# NCERT Solutions for Exercise 11.2 Class 12 Maths Chapter 11- Three Dimensional Geometry

NCERT solutions for exercise 11.2 Class 12 Maths chapter 11 is about the lines in space and the equation of these lines in cartesian and vector form. NCERT solutions for Class 12 Maths chapter 11 exercise 11.2 also covers the topic of the distance between lines. The topics covered in the exercise 11.2 Class 12 Maths are very important if the CBSE Class 12 Maths Previous Paper is considered. The main focus of solving Class 12 Maths chapter 11 exercise 11.2 should be to check whether the concepts are grasped or not. For these first giving multiple tries to solve Class 12th Maths chapter 11 exercise, 11.2 should be given and then if confused in any steps, look at the NCERT solutions for Class 12 Maths chapter 11 exercise 11.2.

• Three Dimensional Geometry 11.1

• Three Dimensional Geometry 11.3

• Three Dimensional Geometry Miscellaneous Exercise

## Question: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.

GIven direction cosines of the three lines;

$L_{1}\ \left ( \frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \right )$ $L_{2}\ \left ( \frac{4}{13}, \frac{12}{13},\frac{3}{13} \right )$ $L_{3}\ \left ( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \right )$

And we know that two lines with direction cosines $l_{1},m_{1},n_{1}$ and $l_{2},m_{2},n_{2}$ are perpendicular to each other, if $l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0$

Hence we will check each pair of lines:

Lines $L_{1}\ and\ L_{2}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{12}{13}\times\frac{4}{13} \right ]+\left [ \frac{-3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times \frac{3}{13} \right ]$

$= \left [ \frac{48}{169} \right ]-\left [ \frac{36}{169} \right ]-\left [ \frac{12}{169} \right ]= 0$

$\therefore$ the lines $L_{1}\ and\ L_{2}$ are perpendicular.

Lines $L_{2}\ and\ L_{3}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{4}{13}\times\frac{3}{13} \right ]+\left [ \frac{12}{13}\times\frac{-4}{13} \right ]+\left [ \frac{3}{13}\times \frac{12}{13} \right ]$

$= \left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]+\left [ \frac{36}{169} \right ]= 0$

$\therefore$ the lines $L_{2}\ and\ L_{3}$ are perpendicular.

Lines $L_{3}\ and\ L_{1}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times\frac{-3}{13} \right ]+\left [ \frac{12}{13}\times \frac{-4}{13} \right ]$

$= \left [ \frac{36}{169} \right ]+\left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]= 0$

$\therefore$ the lines $L_{3}\ and\ L_{1}$ are perpendicular.

Thus, we have all lines are mutually perpendicular to each other.

Question: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).

We have given points where the line is passing through it;

Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and line joining the points (0, 3, 2) and (3, 5, 6).is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are $a_{1},b_{1}, c_{1}$

$(3-1),\ (4-(-1)),\ and\ (-2-2)$ or $2,\ 5,\ and\ -4$

Direction ratios of CD are $a_{2},b_{2}, c_{2}$

$(3-0),\ (5-3)),\ and\ (6-2)$ or $3,\ 2,\ and\ 4$ .

Now, lines AB and CD will be perpendicular to each other if $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =0$

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =\left ( 2\times3 \right ) +\left ( 5\times2 \right )+ \left ( -4\times 4 \right )$

$= 6+10-16 = 0$

Therefore, AB and CD are perpendicular to each other.

Question: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).

We have given points where the line is passing through it;

Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and line joining the points (– 1, – 2, 1) and (1, 2, 5)..is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are $a_{1},b_{1}, c_{1}$

$(2-4),\ (3-7),\ and\ (4-8)$ or $-2,\ -4,\ and\ -4$

Direction ratios of CD are $a_{2},b_{2}, c_{2}$

$(1-(-1)),\ (2-(-2)),\ and\ (5-1)$ or $2,\ 4,\ and\ 4$ .

Now, lines AB and CD will be parallel to each other if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Therefore we have now;

$\frac{a_{1}}{a_{2}} = \frac{-2}{2}=-1$ $\frac{b_{1}}{b_{2}} = \frac{-4}{4}=-1$ $\frac{c_{1}}{c_{2}} = \frac{-4}{4}=-1$

$\therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Hence we can say that AB is parallel to CD.

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

It is given that the line is passing through A (1, 2, 3) and is parallel to the vector $\vec{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}$

We can easily find the equation of the line which passes through the point A and is parallel to the vector $\vec{b}$ by the known relation;

$\vec{r} = \vec{a} +\lambda\vec{b}$ , where $\lambda$ is a constant.

So, we have now,

$\\\mathrm{\Rightarrow \vec{r} = \widehat{i}+2\widehat{j}+3\widehat{k} + \lambda(3\widehat{i}+2\widehat{j}-2\widehat{k})}$

Thus the required equation of the line.

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

Given that the line is passing through the point with position vector $2\widehat{i}-\widehat{j}+4\widehat{k}$ and is in the direction of the line $\widehat{i}+2\widehat{j}-\widehat{k}$ .

And we know the equation of the line which passes through the point with the position vector $\vec{a}$ and parallel to the vector $\vec{b}$ is given by the equation,

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

$\Rightarrow \vec{r} =2\widehat{i}-\widehat{j}+4\widehat{k} + \lambda(\widehat{i}+2\widehat{j}-\widehat{k})$

So, this is the required equation of the line in the vector form.

$\vec{r} =x\widehat{i}+y\widehat{j}+z\widehat{k} = (\lambda+2)\widehat{i}+(2\lambda-1)\widehat{j}+(-\lambda+4)\widehat{k}$

Eliminating $\lambda$ , from the above equation we obtain the equation in the Cartesian form :

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

Hence this is the required equation of the line in Cartesian form.

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

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ ;

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

So, the required line is parallel to the above line.

Therefore we can take direction ratios of the required line as 3k , 5k , and 6k , where k is a non-zero constant.

And we know that the equation of line passing through the point $(x_{1},y_{1},z_{1})$ and with direction ratios a, b, c is written by: $\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}$ .

Therefore we have the equation of the required line:

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

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

The required line equation.

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

Given the Cartesian equation of the line;

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}$

Here the given line is passing through the point $(5,-4,6)$ .

So, we can write the position vector of this point as;

$\vec{a} = 5\widehat{i}-4\widehat{j}+6\widehat{k}$

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

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

Now, we can easily find the required equation of line:

As we know that the line passing through the position vector $\vec{a}$ and in the direction of the vector $\vec{b}$ is given by the relation,

$\vec{r} = \vec{a} + \lambda \vec{b},\ \lambda \epsilon R$

So, we get the equation.

$\vec{r} = 5\widehat{i}-4\widehat{j}+6\widehat{k} + \lambda(3\widehat{i}+7\widehat{j}+2\widehat{k}),\ \lambda \epsilon R$

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

Question:8 Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).

GIven that the line is passing through the $(0,0,0)$ and $(5,-2,3)$

Thus the required line passes through the origin.

$\therefore$ its position vector is given by,

$\vec{a} = \vec{0}$

So, the direction ratios of the line through $(0,0,0)$ 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, $\vec{b} = 5\widehat{i}-2\widehat{j}+3\widehat{k}$

Therefore the equation of the line passing through the point with position vector $\vec{a}$ and parallel to $\vec{b}$ is given by;

$\vec{r} = \vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R$

$\Rightarrow\vec{r} = 0+\lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})$

$\Rightarrow\vec{r} = \lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})$

Now, the equation of the line through the point $(x_{1},y_{1},z_{1})$ and the direction ratios a, b, c is given by;

$\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}$

Therefore the equation of the required line in the Cartesian form will be;

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

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

Question:9 Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

Let the line passing through the points $A(3,-2,-5)$ and $B(3,-2,6)$ is AB;

Then as AB passes through through A so, we can write its position vector as;

$\vec{a} =3\widehat{i}-2\widehat{j}-5\widehat{k}$

Then direction ratios of PQ are given by,

$(3-3)= 0,\ (-2+2) = 0,\ (6+5)=11$

Therefore the equation of the vector in the direction of AB is given by,

$\vec{b} =0\widehat{i}-0\widehat{j}+11\widehat{k} = 11\widehat{k}$

We have then the equation of line AB in vector form is given by,

$\vec{r} =\vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R$

$\Rightarrow \vec{r} = (3\widehat{i}-2\widehat{j}-5\widehat{k}) + 11\lambda\widehat{k}$

So, the equation of AB in Cartesian form is;

$\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}$

or $\frac{x-3}{0} =\frac{y+2}{0} =\frac{z+5}{11}$

Question:10 Find the angle between the following pairs of lines:

(i) $\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k})$ and $\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

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

We have two lines :

$\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k})$ and

$\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})$

The given lines are parallel to the vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ ;

where $\vec{b_{1}}= 3\widehat{i}+2\widehat{j}+6\widehat{k}$ and $\vec{b_{2}}= \widehat{i}+2\widehat{j}+2\widehat{k}$ respectively,

Then we have

$\vec{b_{1}}.\vec{b_{2}} =(3\widehat{i}+2\widehat{j}+6\widehat{k}).(\widehat{i}+2\widehat{j}+2\widehat{k})$

$=3+4+12 = 19$

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

$|\vec{b_{2}}| = \sqrt{1^2+2^2+2^2} = 3$

Therefore we have;

$\cos A = \left | \frac{19}{7\times3} \right | = \frac{19}{21}$

or $A = \cos^{-1} \left ( \frac{19}{21} \right )$

Question:10 Find the angle between the following pairs of lines:

(ii) $\overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k})$ and $\overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

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

We have two lines :

Question:11 Find the angle between the following pair of lines:

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

Given lines are;

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

So, we two vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ which are parallel to the pair of above lines respectively.

$\vec{b_{1}}\ =2\widehat{i}+5\widehat{j}-3\widehat{k}$ and $\vec{b_{2}}\ =-\widehat{i}+8\widehat{j}+4\widehat{k}$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

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

Then we have

$\vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+5\widehat{j}-3\widehat{k}).(-\widehat{i}+8\widehat{j}+4\widehat{k})$

$=-2+40-12 = 26$

and $|\vec{b_{1}}| = \sqrt{2^2+5^2+(-3)^2} = \sqrt{38}$

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

Therefore we have;

$\cos A = \left | \frac{26}{\sqrt{38} \times9} \right | = \frac{26}{9\sqrt{38}}$

or $A = \cos^{-1} \left ( \frac{26}{9\sqrt{38}} \right )$

Question:11 Find the angle between the following pair of lines:

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

Given lines are;

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

So, we two vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ which are parallel to the pair of above lines respectively.

$\vec{b_{1}}\ =2\widehat{i}+2\widehat{j}+\widehat{k}$ and $\vec{b_{2}}\ =4\widehat{i}+\widehat{j}+8\widehat{k}$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

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

Then we have

$\vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+2\widehat{j}+\widehat{k}).(4\widehat{i}+\widehat{j}+8\widehat{k})$

$=8+2+8 = 18$

and $|\vec{b_{1}}| = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3$

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

Therefore we have;

$\cos A = \left | \frac{18}{ 3\times9} \right | = \frac{2}{3}$

or $A = \cos^{-1} \left ( \frac{2}{3} \right )$

Question:12 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 we have to write the given equation of lines in the standard form;

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

Then we have the direction ratios of the above lines as;

$-3,\ \frac{2p}{7},\ 2$ and $\frac{-3p}{7},\ 1,\ -5$ 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, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0$

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

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

$\Rightarrow 11p =70$

$\Rightarrow p =\frac{70}{11}$

Thus, the value of p is $\frac{70}{11}$ .

Question:13 Show that the lines $\frac{x-5}{7}=\frac{y+2}{-3}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

First, we have to write the given equation of lines in the standard form;

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

Then we have the direction ratios of the above lines as;

$7,\ -5,\ 1$ 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, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0$

$\therefore 7(1) + (-5)(2)+1(3) = 7-10+3 = 0$

Therefore the two lines are perpendicular to each other.

Question:14 Find the shortest distance between the lines

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k})$ and $\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})$

So given equation of lines;

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k})$ and $\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})$ in the vector form.

Now, we can find 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 the formula,

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

Now comparing the values from the equation, we obtain

$\vec{a_{1}} = \widehat{i}+2\widehat{j}+\widehat{k}$ $\vec{b_{1}} = \widehat{i}-\widehat{j}+\widehat{k}$

$\vec{a_{2}} = 2\widehat{i}-\widehat{j}-\widehat{k}$ $\vec{b_{2}} = 2\widehat{i}+\widehat{j}+2\widehat{k}$

$\vec{a_{2}} -\vec{a_{1}} =\left ( 2\widehat{i}-\widehat{j}-\widehat{k} \right ) - \left ( \widehat{i}+2\widehat{j}+\widehat{k} \right ) = \widehat{i}-3\widehat{j}-2\widehat{k}$

Then calculating

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

$\vec{b_{1}}\times \vec{b_{2}} = (-2-1)\widehat{i} - (2-2) \widehat{j} +(1+2) \widehat{k} = -3\widehat{i}+3\widehat{k}$

$\Rightarrow \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} =\sqrt{18} =3\sqrt2$

So, substituting the values now in the formula above we get;

$d =\left | \frac{\left ( -3\widehat{i}+3\widehat{k} \right ).(\widehat{i}-3\widehat{j}-2\widehat{k})}{3\sqrt2} \right |$

$\Rightarrow d = \left | \frac{-3.1+3(-2)}{3\sqrt2} \right |$

$d = \left | \frac{-9}{3\sqrt2} \right | = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}$

Therefore, the shortest distance between the two lines is $\frac{3\sqrt2}{2}$ units.

Question:15 Find the shortest distance between the lines

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

We have given two lines:

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

Calculating 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 $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$

by the formula

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

Now, comparing the given equations, we obtain

$x_{1} = -1,\ y_{1} =-1,\ z_{1} =-1$

$a_{1} = 7,\ b_{1} =-6,\ c_{1} =1$

$x_{2} = 3,\ y_{2} =5,\ z_{2} =7$

$a_{2} = 1,\ b_{2} =-2,\ c_{2} =1$

Then calculating determinant

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

$= 4(-6+2)-6(7-1)+8(-14+6)$

$= -16-36-64$

$=-116$

Now calculating the denominator,

$\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2} = \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}$ $= \sqrt{16+36+64}$

$= \sqrt{116} = 2\sqrt{29}$

So, we will substitute all the values in the formula above to obtain,

$d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = \frac{-2\times29}{\sqrt{29}} = -2\sqrt{29}$

Since distance is always non-negative, the distance between the given lines is

$2\sqrt{29}$ units.

Question:16 Find the shortest distance between the lines whose vector equations are $\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k})$ and

$\overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})$

Given two equations of line

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k})$ $\overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})$ in the vector form.

So, we will apply the distance formula for knowing the distance between two lines $\vec{r} =\vec{a_{1}}+\lambda{b_{1}}$ and $\vec{r} =\vec{a_{2}}+\lambda{b_{2}}$

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

After comparing the given equations, we obtain

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

$\vec{a_{2}} = 4\widehat{i}+5\widehat{j}+6\widehat{k}$ $\vec{b_{2}} = 2\widehat{i}+3\widehat{j}+\widehat{k}$

$\vec{a_{2}}-\vec{a_{1}} = (4\widehat{i}+5\widehat{j}+6\widehat{k}) - (\widehat{i}+2\widehat{j}+3\widehat{k})$

$= 3\widehat{i}+3\widehat{j}+3\widehat{k}$

Then calculating the determinant value numerator.

$\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1& -3 &2 \\ 2& 3& 1 \end{vmatrix}$

$= (-3-6)\widehat{i}-(1-4)\widehat{j}+(3+6)\widehat{k} = -9\widehat{i}+3\widehat{j}+9\widehat{k}$

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

$= \sqrt{81+9+81} = \sqrt{171} =3\sqrt{19}$

$\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(-9\widehat{i}+3\widehat{j}+9\widehat{k})(3\widehat{i}+3\widehat{j}+3\widehat{k})$

$= (-9\times3)+(3\times3)+(9\times3) = 9$

Now, after substituting the value in the above formula we get,

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

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

Question:17 Find the shortest distance between the lines whose vector equations are

$\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}$ and $\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}$

Given two equations of the line

$\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}$ $\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}$ in the vector form.

So, we will apply the distance formula for knowing the distance between two lines $\vec{r} =\vec{a_{1}}+\lambda{b_{1}}$ and $\vec{r} =\vec{a_{2}}+\lambda{b_{2}}$

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

After comparing the given equations, we obtain

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

$\vec{a_{2}} = \widehat{i}-\widehat{j}-\widehat{k}$ $\vec{b_{2}} = \widehat{i}+2\widehat{j}-2\widehat{k}$

$\vec{a_{2}}-\vec{a_{1}} = (\widehat{i}-\widehat{j}-\widehat{k}) - (\widehat{i}-2\widehat{j}+3\widehat{k}) = \widehat{j}-4\widehat{k}$

Then calculating the determinant value numerator.

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

$= (-2+4)\widehat{i}-(2+2)\widehat{j}+(-2-1)\widehat{k} = 2\widehat{i}-4\widehat{j}-3\widehat{k}$

That implies,

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

$= \sqrt{4+16+9} = \sqrt{29}$

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

Now, after substituting the value in the above formula we get,

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

Therefore, $\frac{8}{\sqrt{29}}$ units are the shortest distance between the two given lines.

## More About NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Ten solved questions are given prior to the exercise 11.2 Class 12 Maths. And 17 questions are covered in Class 12 Maths chapter 11 exercise 11.2. Broadly speaking Class 12th Maths chapter 11 exercise 11.2 covers questions related to the equation of a line parallel to a given vector and that passes through a given point, the line passing through two given points, the angle between lines, the smallest distance between two lines and distance between the skew lines and parallel lines.

Also Read| Three Dimensional Geometry Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

• Solving the given NCERT examples for topic 11.3 and exercise 11.2 Class 12 Maths help to score well in the exam

• Question of similar type in Class 12th Maths chapter 11 exercise 11.2 can be expected for CBSE board exam

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 11

• NCERT Solutions for Class 12 Maths Chapter 11

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology