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

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).

Answer:

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 $3,2,and4$ .

Now, lines AB and CD will be perpendicular to each other if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$a_1{a}_2+b_1{b}_2+c_1{c}_2=(2\times 3)+(5\times 2)+(-4\times 4)$

$=6+10-16=0$

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).

Answer:

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 $2,4,and4$ .

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

Answer:

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

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

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

So, we have now,

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

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 $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$ .

Answer:

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

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

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

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

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

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

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 $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ .

Answer:

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 .

Answer:

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;

$\overrightarrow{a}=5\hat{i}-4\hat{j}+6\hat{k}$

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, $\overrightarrow{b}=3\hat{i}+7\hat{j}+2\hat{k}$ .

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

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

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\lambda ϵR$

So, we get the equation.

$\overrightarrow{r}=5\hat{i}-4\hat{j}+6\hat{k}+\lambda (3\hat{i}+7\hat{j}+2\hat{k}),\lambda ϵR$

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).

Answer:

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,

$\overrightarrow{a}=\overrightarrow{0}$

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, $\overrightarrow{b}=5\hat{i}-2\hat{j}+3\hat{k}$

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

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},where\lambda ϵR$

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

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

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).

Answer:

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;

$\overrightarrow{a}=3\hat{i}-2\hat{j}-5\hat{k}$

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,

$\overrightarrow{b}=0\hat{i}-0\hat{j}+11\hat{k}=11\hat{k}$

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

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},where\lambda ϵR$

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

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

Answer:

To find the angle A between the pair of lines $\overrightarrow{b_1}and\overrightarrow{b_2}$ we have the formula;

$\cos A=\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|$

We have two lines :

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

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

The given lines are parallel to the vectors $\overrightarrow{b_1}and\overrightarrow{b_2}$ ;

where $\overrightarrow{b_1}=3\hat{i}+2\hat{j}+6\hat{k}$ and $\overrightarrow{b_2}=\hat{i}+2\hat{j}+2\hat{k}$ respectively,

Then we have

$\overrightarrow{b_1}.\overrightarrow{b_2}=(3\hat{i}+2\hat{j}+6\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})$

$=3+4+12=19$

and $|\overrightarrow{b_1}|=\sqrt{3^2+2^2+6^2}=7$

$|\overrightarrow{b_2}|=\sqrt{1^2+2^2+2^2}=3$

Therefore we have;

$\cos A=\left|\frac{19}{7\times 3}\right|=\frac{19}{21}$

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

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

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

Answer:

To find the angle A between the pair of lines $\overrightarrow{b_1}and\overrightarrow{b_2}$ we have the formula;

$\cos A=\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|$

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

Answer:

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 $\overrightarrow{b_1}and\overrightarrow{b_2}$ which are parallel to the pair of above lines respectively.

$\overrightarrow{b_1}=2\hat{i}+5\hat{j}-3\hat{k}$ and $\overrightarrow{b_2}=-\hat{i}+8\hat{j}+4\hat{k}$

To find the angle A between the pair of lines $\overrightarrow{b_1}and\overrightarrow{b_2}$ we have the formula;

$\cos A=\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|$

Then we have

$\overrightarrow{b_1}.\overrightarrow{b_2}=(2\hat{i}+5\hat{j}-3\hat{k}).(-\hat{i}+8\hat{j}+4\hat{k})$

$=-2+40-12=26$

and $|\overrightarrow{b_1}|=\sqrt{2^2+5^2+(-3{)}^2}=\sqrt{38}$

$|\overrightarrow{b_2}|=\sqrt{(-1{)}^2+(8{)}^2+(4{)}^2}=\sqrt{81}=9$

Therefore we have;

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

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

Answer:

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 $\overrightarrow{b_1}and\overrightarrow{b_2}$ which are parallel to the pair of above lines respectively.

$\overrightarrow{b_1}=2\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b_2}=4\hat{i}+\hat{j}+8\hat{k}$

To find the angle A between the pair of lines $\overrightarrow{b_1}and\overrightarrow{b_2}$ we have the formula;

$\cos A=\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\right|$

Then we have

$\overrightarrow{b_1}.\overrightarrow{b_2}=(2\hat{i}+2\hat{j}+\hat{k}).(4\hat{i}+\hat{j}+8\hat{k})$

$=8+2+8=18$

and $|\overrightarrow{b_1}|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$

$|\overrightarrow{b_2}|=\sqrt{(4{)}^2+(1{)}^2+(8{)}^2}=\sqrt{81}=9$

Therefore we have;

$\cos A=\left|\frac{18}{3\times 9}\right|=\frac{2}{3}$

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.

Answer:

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.

Answer:

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

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

Answer:

So given equation of lines;

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

Now, we can find the shortest distance between the lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ , is given by the formula,

$d=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

Now comparing the values from the equation, we obtain

$\overrightarrow{a_1}=\hat{i}+2\hat{j}+\hat{k}$ $\overrightarrow{b_1}=\hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{a_2}=2\hat{i}-\hat{j}-\hat{k}$ $\overrightarrow{b_2}=2\hat{i}+\hat{j}+2\hat{k}$

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

Then calculating

$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|$

$\overrightarrow{b_1}\times \overrightarrow{b_2}=(-2-1)\hat{i}-(2-2)\hat{j}+(1+2)\hat{k}=-3\hat{i}+3\hat{k}$

$\Rightarrow |\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{(-3{)}^2+(3{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

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

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

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

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

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

Question:15 Find the shortest distance between the lines

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

Answer:

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

$d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}}$

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

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=\left|\begin{array}{ccc}4& 6& 8\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|$

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

$=-16-36-64$

$=-116$

Now calculating the denominator,

$\sqrt{(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2+(a_1{b}_2-a_2{b}_1{)}^2}=\sqrt{(-6+2{)}^2+(1+7{)}^2+(-14+6{)}^2}$ $=\sqrt{16+36+64}$