√2 units.

(ii) $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1};\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=4\hat{i}+6\hat{j}+8\hat{k}$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\7&-6&1\\1&-2&1\\\end{array}\right|$

= $-4\hat{i}-6\hat{k}-8\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=-116$

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(-3)^2}$

= √116

⇒ SD = 2√29 units.

(iii) $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(\hat{i}-3\hat{j}+2\hat{k})$ and $\vec{r}=4\hat{i}+5\hat{j}+6\hat{k}+μ(2\hat{i}+3\hat{j}+\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=-9\hat{i}+3\hat{j}+9\hat{k}$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-3&2\\2&3&1\\\end{array}\right|$

= $3\hat{i}+3\hat{k}+3\hat{k}$

$(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=9$

$|\vec{b_1}×\vec{b_2}|=\sqrt{(-9)^2+(3)^2+(9)^2}$

= √171

⇒ SD = 3√19 units.

(iv) $\vec{r}=(6\hat{i}+2\hat{j}+2\hat{k})+λ(\hat{i}-2\hat{j}+2\hat{k})$ and $\vec{r}=-4\hat{i}-\hat{k}+μ(3\hat{i}-2\hat{j}-2\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|$

Now, $\vec{a_2}-\vec{a_1}=-10\hat{i}-2\hat{j}-3\hat{k}$

$(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&2\\3&-2&-2\\\end{array}\right|$

= $8\hat{i}+8\hat{k}+4\hat{k}$

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=108

|\vec{b_1}×\vec{b_2}|=\sqrt{(-9)^2+(3)^2+(9)^2}

= 12

⇒ SD = 9 units.

Question 8. Find the distance between the lines: $\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r}=3\hat{i}+3\hat{j}-5\hat{k}+μ(2\hat{i}+3\hat{j}+6\hat{k})$

Solution:

As we know that the shortest distance between the lines $\vec{r}=\vec{a_1}+λ(\vec{b_1})$ and $\vec{r}=\vec{a_2}+μ(\vec{b_2})$ is:

D= $|\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|$

$(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\\\end{array}\right|$

= $9\hat{i}-14\hat{j}+4\hat{k}$

⇒ $|(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(9)^2+(-14)^2+(4)^2}$

= √293

$\vec{|b|}=7$

⇒ SD = √293/7 units.

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Last Updated : 16 Jun, 2021

Class 12 RD Sharma Solutions- Chapter 28 The Straight Line in Space - Exercise 28.4

Class 12 RD Sharma Solutions - Chapter 29 The Plane - Exercise 29.1

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.5

Question 1. Find the shortest distance between the pair of lines whose vector equation is:

(i) and

(ii) and

(iii) and

(iv) and

(v) and

(vi) and

(vii) and

(viii) and

Question 2. Find the shortest distance between the pair of lines whose cartesian equation is:

(i) and

(ii) and

(iii) and

(i) $(\vec{r}=3\hat{i}+8\hat{j}+3\hat{k}+λ(3\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=-3\hat{i}-7\hat{j}+6\hat{k}+μ(-3\hat{i}+2\hat{j}+4\hat{k})$

(ii) $\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+λ(\hat{i}-2\hat{j}+7\hat{k})$ and $\vec{r}=-\hat{i}-\hat{j}-\hat{k}+μ(7\hat{i}-6\hat{j}+\hat{k}).$

(iii) $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(2\hat{i}+3\hat{j}+4\hat{k})$ and $\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+μ(3\hat{i}+4\hat{j}+5\hat{k})$

(iv) $\vec{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-t)\hat{k}$ and $\vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$

(v) $\vec{r}=(λ-1)\hat{i}+(λ+1)\hat{j}-(1+λ)\hat{k}$ and $\vec{r}=(1-μ)\hat{i}+(2μ-1)\hat{j}+(μ+2)\hat{k}$

(vi) $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+λ(2\hat{i}-5\hat{j}+2\hat{k})$ and $\vec{r}=(\hat{i}+2\hat{j}+\hat{k})+μ(\hat{i}-\hat{j}+\hat{k})$

(vii) $\vec{r}=\hat{i}+\hat{j}+λ(2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2\hat{i}+\hat{j}-\hat{k}+µ(3\hat{i}-5\hat{j}+2\hat{k})$

(viii) $\vec{r}=(8+3λ)\hat{i}-(9+16λ)\hat{j}+(10+7λ)\hat{k}$ and $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+µ(3\hat{i}+8\hat{j}-5\hat{k})$

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

(ii) $\frac{x-1}{2}=\frac{y+1}{3}=z$ and $\frac{x+1}{3}=\frac{y-2}{1};z=2$

(iii) $\frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}$ and $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}$

(iv) $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{-1}$ and $\vec{r}=\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$

(i) $\vec{r}=(\hat{i}-\hat{j})+λ(2\hat{i}+\hat{k})$ and $\vec{r}=2\hat{i}-\hat{j}+µ(\hat{i}+\hat{j}-\hat{k})$

(ii) $\vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})$ and $\vec{r}=(4\hat{i}-\hat{k})+µ(2\hat{i}+3\hat{k})$

(iii) $\frac{x-1}{2}=\frac{y+1}{3}=z$ and $\frac{x+1}{5}=\frac{y-2}{1};z=2$

(iv) $\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}$

(i) $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(-\hat{i}+\hat{j}-\hat{k})$

(ii) $\vec{r}=(\hat{i}+\hat{j})+λ(2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+µ(4\hat{i}-2\hat{j}+2\hat{k})$

(i) $\vec{r}=\hat{i}+2\hat{j}+\hat{k}+λ(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(2\hat{i}+\hat{j}+2\hat{k})$

(ii) $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1};\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

(iii) $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(\hat{i}-3\hat{j}+2\hat{k})$ and $\vec{r}=4\hat{i}+5\hat{j}+6\hat{k}+μ(2\hat{i}+3\hat{j}+\hat{k})$

(iv) $\vec{r}=(6\hat{i}+2\hat{j}+2\hat{k})+λ(\hat{i}-2\hat{j}+2\hat{k})$ and $\vec{r}=-4\hat{i}-\hat{k}+μ(3\hat{i}-2\hat{j}-2\hat{k})$

Question 8. Find the distance between the lines: $\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})$ and $\vec{r}=3\hat{i}+3\hat{j}-5\hat{k}+μ(2\hat{i}+3\hat{j}+6\hat{k})$