NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

Question:1 If a line makes angles ${90}^{\circ },{135}^{\circ },{45}^{\circ }$ with the x, y and z-axes respectively, find its direction cosines.

Answer:

Let the direction cosines of the line be l, m, and n.

So, we have

$l=\cos {90}^{\circ }=0$

$m=\cos {135}^{\circ }=-\frac{1}{\sqrt{2}}$

$n=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

Therefore the direction cosines of the lines are $0,-\frac{1}{\sqrt{2}},and\frac{1}{\sqrt{2}}$.

Question:2 Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:

If the line is making equal angle with the coordinate axes. Then,

Let the common angle made is $\alpha$ with each coordinate axis.

Therefore, we can write;

$l=\cos \alpha ,m=\cos \alpha ,andn=\cos \alpha$

And as we know the relation; $l^2+m^2+n^2=1$

$\Rightarrow {\cos }^2\alpha +{\cos }^2\alpha +{\cos }^2\alpha =1$

$\Rightarrow {\cos }^2\alpha =\frac{1}{3}$

or $\cos \alpha =\pm \frac{1}{\sqrt{3}}$

Thus the direction cosines of the line are $\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},and\pm \frac{1}{\sqrt{3}}$

Question:3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines?

Answer:

Given a line has direction ratios of -18, 12, – 4 then its direction cosines are;

A line having direction ratio -18 has direction cosine:

$\frac{-18}{\sqrt{(-18{)}^2+(12{)}^2+(-4{)}^2}}=\frac{-18}{22}=\frac{-9}{11}$

A line having direction ratio 12 has direction cosine:

$\frac{12}{\sqrt{(-18{)}^2+(12{)}^2+(-4{)}^2}}=\frac{12}{22}=\frac{6}{11}$

A line having direction ratio -4 has direction cosine:

$\frac{12}{\sqrt{(-4{)}^2+(12{)}^2+(-4{)}^2}}=\frac{-4}{22}=\frac{-2}{11}$

Thus, the direction cosines are $\frac{-9}{11},\frac{6}{11},\frac{-2}{11}$ .

Question:4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.

Answer:

We have the points, A (2, 3, 4), B (– 1, – 2, 1), C (5, 8, 7);

And as we can find the direction ratios of the line joining the points $(x_1,y_1,z_1)and(x_2,y_2,z_2)$ is given by $x_2-x_1,y_2-y_1,and{z}_2-z_1.$

The direction ratios of AB are $(-1-2),(-2-3),and(1-4)$ i.e., $-3,-5,and-3$

The direction ratios of BC are $(5-(-1)),(8-(-2)),and(7-1)$ i.e., $6,10,and6$ .

We can see that the direction ratios of AB and BC are proportional to each other and is -2 times.

$\therefore$ AB is parallel to BC. and as point B is common to both AB and BC,

Hence, points A, B and C are collinear.

Question:5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Answer:

Given vertices of the triangle $\mathrm{△}ABC$ (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

$\Rightarrow$ Direction ratios of side AB are $(-1-3),(1-5),and(2-(-4))$ i.e.,

$-4,-4,and6.$

Therefore its direction cosines values are;

$\frac{-4}{\sqrt{(-4{)}^2+(-4{)}^2+(6{)}^2}},\frac{-4}{\sqrt{(-4{)}^2+(-4{)}^2+(6{)}^2}},\frac{6}{\sqrt{(-4{)}^2+(-4{)}^2+(6{)}^2}}$ $or\frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}or\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

Similarly for side BC;

$\Rightarrow$ Direction ratios of side BC are $(-5-(-1)),(-5-1),and(-2-2)$ i.e.,

$-4,-6,and-4.$

Therefore its direction cosines values are;

$\frac{-4}{\sqrt{(-4{)}^2+(-6{)}^2+(-4{)}^2}},\frac{-6}{\sqrt{(-4{)}^2+(-6{)}^2+(-4{)}^2}},\frac{-4}{\sqrt{(-4{)}^2+(-6{)}^2+(-4{)}^2}}$ $or\frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}or\frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

$\Rightarrow$ Direction ratios of side CA are $(-5-3),(-5-5),and(-2-(-4))$ i.e.,

$-8,-10,and2.$

Therefore its direction cosines values are;

$\frac{-8}{\sqrt{(-8{)}^2+(10{)}^2+(2{)}^2}},\frac{-5}{\sqrt{(-8{)}^2+(10{)}^2+(2{)}^2}},\frac{2}{\sqrt{(-8{)}^2+(10{)}^2+(2{)}^2}}$ $or\frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}or\frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$

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.

Answer:

Given direction cosines of the three lines;

$L_1(\frac{12}{13},\frac{-3}{13},\frac{-4}{13})$ $L_2(\frac{4}{13},\frac{12}{13},\frac{3}{13})$ $L_3(\frac{3}{13},\frac{-4}{13},\frac{12}{13})$

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 $L_{1}and{L}_2$ ;

$l_1{l}_2+m_1{m}_2+n_1{n}_2=[\frac{12}{13}\times \frac{4}{13}]+[\frac{-3}{13}\times \frac{12}{13}]+[\frac{-4}{13}\times \frac{3}{13}]$

$\Rightarrow$$\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 $L_{2}and{L}_3$ ;

$l_1{l}_2+m_1{m}_2+n_1{n}_2=[\frac{4}{13}\times \frac{3}{13}]+[\frac{12}{13}\times \frac{-4}{13}]+[\frac{3}{13}\times \frac{12}{13}]$

$\Rightarrow$$\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 $L_{3}and{L}_1$ ;

$l_1{l}_2+m_1{m}_2+n_1{n}_2=[\frac{3}{13}\times \frac{12}{13}]+[\frac{-4}{13}\times \frac{-3}{13}]+[\frac{12}{13}\times \frac{-4}{13}]$

$\Rightarrow$$\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 that 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).

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 $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

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

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

(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:

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

(ii) 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 :

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

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

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

where $\overrightarrow{b_1}=\hat{i}-\hat{j}-2\hat{k}$ and $\overrightarrow{b_2}=3\hat{i}-5\hat{j}-4\hat{k}$ respectively,

Then we have

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

$=3+5+8=16$

and $|\overrightarrow{b_1}|=\sqrt{1^2+(-1{)}^2+(-2{)}^2}=\sqrt{6}$

$|\overrightarrow{b_2}|=\sqrt{3^2+(-5{)}^2+(-4{)}^2}=\sqrt{50}=5\sqrt{2}$

Therefore we have;

$\cos A=\left|\frac{16}{\sqrt{6}\times 5\sqrt{2}}\right|=\frac{16}{10\sqrt{3}}$

or $A={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right)$

Question:9 Find the angle between the following pair of lines:

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

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

Answer:

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

(ii) 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 have 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:10 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 $\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$