NCERT Solutions Maths Class 12 Chapter 11 Three Dimensional Geometry: Questions with Answers PDF

Exercise 11.1

 

Q4. Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

A.4. Given,

A(2,3,4), B(-1,-2,1), C(5,8,7)

Direction ratio of AB=  $\begin{array}{l}\left(-1-2\right),\left(-2-3\right),\left(1-4\right)\\ =\left(3,-5,-3\right)\end{array}$

Where, a₁=3, b₁=-5, c₁=-3

Direction ratio of BC=  $\begin{array}{l}\left(5-\left(-1\right)\right),\left(8-\left(-2\right)\right),\left(7-1\right)\\ =\left(6,10,6\right)\end{array}$

Where, a₂=6, b₂=10, c₂=6

Now,

$\begin{array}{l}\frac{a_2}{a_1}=\frac{6}{-3}=-2\\ \frac{b_2}{b_1}=\frac{10}{-5}=-2\\ \frac{c_2}{c_1}=\frac{6}{-3}=-2\end{array}$

Here, direction ratio of two-line segments are proportional.

So, A, B, C are collinear.

 

Q5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2)

A.5. The vertices of U ABC are A(3,5,-4), B(-1,1,2) and C(-5,-5,-2)

Direction ratio of side AB  $\begin{array}{l}=\left(-1-3\right)\left(1-5\right)\left(2-(-4)\right)\\ =\left(-4,-4,6\right)\end{array}$

Direction cosine of AB,

 

Example 11.2

Q1. 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.

A.1. Two lines with direction cosines l, m, n and l₂, m₂, n₂ are perpendicular to each other, if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$ .

Now, for the 3 lines with direction cosine,

$\begin{array}{l}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}and\frac{4}{13},\frac{12}{13},\frac{3}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{12}{13}\times \frac{4}{13}+\frac{\left(-3\right)}{13}\times \frac{12}{13}+\frac{\left(-4\right)}{13}\times \frac{3}{13}\\ =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ =0\end{array}$

Hence, the lines are perpendicular.

For lines with direction cosines,

$\begin{array}{l}\frac{4}{13},\frac{12}{13},\frac{3}{13}and\frac{3}{13},\frac{-4}{13},\frac{12}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{4}{13}\times \frac{3}{13}+\frac{12}{13}\times \frac{\left(-4\right)}{13}+\frac{3}{13}\times \frac{12}{13}\\ =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

For the lines with direction cosines,

$\begin{array}{l}\frac{3}{13},\frac{-4}{13},\frac{12}{13}and\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{3}{13}\times \frac{12}{13}+\frac{\left(-4\right)}{13}\times \frac{\left(-3\right)}{13}+\frac{12}{13}\times \frac{\left(-4\right)}{13}\\ =\frac{36+12-48}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

Therefore, all the lines are 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).

A2. Let AB be the line joining the points  $\left(1,-1,2\right)$ and  $\left(3,4,-2\right)$ and CD be the line joining the point  $\left(0,3,2\right)$ and  $\left(3,5,6\right)$ .

The direction ratios, a, b, c of AB are  $\begin{array}{l}\left(3-1\right),\left(4-\left(-1\right)\right),\left(-2-2\right)\\ =\left(2,5,-4\right)\end{array}$

The direction ratios  $a_2,b_2,c_2$ of CD are  $\begin{array}{l}\left(3-0\right),\left(5-3\right),\left(6-2\right)\\ =\left(3,2,4\right)\end{array}$

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

$\begin{array}{l}=2\times 3+5\times 2+\left(-4\right)\times 4\\ =6+10-16\\ =0\end{array}$

$\therefore$ Therefore, AB and CD are perpendicular to each other.

 

Q3. Show that the line through points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

A.3. Let AB be the line through the point  $\left(4,7,8\right)$ and  $\left(2,3,4\right)$ and CD be line through the point  $\left(-1,-2,1\right)$ and  $\left(1,2,5\right)$

Direction cosine,  $a_1,b_1,c_1$ of AB are

$\begin{array}{l}=\left(2-4\right),\left(3-7\right),\left(4-8\right)\\ =\left(-2,-4,-4\right)\end{array}$

Direction cosine,  $a_2,b_2,c_2$ of CD are

$\begin{array}{l}=\left(1-\left(-1\right)\right),\left(2-\left(-2\right)\right),\left(5-1\right)\\ =\left(2,4,4\right)\end{array}$

AB will be parallel to CD only

If

$\begin{array}{l}\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\ \Rightarrow \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}\\ \Rightarrow -1=-1=-1\end{array}$

Here,  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Therefore, AB is parallel to CD.

 

Q4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector   $3\hat{i}+2\hat{j}-2\hat{k}$

A.4. Given,

The line passes through the point  $A\left(1,2,3\right)$ .

Position vector of A,

$\stackrel{\to }{a}=\hat{i}+2\hat{j}+3\hat{k}$

Let  $\stackrel{\to }{b}=3\hat{i}+2\hat{j}-2\hat{k}$

The line which passes through point  $\stackrel{\to }{a}$ and parallel to  $\stackrel{\to }{b}$ is given by,

$\begin{array}{l}\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\\ =\hat{i}+2\hat{j}+3\hat{k}+\lambda \left(3\hat{i}+2\hat{j}-2\hat{k}\right)\end{array}$ , where  $\lambda$ is constant

 

Q5. 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   $\hat{i}+2\hat{j}-\hat{k}$

A5. The line passes through the point with position vector,  $\stackrel{\to }{a}=2\hat{i}-\hat{j}+4\hat{k}-----(1)$

The given vector:  $\stackrel{\to }{b}=\hat{i}+2\hat{j}-\hat{k}-----(2)$

The line which passes through a point with position vector  $\stackrel{\to }{a}$ and parallel to  $\stackrel{\to }{b}$ is given by,

$\begin{array}{l}\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\\ \stackrel{\to }{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right)\end{array}$

$\therefore$ This is required equation of the line in vector form.

Now,

Let  $\begin{array}{l}\stackrel{\to }{r}=x\hat{i}-y\hat{j}+z\hat{k}\\ \Rightarrow x\hat{i}-y\hat{j}+z\hat{k}=\left(\lambda +2\right)\hat{i}+\left(2\lambda -1\right)\hat{j}+\left(-\lambda +4\right)\hat{k}\end{array}$

Comparing the coefficient to eliminate  $\lambda$ ,

$\begin{array}{l}x=\lambda +2,x_1=2,a=1\\ y=2\lambda -1,y_1=-1,b=2\\ z=-\lambda +4,z_1=4,c=-1\end{array}$

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

 

Q6. Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by   $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

A.6. Given,

The point  $\left(-2,4,-5\right)$ .

The Cartesian equation of a line through a point  $\left(x_1,y_1,z_1\right)$ and having direction ratios a, b, c is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Now, given that

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

to point  $\left(-2,4,-5\right)$

Here, the point  $\left(x_1,y_1,z_1\right)$ is  $\left(-2,4,-5\right)$ and the direction ratio is given by  $a=3,b=5,c=6$

$\therefore$ The required Cartesian equation is

$\begin{array}{l}\frac{x-\left(-2\right)}{3}=\frac{y-4}{5}=\frac{z-\left(-5\right)}{6}\\ \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\end{array}$

 

Q7. The Cartesian equation of a line is    $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ Write its vector form.

A.7. Given,

Cartesian equation,

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

The given line passes through the point  $\left(5,-4,6\right)$

i.e. position vector of  $\stackrel{\to }{a}=5\hat{i}-4\hat{j}+6\hat{k}$

Direction ratio are 3, 7 and 2.

Thus, the required line passes through the point  $\left(5,-4,6\right)$ and is parallel to the vector  $3\hat{i}+7\hat{j}+2\hat{k}$ .

Let  $\stackrel{\to }{r}$ be the position vector of any point on the line, then the vector equation of the line is given by,

$\stackrel{\to }{r}=\left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\lambda \left(3\hat{i}+7\hat{j}+2\hat{k}\right)$

 

Q8. Find the vector and Cartesian equations of the line that passes through the origin and (5, −2, 3).

A.8. The required line passes through the origin. Therefore, its position vector is given by,

$\stackrel{\to }{a}=0.........(1)$

The direction ratios of the line through origin and  $\left(5,-2,3\right)$ are

$\left(5-0\right)=5,\left(-2-0\right)=-2,\left(3-0\right)=3$

The line is parallel to the vector given by the equation,   $\stackrel{\to }{b}=5\hat{i}-2\hat{j}+3\hat{k}$

The equation of the line in vector form through a point with position vector   $\stackrel{\to }{a}$ and parallel to   $\stackrel{\to }{b}$ is,   $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b},\lambda \in R$

$\begin{array}{l}\Rightarrow \stackrel{\to }{r}=\stackrel{\to }{0}+\lambda \left(5\hat{i}-2\hat{j}+3\hat{k}\right)\\ \Rightarrow \stackrel{\to }{r}=\lambda \left(5\hat{i}-2\hat{j}+3\hat{k}\right)\end{array}$

The equation of the line through the point  $(x_1, y_1, z_1)$ and 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 is

$\begin{array}{l}\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\\ \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\end{array}$

 

Q9. Find vector and Cartesian equations of the line that passes through the points  $\left(3,-2,-5\right),\left(3,-2,6\right).$

A.9. Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

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

The direction ratios of PQ are given by,

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

The equation of the vector in the direction of PQ is

$\stackrel{\to }{b}=0.\hat{i}-0.\hat{j}+11\hat{k}=11\hat{k}$

The equation of PQ in vector form is given by,   $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b},\lambda \in R$

$\Rightarrow \stackrel{\to }{r}=\left(5\hat{i}-2\hat{j}-5\hat{k}\right)+11\lambda \hat{k}$

The equation of PQ in Cartesian form is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ i.e,

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

 

Q10. Find the angle between the following pairs of lines:

$\begin{array}{l}(i)\stackrel{\to }{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda \left(3\hat{i}-2\hat{j}+6\hat{k}\right)\&\stackrel{\to }{r}=7\hat{i}-6\hat{k}+\mu \left(\hat{i}+2\hat{j}+2\hat{k}\right)\\ (ii)\stackrel{\to }{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda \left(\hat{i}-\hat{j}-2\hat{k}\right)\&\stackrel{\to }{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu \left(\widehat{3i}-5\hat{j}-4\hat{k}\right)\end{array}$

A.10.

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,   $cosQ=\left|\frac{{\stackrel{\to }{b}}_1.{\stackrel{\to }{b}}_2}{\left|{\stackrel{\to }{b}}_1\right|\left|{\stackrel{\to }{b}}_2\right|}\right|$

The given lines are parallel to the vectors,   ${\stackrel{\to }{b}}_1=3\hat{i}+2\hat{j}+6\hat{k}\&{\stackrel{\to }{b}}_2=\hat{i}+2\hat{j}+2\hat{k}$ , respectively.

 

Q11. Find the angle between the following pair of lines

$\begin{array}{l}(i)\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\&\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\\ (ii)\frac{x}{y}=\frac{y}{2}=\frac{z}{1}\&\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\end{array}$

A.11. (i) Let   ${\stackrel{\to }{b}}_1$  and   ${\stackrel{\to }{b}}_2$  be the vectors parallel to the pair of lines,   $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\&\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ , respectively. $\begin{array}{l}{\stackrel{}{}}_{}\end{array}$

If Q is the angle between the given pair of lines, then  $cosQ=\left|\frac{{\stackrel{\to }{b}}_1.{\stackrel{\to }{b}}_2}{\left|{\stackrel{\to }{b}}_1\right|\left|{\stackrel{\to }{b}}_2\right|}\right|$

$\begin{array}{l}\Rightarrow cosQ=\frac{18}{3\times 9}=\frac{2}{3}\\ \Rightarrow Q=co{s}^{-1}\left(\frac{2}{3}\right)\end{array}$

 

Q12. Find the values of p so that the lines   $\frac{1-x}{3}=\frac{7y-14}{2}p=\frac{z-3}{2}and\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$   are at right angles.

A.12.  $\frac{1-x}{3}=\frac{7y-14}{2\rho }=\frac{z-3}{2}and\frac{7-7x}{3\rho }=\frac{y-5}{1}=\frac{6-z}{5}$

The standard form of a pair of Cartesian lines is;

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}and\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}-----(1)$

So,

$\begin{array}{l}\frac{-\left(x-1\right)}{3}=\frac{7\left(y-5\right)}{2\rho }=\frac{z-3}{2}and\frac{-7\left(x-1\right)}{3\rho }=\frac{y-5}{1}=\frac{-\left(z-6\right)}{5}\\ \frac{x-1}{-3}=\frac{y-2}{2\rho /7}=\frac{z-3}{2}and\frac{x-1}{-3\rho /7}=\frac{y-5}{1}=\frac{z-6}{-5}-----(2)\end{array}$

Comparing (1) and (2) we get

$\begin{array}{l}{a}_1=-3,b_1=\frac{2\rho }{7},c_1=2\\ a_2=\frac{-3\rho }{7},b_2=1,c_2=-5\end{array}$

Now, both the lines are at right angles

So,  $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}\Rightarrow \left(-3\right)\times \frac{\left(-3\rho \right)}{7}+\frac{2\rho }{7}\times 1+2\times \left(-5\right)=0\\ \Rightarrow \frac{9\rho }{7}+\frac{2\rho }{7}+\left(-10\right)=0\\ \Rightarrow \frac{9\rho +2\rho }{7}=10\\ \Rightarrow 11\rho =70\\ \rho =\frac{70}{11}\end{array}$

$\therefore$ The value of  $\rho$ 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.

A.13.  $\begin{array}{l}\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\\ \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\end{array}$

Direction ratios of given lines are (7,-5,1) and (1,2,3).

i.e.,  $\begin{array}{l}{a}_1=7,b_1=-5,c_1=1\\ a_2=1,b_2=2,c_2=3\end{array}$

Now,

$\begin{array}{l}=a_1{a}_2+b_1{b}_2+c_1{c}_2\\ =7\times 1+\left(-5\right)\times 2+1\times 3\\ =7-10+3\\ =10-10\\ =0\end{array}$

$\therefore$ These two lines are perpendicular to each other.

 

Q14. Find the shortest distance between the  $\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)and\stackrel{\to }{r}=2\hat{i}+\hat{j}+\hat{k}+\mu \left(2\hat{i}+\hat{j}+2\hat{k}\right)$ lines 

A.14.  $\begin{array}{l}\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)and-----(1)\\ \stackrel{\to }{r}=2\hat{i}-\hat{j}-\hat{k}+\mu \left(2\hat{i}+\hat{j}+2\hat{k}\right)-----(2)\end{array}$

Solution. Comparing (1) and (2) with  $\stackrel{\to }{r}={\stackrel{\to }{a}}_1+\lambda {\stackrel{\to }{b}}_1$ and  $\stackrel{\to }{r}={\stackrel{\to }{a}}_2+\mu {\stackrel{\to }{b}}_2$ respectively.

We get,

$\begin{array}{l}{a}_1=\hat{i}+2\hat{j}+\hat{k},b_1=\hat{i}-\hat{j}+\hat{k}\\ a_2=2\hat{i}-\hat{j}-\hat{k},b_2=2\hat{i}+\hat{j}+2\hat{k}\end{array}$

Therefore,

$\begin{array}{l}{\stackrel{\to }{a}}_2-{\stackrel{\to }{a}}_1=\hat{i}-3\hat{j}-2\hat{k}\\ {\stackrel{\to }{b}}_1\times {\stackrel{\to }{b}}_2=\left(\hat{i}-\hat{j}+\hat{k}\right)\times \left(2\hat{i}+\hat{j}+2\hat{k}\right)\end{array}$

 

Q15. Find the shortest distance between the lines   $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}and\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ .

A.15.  $\begin{array}{l}\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\\ \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\end{array}$

Shortest distance between two lines is given by,

$\begin{array}{l}{x}_1=-1,y_1=-1,z_1=-1\\ a_1=7,b_1=-6,c_1=1\\ x_2=3,y_2=5,z_2=7\\ a_2=1,b_2=-2,c_2=1\end{array}$

Then,

$\begin{array}{l}=4\left(-6+2\right)-6\left(7-1\right)+8\left(-14+6\right)\\ =-16-36-64\\ =-116\end{array}$

 

Q16. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l}\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}+3\hat{j}+2\hat{k}\right)\\ and\stackrel{\to }{r}=4\hat{i}+5\hat{j}+6\hat{k}+\mu \left(2\hat{i}+3\hat{j}+\hat{k}\right)\end{array}$

A.16.  $\begin{array}{l}\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}-3\hat{j}+2\hat{k}\right)and-----(1)\\ \stackrel{\to }{r}=4\hat{i}+5\hat{j}+6\hat{k}+\mu \left(2\hat{i}+3\hat{j}+\hat{k}\right)-----(2)\end{array}$

Here, comparing (1) and (2) with  $\stackrel{\to }{r}={\stackrel{\to }{a}}_1+\lambda {\stackrel{\to }{b}}_1$ and  $\stackrel{\to }{r}={\stackrel{\to }{a}}_2+\mu {\stackrel{\to }{b}}_2$ , we have

$\begin{array}{l}{a}_1=\hat{i}+2\hat{j}+3\hat{k},b_1=\hat{i}-3\hat{j}+2\hat{k}\\ a_2=4\hat{i}+5\hat{j}+6\hat{k},b_2=2\hat{i}+3\hat{j}+\hat{k}\end{array}$

 

Q17. Find the shortest distance between the lines whose vector equations are 

$\begin{array}{l}\stackrel{\to }{r}=\left(1-t\right)\hat{i}+\left(t-2\right)\hat{j}=\left(3-2t\right)\hat{k}and\\ \stackrel{\to }{r}=\left(s+1\right)\hat{i}+\left(2s-1\right)\hat{j}=\left(2s+1\right)\hat{k}\end{array}$

A.17.

$\begin{array}{l}\stackrel{\to }{r}=\left(1-t\right)\hat{i}+\left(t-2\right)\hat{j}+\left(3-2t\right)\hat{k}\\ \stackrel{\to }{r}=\hat{i}-t\hat{i}+t\hat{j}-2\hat{j}+3\hat{k}-2t\hat{k}\\ \stackrel{\to }{r}=\hat{i}-2\hat{j}+3\hat{k}+t\left(-\hat{i}+\hat{j}-2\hat{k}\right)-----(1)\end{array}$

$\begin{array}{l}\Rightarrow \stackrel{\to }{r}=\left(s+1\right)\hat{i}+\left(2s-1\right)\hat{j}-\left(2s+1\right)\hat{k}\\ \stackrel{\to }{r}=s\hat{i}+\hat{i}+2s\hat{j}-\hat{j}-2s\hat{k}-\hat{k}\\ \stackrel{\to }{r}=\hat{i}-\hat{j}-\hat{k}+s\left(\hat{i}+2\hat{j}-2\hat{k}\right)-----(2)\end{array}$

Here,

 

EXERCISE 11.3

Q1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$\begin{array}{}\end{array}\left(a\right)z=2$

$\left(b\right)x+y+z=1$

$\left(c\right)2x+3y-z=5$

$\left(d\right)5y+8=0$

A.1. (a) The equation of the plane is   $z =2or0x +0y + z =2..........\left(1\right)$

The direction ratios of normal are  $0,0,and1.$

$\therefore \surd 02+02 +12 =1$

Dividing both sides of equation (1) by 1, we obtain

$0.x+0.y+1.z=2$

This is of the form   $lx + my + nz = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b)   $x + y + z =1..........\left(1\right)$

The direction ratios of normal are 1, 1, and 1.

 

Q2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector   $3\hat{i}+5\hat{j}-6\hat{k}$

 

Q3. Find the Cartesian equation of the following planes:

$\begin{array}{l}(a)\stackrel{\to }{r}.\left(\hat{i}+\hat{j}-\hat{k}\right)=2\\ (b)\stackrel{\to }{r}.\left(2\hat{i}+3\hat{j}-4\hat{k}\right)=1\\ (c)\stackrel{\to }{r}.\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]=15\end{array}$

A.3. (a) It is given that equation of the plane is

$\stackrel{\to }{r}.\left(\hat{i}+\hat{j}-\hat{k}\right)=2$

For any arbitrary point  $P\left(x, y, z\right)$ on the plane, position vector   $\stackrel{\to }{r}$ I s given by,   $\stackrel{\to }{r}=x\hat{i}+y\hat{j}-z\hat{k}$

Substituting the value of   $\stackrel{\to }{r}$ in equation (1), we obtain

$\left(x\hat{i}+y\hat{j}-z\hat{k}\right).\left(\hat{i}+\hat{j}-\hat{k}\right)=2$

$\Rightarrow x+y-z=2$

This is the Cartesian equation of the plane.

(b)   $\stackrel{\to }{r}.\left(2\hat{i}+3\hat{j}-4\hat{k}\right)=1$

For any arbitrary point  $P\left(x, y, z\right)$ on the plane, position vector   $\stackrel{\to }{r}$ is given by,   $\stackrel{\to }{r}=\left(x\hat{i}+y\hat{j}-z\hat{k}\right)$

Substituting the value of   $\stackrel{\to }{r}$ in equation (1), we obtain

$\left(x\hat{i}+y\hat{j}-z\hat{k}\right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)=1$

$\Rightarrow 2x+3y-4z=1$

This is the Cartesian equation of the plane.

(c)   $\stackrel{\to }{r}.\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]=15$

For any arbitrary point  $P\left(x, y, z\right)$ on the plane, position vector   $\stackrel{\to }{r}$ is given by,   $\stackrel{\to }{r}=\left(x\hat{i}+y\hat{j}-z\hat{k}\right)$

Substituting the value of   $\stackrel{\to }{r}$ in equation (1), we obtain

$\begin{array}{l}\left(x\hat{i}+y\hat{j}-z\hat{k}\right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]=15\\ \Rightarrow (s-2t)x+(3-t)y+(2s+t)z=15\end{array}$

This is the Cartesian equation of the given plane.

 

Q4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:

$\begin{array}{}\end{array}\left(a\right) 2x+3y+4z-12=0$

$<br>\mathrm{(b})3y+4z-6=0$

$\left(c\right)x+y+z=1$

$\left(d\right)5y+8=0$

A.4. (a) Let the coordinates of the foot of perpendicular P from the origin to the plane be  $(x_1, y_1, z_1).$

$\begin{array}{l}2x + 3y + 4z - 12 = 0\hfill \end{array}$

$\begin{array}{l}\Rightarrow 2x + 3y + 4z = 12 \dots \left(1\right)\hfill \end{array}$

 

Q5. Find the vector and Cartesian equations of the planes

(a) that passes through the point  $\left(1,0,-2\right)$ and the normal to the plane is   $\hat{i}-\hat{j}-\hat{k}$ .

(b) that passes through the point  $\left(1,4,6\right)$ and the normal vector to the plane is   $\hat{i}-2\hat{j}+\hat{k}$ .

A.5. (a) The position vector of point  $\left(1,0,-2\right)$ is   $\stackrel{\to }{a}=\hat{i}-2\hat{k}$

The normal vector N perpendicular to the plane is   $\stackrel{\to }{N}=\hat{i}+\hat{j}-\hat{k}$

The vector equation of the plane is given by,   $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\stackrel{\to }{N}=0$

$\Rightarrow \left[\stackrel{\to }{r}-\left(\hat{i}-2\hat{k}\right)\right].\left(\hat{i}+\hat{j}-\hat{k}\right)=0.........(1)$

$\stackrel{\to }{r}$ is the position vector of any point  $\left(x, y, z\right)$ in the plane.

$\therefore \stackrel{\to }{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left[\left(x\hat{i}+y\hat{j}+z\hat{k}\right)-\left(\hat{i}-2\hat{k}\right)\right].\left(\hat{i}+\hat{j}-\hat{k}\right)=0\\ \Rightarrow \left[(x-1)\hat{i}+y\hat{j}+(z+2)\hat{k}\right].\left(\hat{i}+\hat{j}-\hat{k}\right)=0\\ \Rightarrow (x-1)+y-(z+2)=0\\ \Rightarrow x+y-z-3=0\\ \Rightarrow x+y-z=3\end{array}$

This is the Cartesian equation of the required plane.

(b) The position vector of the point  $\left(1,4,6\right)$ is   $\stackrel{\to }{a}=\hat{i}+4\hat{j}+6\hat{k}$

The normal vector   $\stackrel{\to }{N}$ perpendicular to the plane is   $\stackrel{\to }{N}=\hat{i}-2\hat{j}+\hat{k}$

The vector equation of the plane is given by,   $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\stackrel{\to }{N}=0$

$\Rightarrow \left[\stackrel{\to }{r}-\left(\hat{i}+4\hat{j}+6\hat{k}\right)\right].\left(\hat{i}-2\hat{j}+\hat{k}\right)=0.........(1)$

$\stackrel{\to }{r}$ is the position vector of any point  $P\left(x, y, z\right)$ in the plane.

$\therefore \stackrel{\to }{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left[\left(x\hat{i}+y\hat{j}+z\hat{k}\right)-\left(\hat{i}+4\hat{j}+6\hat{k}\right)\right].\left(\hat{i}-2\hat{j}+\hat{k}\right)=0\\ \Rightarrow \left[(x-1)\hat{i}+(y-4)\hat{j}+(z-6)\hat{k}\right].\left(\hat{i}-2\hat{j}+\hat{k}\right)=0\\ \Rightarrow (x-1)+2(y-4)+(z-6)=0\\ \Rightarrow x-2y+z+1=0\end{array}$

This is the Cartesian equation of the required plane.

 

Q6. Find the equations of the planes that passes through three points:

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

A.6. We know that through three collinear points  $A,B,C$ i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are  $A\left(1,1,-1\right),B\left(6,4,-5\right),andC\left(-4,-2,3\right).$

$\begin{array}{l}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 6\hfill & \hfill 4\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=(12-10)-(18-20)-(-12+16)\\ \end{array}$

$\begin{array}{l}=2+2-4\\ =0\end{array}$

Since  $A,B,C$ are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are  $A\left(1,1,0\right),B\left(1,2,1\right),andC\left(-2,2,-1\right).$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=(-2-2)-(2+2)=-8\ne 0$

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points,   $(x_1, y_1, z_1),(x_2, y_2, z_2)\&(x_3, y_3, z_3)$ , is

$\begin{array}{l}\left|\begin{array}{ccc}\hfill x-x_1\hfill & \hfill y-y_1\hfill & \hfill z-z_1\hfill \\ \hfill x_2-x_1\hfill & \hfill y_2-y_1\hfill & \hfill z_2-z_1\hfill \\ \hfill x_3-x_1\hfill & \hfill y_3-y_1\hfill & \hfill z_3-z_1\hfill \end{array}\right|=0\\ \Rightarrow \left|\begin{array}{ccc}\hfill x-1\hfill & \hfill y-1\hfill & \hfill z\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -3\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right|=0\\ \Rightarrow (-2)(x-1)-3(y-1)+3z=0\\ \Rightarrow -2x-3y+3z+2+3=0\\ \Rightarrow -2x-3y+3z=-5\\ \Rightarrow 2x+3y-3z=5\end{array}$

This is the Cartesian equation of the required plane.

 

Q7. Find the intercepts cut off by the plane  $2x+y-z=5$

A.7. $2x+y-z=5 .........\left(1\right)$

Dividing both sides of equation (1) by 5, we obtain

$\begin{array}{l}\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1\\ \Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1..........(2)\end{array}$

It is known that the equation of a plane in intercept form is  $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ , where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

$a=\frac{5}{2},b=5andc=-5$

Thus, the intercepts cut off by the plane are  $\frac{5}{2}$  $,5and-5.$

 

Q8. Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.

A.8. The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

 

Q9. Find the equation of the plane through the intersection of the planes  $3x-y+2z-4=0andx+$  $y+z-2=0$ and the point  $\left(2,2,1\right).$

A.9. The equation of any plane through the intersection of the planes,

$3x - y +2z ­-4=0and x + y + z -2=0,$ is

$\left(3x - y +2z -4\right)+\alpha \left(x + y + z -2\right)=0,where,\alpha \in R..........(1)$

The plane passes through the point  $\left(2,2,1\right).$ Therefore, this point will satisfy equation (1).

$\begin{array}{l}\therefore (3\times 2-2+2\times 1-4)+\alpha (2+2+1-2)=0\\ \Rightarrow 2+3\alpha =0\\ \Rightarrow \alpha =-\frac{2}{3}\end{array}$

Substituting  $\alpha =-\frac{2}{3}$ in equation (1), we obtain

$\begin{array}{l}(3x-y+2z-4)-\frac{2}{3}(x+y+z-2)=0\\ \Rightarrow 3(3x-y+2z-4)-2(x+y+z-2)=0\\ \Rightarrow (9x-3y+6z-12)-2(x+y+z-2)=0\\ \Rightarrow 7x-5y+4z-8=0\end{array}$

 

Q10. Find the vector equation of the plane passing through the intersection of the planes   $\stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-2\hat{k}\right)=7,\stackrel{\to }{r}.\left(2\hat{i}+5\hat{j}+3\hat{k}\right)=9$  and through the point  $\left(2,1,3\right).$

A.10. The equations of the planes are   $\stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-2\hat{k}\right)=7,\stackrel{\to }{r}.\left(2\hat{i}+5\hat{j}+3\hat{k}\right)=9$

$\begin{array}{l}\Rightarrow \stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-2\hat{k}\right)-7=0..........(1)\\ \Rightarrow \stackrel{\to }{r}.\left(2\hat{i}+5\hat{j}+3\hat{k}\right)-9=0..........(2)\end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l}\left[\stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-2\hat{k}\right)-7\right]+\lambda \left[\stackrel{\to }{r}.\left(2\hat{i}+5\hat{j}+3\hat{k}\right)-9\right]=0,where,\lambda \in R\\ \stackrel{\to }{r}.\left[\left(2\hat{i}+2\hat{j}-2\hat{k}\right)+\lambda \left(2\hat{i}+5\hat{j}+3\hat{k}\right)\right]=9\lambda +7\\ \stackrel{\to }{r}.\left[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(3\lambda -3)\hat{k}\right]=9\lambda +7..........(3)\end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\stackrel{\to }{r}=2\hat{i}+1\hat{j}+3\hat{k}$

Substituting in equation (3), we obtain

$\begin{array}{l}\left(2\hat{i}+\hat{j}+3\hat{k}\right).\left[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(3\lambda -3)\hat{k}\right]=9\lambda +7\\ \Rightarrow 2(2+2\lambda )+1(2+5\lambda )+3(3\lambda -3)=9\lambda +7\\ \Rightarrow 4+4\lambda +2+5\lambda +9\lambda -9=9\lambda +7\\ \Rightarrow 18\lambda -3=9\lambda +7\\ \Rightarrow 9\lambda =10\\ \Rightarrow \lambda =\frac{10}{9}\end{array}$

Substituting  $\lambda =\frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l}\stackrel{\to }{r}.\left(\frac{38}{9}\hat{i}+\frac{68}{9}\hat{j}+\frac{3}{9}\hat{k}\right)=17\\ \Rightarrow \stackrel{\to }{r}.\left(38\hat{i}+68\hat{j}+3\hat{k}\right)=153\end{array}$

This is the vector equation of the required plane.

 

Q11. Find the equation of the plane through the line of intersection of the planes  $x+y+z=1and2x+3y+4z=5$ which is perpendicular to the plane  $x-y+z=0$

A.11. The equation of the plane through the intersection of the planes,  $x+y+z=1and2x+3y+4z=5$ , is

   $\left(x+y+z-1\right) +\lambda \left(2x+3y+4z-5\right)$

$\Rightarrow \left(2\lambda +1\right)x+\left(3\lambda +1\right)y+\left(4\lambda +1\right)z-\left(5\lambda +1\right)=0 ..........\left(1\right)$

The direction ratios,   $a_1, b_1, c_1,$ of this plane are  $\left(2\lambda +1\right),\left(3\lambda +1\right),and\left(4\lambda +1\right).$

The plane in equation (1) is perpendicular to  $x-y+z=0$

Its direction ratios,   $a_2, b_2, c_2,$ are  $1,-1,and1$ .

Since the planes are perpendicular,

$\begin{array}{l}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ \Rightarrow \left(2\lambda +1\right)-\left(3\lambda +1\right)+\left(4\lambda +1\right)=0\\ \Rightarrow 3\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{3}\end{array}$

Substituting  $\lambda =-\frac{1}{3}$ in equation (1), we obtain

$\begin{array}{l}\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3}=0\\ \Rightarrow x-z+2=0\end{array}$

This is the required equation of the plane.

 

Q12. Find the angle between the planes whose vector equations are   $\stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-3\hat{k}\right)=5\&\stackrel{\to }{r}.\left(3\hat{i}-3\hat{j}+5\hat{k}\right)=3$

A.12. The equations of the given planes are   $\stackrel{\to }{r}.\left(2\hat{i}+2\hat{j}-3\hat{k}\right)=5\&\stackrel{\to }{r}.\left(3\hat{i}-3\hat{j}+5\hat{k}\right)=3$

It is known that if n1 and n2 are normal to the planes,   $\stackrel{\to }{r}.\stackrel{\to }{n_1}=d_1\&\stackrel{\to }{r}.\stackrel{\to }{n_2}=d_2$  then the angle between them, Q, is given by,

$cosQ=\left|\frac{\stackrel{\to }{n_1}.\stackrel{\to }{n_2}}{\left|\stackrel{\to }{n_1}\right|\left|\stackrel{\to }{n_2}\right|}\right|..........(1)$

 

Q13. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

$\begin{array}{}\end{array}\left(a\right)7x+5y+6z+30=0and3x-y-10z+4=0$

$\left(b\right)2x+y+3z-2=0andx-2y+5=0$

$\mathrm{(}c)2x-2y+4z+5=0and3x-3y+6z-1$

$\left(d\right)2x-y+3z-1=0and2x-y+3z+3=0$

$\left(e\right)4x+8y+z-8=0andy+z-4=0$

A.13. The direction ratios of normal to the plane,  $L_1:a_{1}x+b_{1}y+c_{1}z=0.$ , are   $a_1, b_1, c_1$ and   $L_2:a_{2}x+b_{2}y+c_{2}z=0.$

$\begin{array}{l}{L}_1<br>L_2,if\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\ L_1\perp L_2,if{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\end{array}$

The angle between  $L_1\&L_2$ is given by,

(b) The equations of the planes are  $2x+y+3z-2=0andx-2y+5=0$

   $\begin{array}{l}{a}_1=2,b_1=1,c_1=3\&a_2 =, b_2 =-2, c_2 =\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 1+1\times (-2)+3\times 0=0\end{array}$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are  $2x-2y+4z+5=0and3x-3y+6z-1$

Here,   $a_1=2,b_1=-2,c_1=4\&a_2 =, b_2 =-3, c_2 =6$

$a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 3+1\times (-2)(-3)+4\times 6=6+6+24=36\ne 0$

Thus, the given planes are not perpendicular to each other.

$\begin{array}{l}\frac{a_1}{a_2}=\frac{2}{3},\frac{b_1}{b_2}=\frac{-2}{-3}=\frac{2}{3}\&\frac{c_1}{c_2}=\frac{4}{6}=\frac{2}{3}\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Thus, the given planes are parallel to each other

(d) The equations of the planes are and  $2x-y+3z-1=0and2x-y+3z+3=0$

$\begin{array}{l}{a}_1=2,b_1=-1,c_1=3\&a_2 =, b_2 =-1, c_2 =3\\ \frac{a_1}{a_2}=\frac{2}{2}=1,\frac{b_1}{b_2}=\frac{-1}{-1}=1\&\frac{c_1}{c_2}=\frac{3}{3}=1\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Thus, the given lines are parallel to each other

(e) The equations of the given planes are   $4x+8y+z-8=0andy+z-4=0$  $\begin{array}{l}{a}_1=4,b_1=8,c_1=1\&a_2 =, b_2 =1, c_2 =1\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=4\times 0+8\times 1+1=9\ne 0\end{array}$

Therefore, the given lines are not perpendicular to each other.

$\begin{array}{l}\frac{a_1}{a_2}=\frac{4}{0},\frac{b_1}{b_2}=\frac{8}{1}=8\&\frac{c_1}{c_2}=\frac{1}{1}=1\\ \therefore \frac{a_1}{a_2}\ne \frac{b_1}{b_2}\ne \frac{c_1}{c_2}\end{array}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

 

Q14. In the following cases find the distances of each of the given points from the corresponding given plane:

$\begin{array}{}\end{array}\left(a\right)Point\left(0,0,0\right)Plane3x-4y+12z=3$

$\left(b\right)Point\left(3,-2,1\right) Plane2x-y+2z+3= 0$

$<br>c)Point\left(2,3,-5\right) Planex+2y-2z=9$

$\left(d\right)Point\left(-6,0,0\right) Plane2x-3y+6z-2=0$

 

Miscellaneous Exercise

Q1. Show that the line joining the origin to the point  $\left(2,1,1\right)$ is perpendicular to the line determined by the points  $\left(3,5,-1\right),\left(4,3,-1\right).$

A.1. Let OA be the line joining the origin,  $O\left(0,0,0\right),$ and the point,  $A\left(2,1,1\right).$

Also, let BC be the line joining the points,  $B\left(3,5,-1\right)andC\left(4,3,-1\right).$

The direction ratios of  $OAare2,1,and1andofBCare\left(4-3\right)=1,\left(3-5\right)=-2,and\left(-1+1\right)=0$

OA is perpendicular to  $BC,if a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\therefore a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =2\times 1+1\left(-2\right)+1\times 0=2-2=0$

Thus, OA is perpendicular to BC.

 

Q2. If   $l_1, m_1, n_1 and l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are   $m_1{n}_2 - m_2{n}_1, n_1{l}_2 - n_2{l}_1, l_1{m}_2 ­- l_2{m}_1.$

A.2. It is given that   $l_1, m_1, n_1 and l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l}{l}_1{l}_2+m_1{m}_2+n_1 n_2=0..........(1)\\ {l_1}^2+{m_1}^2+n_1{ }^2=1..........(2)\\ {l_2}^2+{m_2}^2+n_2{ }^2=1..........(3)\end{array}$

Let   $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines   $l_1, m_1, n_1 and l_2, m_2, n_2.$

$\begin{array}{l}\therefore l{l}_1+ m{m}_1+ n{n}_1 =0\\ l l_2+m m_2+n n_2=0\\ \therefore \frac{l}{m_1{n}_2-m_2{n}_1}=\frac{m}{n_1{l}_2-n_2{l}_1}=\frac{n}{l_1{m}_2-l_2{m}_1}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_1\right)}^2}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_2\right)}^2}\\ =\frac{l^2+m^2+n^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_2\right)}^2}..........(4)\end{array}$

$l, m, n$ are the direction cosines of the line.

$\therefore l^{2 }+ m^2 + n^2 =1\dots \left(5\right)$

It is known that,

$\begin{array}{l}\left({l_1}^2+{m_1}^2+n_1{ }^2\right)\left({l_2}^2+{m_2}^2+n_2{ }^2\right)-{\left(l_1{l}_2+m_1{m}_2+n_1 n_2\right)}^2\\ ={\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2\\ From,(1),(2)\&(3),we.obtain\\ \Rightarrow 1.1-0={\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2\\ \therefore {\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2=1..........(6)\end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_1\right)}^2}=1$

Thus, the direction cosines of the required line are   $m_1{n}_2-m_2{n}_1,n_1{l}_2-n_2{l}_1,l_1{m}_2-l_2{m}_1$

 

Q3. Find the angle between the lines whose direction ratios are   $a, b, c and b - c, c - a, a - b.$

A.3. The angle Q between the lines with direction cosines,   $a, b, c and b - c, c - a,$

$a - b,$ is given by,

$\begin{array}{l}\\ \end{array}$

Thus, the angle between the lines is 90°

 

Q4. Find the equation of the line parallel to x-axis and passing through the origin.

A.4. The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by  $\left(a,0,0\right),where a \in R.$ Direction ratios of  $OAare\left(a -0\right)= a,0,0$

The equation of OA is given by,

$\begin{array}{l}\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\\ \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

 

Q5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

A.5. The coordinates of  $A,B,C,andDare\left(1,2,3\right),\left(4,5,7\right),\left(­-4,3,-6\right),$ and

$\left(2,9,2\right)$ respectively.

The direction ratios of  $ABare\left(4-1\right)=3,\left(5-2\right)=3,and\left(7-3\right)=4$

The direction ratios of  $CDare\left(2-\left(-4\right)\right)=6,\left(9-3\right)=6,and\left(2-\left(-6\right)\right)=8$

It can be seen that,  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, the angle between  $ABandCDiseither0°or180°.$

 

Q6. .If the lines   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$  are perpendicular, find the value of k

A.6. The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are  $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if   $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k\times 1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

 

Q7. Find the vector equation of the line passing through  $\left(1,2,3\right)$ and perpendicular to the plane

$\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}-5\hat{k}\right)+9=0$  

A.7. The position vector of the point  $\left(1,2,3\right)$ is    $\stackrel{\to }{r}=\hat{i}+2\hat{j}+3\hat{k}$

The direction ratios of the normal to the plane,   $\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}-5\hat{k}\right)+9=0$ , are  $1,2,and-5$ and the normal vector is   $\stackrel{\to }{N}=\left(\hat{i}+2\hat{j}-5\hat{k}\right)$

The equation of a line passing through a point and perpendicular to the given plane is given by,

$\stackrel{\to }{l}=\stackrel{\to }{r}+\lambda \stackrel{\to }{N},\lambda \in R\Rightarrow \stackrel{\to }{l}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}-5\hat{k}\right)$

 

Q8. .Find the equation of the plane passing through  $\left(a, b, c\right)$ and parallel to the plane   $\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=2$

A.8. Any plane parallel to the plane,   $\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=2$ , is of the form   $\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=\lambda .........(1)$           

The plane passes through the point (a, b, c). Therefore, the position vector   $\stackrel{\to }{r}$  of this point is   $\stackrel{\to }{r}=a\hat{i}+b\hat{j}+c\hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left(a\hat{i}+b\hat{j}+c\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)=\lambda \\ \Rightarrow a+b+c=\lambda \end{array}$

Substituting   $\lambda =a+b+c$ in equation (1), we obtain

$\stackrel{\to }{r}=\left(\hat{i}+\hat{j}+\hat{k}\right)=a+b+c.........(2)$

This is the vector equation of the required plane.

Substituting   $\stackrel{\to }{r}=x\hat{i}+y\hat{j}+z\hat{k}$  in equation (2), we obtain

$\begin{array}{l}\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)=a+b+c\\ \Rightarrow x+y+z=a+b+c\end{array}$

 

Q10. Find the coordinates of the point where the line through  $\left(5,1,6\right)and\left(3,4,1\right)$ crosses the YZ-plane.

A.10. It is known that the equation of the line passing through the points,  $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is   $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passing through the points,  $\left(5,1,6\right)and\left(3,4,1\right),$ is given by,

$\begin{array}{l}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$

Any point on the line is of the form  $\left(5-2k,3k +1,6-5k\right).$

The equation of  $YZ-planeis x =0$

Since the line passes through YZ-plane,

$5-2k =0$

$\begin{array}{l}\Rightarrow k=\frac{5}{2}\\ \Rightarrow 3k+1=3\times \frac{5}{2}+1=\frac{17}{2}\\ 6-5k=6-5\times \frac{5}{2}=\frac{-13}{2}\end{array}$

Therefore, the required point is   $\left(0,\frac{17}{2},\frac{-13}{2}\right)$ .

 

Q11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

A.11. It is known that the equation of the line passing through the points,  $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is   $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passing through the points,  $\left(5,1,6\right)and\left(3,4,1\right),$ is given by,

$\begin{array}{l}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$

Any point on the line is of the form  $\left(5-2k,3k +1,6-5k\right).$

Since the line passes through ZX-plane,

$\begin{array}{l}3k+1=0\\ \Rightarrow k=-\frac{1}{3}\\ \Rightarrow 5-2k=5-2\left(-\frac{1}{3}\right)=\frac{17}{3}\\ 6-5k=6-5\left(-\frac{1}{3}\right)=\frac{23}{3}\end{array}$

Therefore, the required point is  $\left(\frac{17}{3},0,\frac{23}{3}\right)$

Q12. Find the coordinates of the point where the line through  $\left(3,-4,-5\right)and\left(2,-3,1\right)$ crosses the plane  $2x + y + z =7).$

A.12. It is known that the equation of the line through the points,  $(x_1, y_1, z_1)and(x_2, y_2, z_2)$ , is

$\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Since the line passes through the points,  $\left(3,-4,-5\right)and\left(2,-3,1\right)$ , its equation is given by,

$\begin{array}{l}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k(say)\\ \Rightarrow x=3-k,y=k-4,z=6k-5\end{array}$

Therefore, any point on the line is of the form  $\left(3- k, k -4,6k -5\right).$

This point lies on the plane,  $2x + y + z =7$

$\begin{array}{l}\therefore 2 \left(3- k\right) + \left(k -4\right) + \left(6k -5\right) = 7\hfill \end{array}$

$\begin{array}{l}5k - 3 = 7\hfill \end{array}$

$\begin{array}{l}k = 2\hfill \end{array}$

Hence, the coordinates of the required point are  $\left(3-2,2-4,6\times 2-5\right)i.e.,$  $\left(1,-2,7\right).$

 

Q13. Find the equation of the plane passing through the point  $\left(-1,3,2\right)$ and perpendicular to each of the planes   $x +2y +3z =5and3x +3y + z =0.$

A.13. The equation of the plane passing through the point  $a \left(x +1\right)+ b \left(y -3\right)+ c \left(z -2\right)=0\dots \left(1\right)$ where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes,   $a_{1}x+b_{1}y+c_{1}z+d_1=0\&a_{2}x+b_{2}y+c_{2}z+d_2=0$ are perpendicular, if  $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Plane (1) is perpendicular to the plane,   $x +2y +3z =5$

$\begin{array}{l}a.1 + b .2 + c.3 = 0\hfill \end{array}$

$\begin{array}{l}a + 2b + 3c = 0 ....\left(2\right)\hfill \end{array}$

Also, plane (1) is perpendicular to the plane,  $3x +3y + z =0$

$\begin{array}{l}a .3 + b.3 + c.1 = 0\hfill \end{array}$

$\begin{array}{l}3a + 3b + c = 0 .....\left(3\right)\hfill \end{array}$

From equations (2) and (3), we obtain

$\begin{array}{l}\frac{a}{2\times 1-3\times 3}=\frac{b}{3\times 3-1\times 1}=\frac{c}{1\times 3-2\times 3}\\ \Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k(say)\\ \Rightarrow a=-7k,b=8k,c=-3k\end{array}$

Substituting the values of a, b, and c in equation (1), we obtain

$\begin{array}{l}-7k(x+1)+8k(y-3)-3k(z-2)=0\\ \Rightarrow (-7x-7)+(8y-24)-3z+6=0\\ \Rightarrow -7x+8y-3z-25=0\\ \Rightarrow 7x-8y+3z+25=0\end{array}$

This is the required equation of the plane.

 

Q14. If the points  $\left(1,1, p\right)and\left(-3,0,1\right)$ be equidistant from the plane   $\stackrel{\to }{r}.\left(3\hat{i}+4\hat{j}-12\hat{k}\right)+13=0$   then find the value of p

A.14. The position vector through the point  $\left(1,1, p\right)$ is   $\stackrel{\to }{a_1}=\hat{i}+\hat{j}+p\hat{k}$

Similarly, the position vector through the point  $\left(-3,0,1\right)$ is

$\stackrel{\to }{a_2}=-4\hat{i}+\hat{k}$

The equation of the given plane is   $\stackrel{\to }{r}.\left(3\hat{i}+4\hat{j}-12\hat{k}\right)+13=0$

It is known that the perpendicular distance between a point whose position vector is   $\stackrel{\to }{a}$  and the plane,   $\stackrel{\to }{r}.\stackrel{\to }{N}=d$ is given by,   $D=\frac{\left|\stackrel{\to }{a}.\stackrel{\to }{N}-d\right|}{\left|\stackrel{\to }{N}\right|}$

 

Q15. Find the equation of the plane passing through the line of intersection of the planes     $\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=1\&\stackrel{\to }{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)+4=0$  and parallel to x - axis.

A.15. Equation of one plane is

$\begin{array}{l}\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)=1\\ \Rightarrow \stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)-1=0\\ \stackrel{\to }{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)+4=0\end{array}$

The equation of any plane passing through the line of intersection of these planes is

$\begin{array}{l}\left[\stackrel{\to }{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)-1\right]+\lambda \left[\stackrel{\to }{r}.\left(2\hat{i}+3\hat{j}-\hat{k}\right)+4\right]=0\\ \stackrel{\to }{r}.\left[\left(2\lambda +1\right)\hat{i}+\left(3\lambda +1\right)\hat{j}+\left(1-\lambda \right)\hat{k}\right]+\left(4\lambda -1\right)=0.....(1)\end{array}$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

$\begin{array}{l}\therefore 1.\left(2\lambda +1\right)+0\left(3\lambda +1\right)+0\left(1-\lambda \right)=0\\ \Rightarrow 2\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{2}\end{array}$

Substituting λ = -1/2 in equation (1), we obtain

$\begin{array}{l}\Rightarrow \stackrel{\to }{r}.\left[-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}\right]+\left(-3\right)=0\\ \Rightarrow \stackrel{\to }{r}\left(\hat{j}-3\hat{k}\right)+6=0\end{array}$

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

 

Q16. If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

A.16. The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x₁, y₁ z₁) is

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$ where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

$\begin{array}{l}1\left(x-1\right)+2\left(y-2\right)-3\left(z+3\right)0\hfill \end{array}$

$\begin{array}{l}\Rightarrow x+2y-3z-14 = 0\hfill \end{array}$

 

Q17. Find the equation of the plane which contains the line of intersection of the planes   $\stackrel{\to }{r}\left(\hat{i}+2\hat{j}+3\hat{k}\right)-4=0,\stackrel{\to }{r}.\left(2\hat{i}+ht\hat{j}-\hat{k}\right)+5=0$   and which is perpendicular to the plane   $\stackrel{\to }{r}.\left(5\hat{i}+3\hat{j}-6\hat{k}\right)+8=0$ .

A.17. The equations of the given planes are

$\begin{array}{l}\stackrel{\to }{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)-4=0.....(1)\\ \stackrel{\to }{r}.\left(2\hat{i}+\hat{j}-\hat{k}\right)+5=0......(2)\end{array}$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$\begin{array}{l}\left[\stackrel{\to }{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)-4\right]+\lambda \left[\stackrel{\to }{r}.\left(2\hat{i}+\hat{j}-\hat{k}\right)+5\right]=0\\ \stackrel{\to }{r}.\left[\left(2\lambda +1\right)\hat{i}+\left(\lambda +2\right)\hat{j}+\left(3-\lambda \right)\hat{k}\right]+\left(5\lambda -4\right)=0.....(3)\end{array}$

The plane in equation (3) is perpendicular to the plane,   $\stackrel{\to }{r}.\left(5\hat{i}+3\hat{j}-6\hat{k}\right)+8=0$

$\begin{array}{l}\therefore 5\left(2\lambda +1\right)+3\left(\lambda +2\right)-6\left(3-\lambda \right)=0\\ \Rightarrow 19\lambda -7=0\\ \Rightarrow \lambda =\frac{7}{19}\end{array}$

Substituting λ = 7/19 in equation (3), we obtain

$\begin{array}{l}\Rightarrow \stackrel{\to }{r}.\left(\frac{13}{19}\hat{i}+\frac{45}{19}\hat{j}+\frac{50}{19}\hat{k}\right)\frac{-41}{19}=0\\ \Rightarrow \stackrel{\to }{r}.\left(33\hat{i}+45\hat{j}+50\hat{k}\right)-41=0\end{array}$

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting   $\stackrel{\to }{r}=x\hat{i}+y\hat{j}+z\hat{k}$  in equation (3).

$\begin{array}{l}\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(33\hat{i}+45\hat{j}+50\hat{k}\right)-41=0\\ \Rightarrow 33x+45y+50z-41=0\end{array}$

The equation of the given line is

$\stackrel{\to }{r}=2\hat{i}-\hat{j}+2\hat{k}+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right)$            .....(1)

The equation of the given plane is

$\stackrel{\to }{r}.\left(\hat{i}-\hat{j}+\hat{k}\right)=5$           .......(2)

Substituting the value of   $\stackrel{\to }{r}$  from equation (1) in equation (2), we obtain

$\begin{array}{l}\left[2\hat{i}-\hat{j}+2\hat{k}+\lambda \left(3\hat{i}+4\hat{j}+2\hat{k}\right)\right].\left(\hat{i}-\hat{j}+\hat{k}\right)=5\\ \Rightarrow \left[\left(3\lambda +2\right)\hat{i}+\left(4\lambda -1\right)\hat{j}+\left(2\lambda +2\right)\hat{k}\right].\left(\hat{i}-\hat{j}+\hat{k}\right)=5\\ \Rightarrow \left(3\lambda +2\right)-\left(4\lambda -1\right)+\left(2\lambda +2\right)=5\\ \Rightarrow \lambda =0\end{array}$

Substituting this value in equation (1), we obtain the equation of the line as

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

This means that the position vector of the point of intersection of the line and the plane is   $\stackrel{\to }{r}=2\hat{i}-\hat{j}+2\hat{k}$

This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). The point is (−1, −5, −10).

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

 

Q19. Find the vector equation of the line passing through (1, 2, 3)  and parallel to the plane   $\stackrel{\to }{r}=\left(\hat{i}-\hat{j}+2\hat{k}\right)=5$  and   $\stackrel{\to }{r}.\left(3\hat{i}+\hat{j}+\hat{k}\right)=6$

A.19. Let the required line be parallel to vector   $\stackrel{\to }{b}$ given by,

$\stackrel{\to }{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

The position vector of the point (1, 2, 3) is   $\stackrel{\to }{a}=\hat{i}+2\hat{j}+3\hat{k}$

The equation of line passing through (1, 2, 3) and parallel to    $\stackrel{\to }{b}$ is given by,

$\begin{array}{l}\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\\ \Rightarrow \stackrel{\to }{r}\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)........(1)\end{array}$

The equations of the given planes are

$\begin{array}{l}\stackrel{\to }{r}=\left(\hat{i}-\hat{j}+2\hat{k}\right)=5........(2)\\ \stackrel{\to }{r}.\left(3\hat{i}+\hat{j}+\hat{k}\right)=6........(3)\end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\begin{array}{l}\Rightarrow \left(\hat{i}-\hat{j}+2\hat{k}\right).\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0\\ \Rightarrow \lambda \left(b_1-b_2+2b_3\right)=0\\ \Rightarrow b_1-b_2+2b_3=0..........(4)\end{array}$

Similarly,  $\left(3\hat{i}+\hat{j}+\hat{k}\right).\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)=0$

From equations (4) and (5), we obtain

$\begin{array}{l}\frac{b_1}{\left(-1\right)\times 1-1\times 2}=\frac{b_2}{2\times 3-1\times 1}=\frac{b_3}{1\times 1-3\times \left(-1\right)}\\ \Rightarrow \frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}\end{array}$

Therefore, the direction ratios of    $\stackrel{\to }{b}$ are −3, 5, and 4.

$\therefore \stackrel{\to }{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=-3\hat{i}+5\hat{j}+4\hat{k}$

Substituting the value of    $\stackrel{\to }{b}$  in equation (1), we obtain

$\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda \left(-3\hat{i}+5\hat{j}+4\hat{k}\right)$

This is the equation of the required line.

 

Q20. Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: 

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and  $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$  

A.20. Let the required line be parallel to the vector   $\stackrel{\to }{b}$  given by,   $\stackrel{\to }{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

The position vector of the point (1, 2, − 4) is   $\stackrel{\to }{a}=\hat{i}+2\hat{j}-4\hat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector    $\stackrel{\to }{b}$  is

$\begin{array}{l}\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}\\ \Rightarrow \stackrel{\to }{r}=\left(\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right).......(1)\end{array}$

The equations of the lines are

$\begin{array}{l}\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}........(2)\\ \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}........(3)\end{array}$

Line (1) and line (2) are perpendicular to each other.

$\therefore 3b_1-16b_2+7b_3=0........(4)$

Also, line (1) and line (3) are perpendicular to each other.

$\therefore 3b_1+8b_2-5b_3=0........(5)$

From equations (4) and (5), we obtain

$\begin{array}{l}\frac{b_1}{\left(-16\right)\left(-5\right)-8\times 7}=\frac{b_2}{7\times 3-3\left(-5\right)}=\frac{b_3}{3\times 8-3\left(-16\right)}\\ \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\\ \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\end{array}$

Direction ratios of     $\stackrel{\to }{b}$  are 2, 3, and 6.

$\therefore \stackrel{\to }{b}=2\hat{i}+3\hat{j}+6\hat{k}$

Substituting   $\stackrel{\to }{b}=2\hat{i}+3\hat{j}+6\hat{k}$  in equation (1), we obtain

$\stackrel{\to }{r}=\left(\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$

This is the equation of the required line.

 

Q21. Prove that if a plane has the intercepts a,b,c and is at a distance of P units from the origin, then   $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}$

A.21. The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The distance (p) of the plane from the origin is given by,

choose the correct answer in Exercise Q. 22 and 23.

 

Q22. Distance between the two planes:  $2x+3y+4z=4and4x+6y+8z=12$ is

(A) 2 units (B) 4 units (C) 8 units (D) 2/√29 units

A.22. The equations of the planes are

$\begin{array}{l}2x + 3y + 4z = 4\hfill \end{array}$

$\begin{array}{l}4x + 6y + 8z = 12\hfill \end{array}$

$\begin{array}{l}\Rightarrow 2x + 3y + 4z = 6\hfill \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes,   $ax + by + cz = d_1 and ax + by + cz = d_2,$ is given by,

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

 

Q23. The planes:  $2x - y +4z =5and5x -2.5y +10z =6$ are

(A) Perpendicular (B) Parallel (C) intersect y-axis

(D) passes through  (0,0,5/4)

A.23. The equations of the planes are

$\begin{array}{l}2x - y + 4z = 5 \dots \left(1\right)\hfill \end{array}$

$\begin{array}{l}5x - 2.5y + 10z = 6 \dots \left(2\right)\hfill \end{array}$

It can be seen that,

$\begin{array}{l}\frac{a_1}{a_2}=\frac{2}{5}\\ \frac{b_1}{b_2}=\frac{-1}{-2.5}=\frac{2}{5}\\ \frac{c_1}{c_2}=\frac{4}{10}=\frac{2}{5}\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.