Exercise 28.1

Question 1

Find the vector and cartesian equations of the line through the point $(5, 2, −4)$ and which is parallel to the vector $3\hat{i}+2\hat{j}-8\hat{k}$

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

$\vec{a}=5\hat{i}+2\hat{j}-4\hat{k}$

$\vec{b}=3\hat{i}+2\hat{j}-8\hat{k}$

Vector equation of the required line is given by

$\vec{r}=(5\hat{i}+2\hat{j}-4\hat{k})+\lambda(3\hat{i}+2\hat{j}-8\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5\hat{i}+2\hat{j}-4\hat{k})+\lambda(3\hat{i}+2\hat{j}-8\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5+3\lambda)\hat{i}+(2+2\lambda)\hat{j}+(-4-8\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=5+3\lambda,~~y=2+2\lambda~~z=-4-8\lambda$

$\Rightarrow \dfrac{x-5}{3}=\lambda,~\dfrac{y-2}{2}=\lambda,~\dfrac{z+4}{-8}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-5}{3}=\dfrac{y-2}{2}=\dfrac{z+4}{-8}=\lambda$

Question 2

Find the vector equation of the line passing through the points $(−1, 0, 2)$ and $(3, 4, 6)$ .

Sol :

We know that the vector equation of a line passing through the points with position vectors $\vec{a}$ and parallel to $(\vec{b}-\vec{a})$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$ , where $\lambda$ is some scalar.

Here,

$\vec{a}=-1\hat{i}+0\hat{j}+2\hat{k}$

$\vec{b}=3\hat{i}+4\hat{j}+6\hat{k}$

Vector equation of the required line is given by

$\Rightarrow \vec{r}=(-1\hat{i}+0\hat{j}+2\hat{k})+\lambda{(3\hat{i}+4\hat{j}+6\hat{k})-(-1\hat{i}+0\hat{j}+2\hat{k})}$

$\Rightarrow \vec{r}=(-1\hat{i}+0\hat{j}+2\hat{k})+\lambda(4\hat{i}+4\hat{j}+4\hat{k})$

Here $\lambda$ is a parameter.

Question 3

Find the vector equation of a line which is parallel to the vector $2\hat{i}-\hat{j}+3\hat{k}$ and which passes through the points $(5,-2,4)$ . Also , reduce it to cartesian form.

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

$\vec{a}=5\hat{i}-2\hat{j}+4\hat{k}$

$\vec{b}=2\hat{i}-1\hat{j}+3\hat{k}$

Vector equation of the required line is given by

$\vec{r}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-1\hat{j}+3\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-1\hat{j}+3\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5+2\lambda)\hat{i}+(-2-1\lambda)\hat{j}+(4+3\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=5+2\lambda,~~y=-2-1\lambda~~z=4+3\lambda$

$\Rightarrow \dfrac{x-5}{2}=\lambda,~\dfrac{y+2}{-1}=\lambda,~\dfrac{z-4}{3}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-5}{2}=\dfrac{y+2}{-1}=\dfrac{z-4}{3}=\lambda$

Question 4

A line passes through the point with position vector $2\hat{i}-3\hat{j}+4\hat{k}$ and is in the direction of $3\hat{i}+4\hat{j}-5\hat{k}$ . Find equations of the line in vector and cartesian form.

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

$\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$

$\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}$

Vector equation of the required line is given by

$\vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+4\hat{j}-5\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+4\hat{j}-5\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2+3\lambda)\hat{i}+(-3+4\lambda)\hat{j}+(4-5\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=2+3\lambda,~~y=-3+4\lambda~~z=4-5\lambda$

$\Rightarrow \dfrac{x-2}{3}=\lambda,~\dfrac{y+3}{4}=\lambda,~\dfrac{z-4}{-5}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-4}{-5}=\lambda$

Question 5

ABCD is a parallelogram. The position vectors of the points A,B,C are respectively $4\hat{i}+5\hat{j}-10\hat{k}$ , $2\hat{i}-3\hat{j}+4\hat{k}$ and $-\hat{i}+2\hat{j}+\hat{k}$ . Find the vector equation of the line BD. Also, reduce it to cartesian form.

Sol :

We know that the position vector of the mid-point of $\vec{a}$ and $\vec{b}$ is $\dfrac{\vec{a}+\vec{b}}{2}$ .

Let the position vector of point D be $x\hat{i}+y\hat{j}+z\hat{k}$

Position vector of mid-point of A and C = Position vector of mid-point of B and D

$\Rightarrow \dfrac{(4\hat{i}+5\hat{j}-10\hat{k})+(-\hat{i}+2\hat{j}+\hat{k})}{2}=\dfrac{(2\hat{i}-3\hat{j}+4\hat{k})+(x\hat{i}+y\hat{j}+z\hat{k})}{2}$

$\Rightarrow \dfrac{3}{2}\hat{i}+\dfrac{7}{2}\hat{j}-\dfrac{9}{2}\hat{k}=\bigg(\dfrac{x+2}{2}\bigg)\hat{i}+\bigg(\dfrac{-3+y}{2}\bigg)\hat{j}+\bigg(\dfrac{4+z}{2}\bigg)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\dfrac{x+2}{2}=\dfrac{3}{2}$

$\Rightarrow x=1$

$\dfrac{-3+y}{2}=\dfrac{7}{2}$

$\Rightarrow y=10$

$\dfrac{4+z}{2}=-\dfrac{9}{2}$

$\Rightarrow z=-13$

Position vector of points $D=\hat{i}+10\hat{j}-13\hat{k}$

The vector equation of line BD passing through the points with position vectors $\vec{a}(B)$ and $\vec{b}(D)$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$

Here,

$\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$

$\vec{b}=\hat{i}+10\hat{j}-13\hat{k}$

Vector equation of the required line is given by

$\vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(\hat{i}+10\hat{j}-13\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(\hat{i}+10\hat{j}-13\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2-\lambda)\hat{i}+(-3+13\lambda)\hat{j}+(4-17\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=2-\lambda,~~y=-3+13\lambda~~z=4-17\lambda$

$\Rightarrow \dfrac{x-2}{-1}=\lambda,~\dfrac{y+3}{13}=\lambda,~\dfrac{z-4}{-17}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-2}{-1}=\dfrac{y+3}{13}=\dfrac{z-4}{-17}=\lambda$

Question 6

Find in vector form as well as in cartesian form, the equation of the line passing through the points A $(1,2,-1)$ and B $(2,1,1)$

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

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

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

Vector equation of the required line is given by

$\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(2\hat{i}+\hat{j}+\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(2\hat{i}+\hat{j}+\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

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

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=1+\lambda,~~y=2-\lambda~~z=-1+2\lambda$

$\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y-2}{-1}=\lambda,~\dfrac{z+1}{2}=\lambda$

Hence, the cartesian form of (1) is

$Rightarrow \dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z+1}{2}=\lambda$

Question 7

Find in vector form as well as in cartesian form, the equation of the line passing through the points $(1,2,3)$ and parallel to the vector $\hat{i}+-2\hat{j}+3\hat{k}$ . Reduce the corresponding equation in cartesian form.

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

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

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

Vector equation of the required line is given by

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)\hat{i}+(2-2\lambda)\hat{j}+(3+3\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=1+\lambda,~~y=2-2\lambda~~z=3+3\lambda$

$\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y-2}{-2}=\lambda,~\dfrac{z-3}{3}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-1}{1}=\dfrac{y-2}{-2}=\dfrac{z-3}{3}=\lambda$

Question 8

Find the vector equation of a line passing through $(2, −1, 1)$ and parallel to the line whose equations are $\dfrac{x-3}{2}=\dfrac{y+1}{7}+\dfrac{z-2}{-3}$

Sol :

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

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

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

Vector equation of the required line is given by

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})$ …..1

Here, $\lambda$ is a parameter

Question 9

The cartesian equations of a line are $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$ . Find a vector equation for the line.

Sol :

The cartesian equation of the given line is $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$ .

It can be re-written as

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

Thus, the given line passes through the point having position vector

$\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}$ and is parallel to the vector $\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}$

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Vector equation of the required line is given by

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

Here, $\lambda$ is a parameter

Question 10

Find the cartesian equation of a line passing through $(1, −1, 2)$ and parallel to the line whose equations are $\dfrac{x-3}{1}=\dfrac{y-1}{2}=\dfrac{z+1}{-2}$ . Also, reduce the equation obtained in vector form.

Sol :

We know that the cartesian equation of a line passing through a point with position vector $\vec{a}$ and parallel to the vector $\vec{m}$ is $\dfrac{x-x_1}{a}=\dfrac{y-y_2}{b}=\dfrac{z-z_3}{c}$

Here,

$\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

$\vec{m}=a\hat{i}+b\hat{j}+c\hat{k}$

Here,

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

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

Cartesian equation of the required line is

$\Rightarrow \dfrac{x-1}{1}=\dfrac{y-(-1)}{2}=\dfrac{z-2}{-2}$

$\Rightarrow \dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z-2}{-2}$

We know that the cartesian equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{m}$ is $\vec{r}=\vec{a}+\lambda\vec{m}$

Here, the line is passing through the point $(1,1,-2)$ and its direction ratios are proportional to 1,2,-2

Vector equation of the required line is

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

Question 11

Find the direction cosines of the line $\dfrac{4-x}{2}=\dfrac{y}{6}=\dfrac{1-z}{3}$ . Also, reduce it to vector form.

Sol :

The cartesian equation of the given line is

$\dfrac{4-x}{2}=\dfrac{y}{6}=\dfrac{1-z}{3}$

It can be re-written as

$\dfrac{x-4}{-2}=\dfrac{y-0}{6}=\dfrac{z-1}{-3}$

This shows that the given line passes through the point $(4,0,1)$ and its direction ratios are proportional to -2 , 6 , -3 .

So , the direction ratios are

$\Rightarrow \dfrac{-2}{\sqrt{(-2)^2+(6)^2+(-3)^2}},~~\dfrac{6}{\sqrt{(-2)^2+(6)^2+(-3)^2}}~~\dfrac{-3}{\sqrt{(-2)^2+(6)^2+(-3)^2}}$

$\dfrac{-2}{7},~~\dfrac{6}{7}~~,\dfrac{-3}{7}$

Thus,the given line passing through a point with position vector $\vec{a}=4\hat{i}+\hat{k}$ and parallel to $\vec{b}=-2\hat{i}+6\hat{j}-3\hat{k}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

$\vec{a}=4\hat{i}+\hat{k}$

$\vec{b}=-2\hat{i}+6\hat{j}-3\hat{k}$

Vector equation of the required line is

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

Here, $\lambda$ is a parameter.

Question 12

The cartesian equations of a line are x = ay,+ b , z = cy + d .Find its direction ratios and reduce it to vector form.

Sol :

The cartesian equation of the given line is

x = ay + b , z = cy + d

It can be re-written as

$\dfrac{x-b}{a}=\dfrac{y-0}{1}=\dfrac{z-d}{c}$

Thus, the given line passes through the point $(b~,0~,d)$ and its direction ratios are proportional to a,1,c . It is also parallel to the vector $\vec{b}=a\hat{i}+\hat{j}+c\hat{k}$

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Vector equation of the required line is

$\vec{r}=(b\hat{i}+0\hat{j}+d\hat{k})+\lambda(a\hat{i}+j\hat{j}+c\hat{k})$

Here, $\lambda$ is a parameter.

Question 13

Find the vector equation of a line passing through the point with position vector $\hat{i}-2\hat{j}-3\hat{k}$ and parallel to the line joining the points with position vector $\hat{i}-\hat{j}+4\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}. Also, find the cartesian equivalent of this equation. Sol : We know that the vector equation of a line passing through a point with position vector [latex]\vec{a}$ and parallel to the vector $(\vec{b}-\vec{a})$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$

Here,

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

$(\vec{b}-\vec{a})=(2\hat{i}+\hat{j}+2\hat{k})-(\hat{i}-\hat{j}+4\hat{k})$

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

Vector equation of the required line is

$\vec{r}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})$ ….1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form , we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)\hat{i}+(-2+2\lambda)\hat{j}+(-3-2\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=1+\lambda,~~y=-2+2\lambda~~z=-3-2\lambda$

$\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y+2}{1}=\lambda,~\dfrac{z+3}{-2}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z+3}{-2}=\lambda$

Question 14

Find the points on the line $\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}$ at a distance of 5 units from the point p $(1,~3,~3)$

Sol :

The coordinates of any point on the line $\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}$ are given by

$\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}=\lambda$

$\Rightarrow x=3\lambda-2,~~y=2\lambda-1,~~z=2\lambda+3$ ….1

Let the coordinates of the desired point be $(3\lambda-2,~~2\lambda-1,~~2\lambda+3)[/lambda] The distance between this point and (1, 3, 3) is 5 units. [latex]\sqrt{(3\lambda-2-1)^2+(2\lambda-1-3)^2+(2\lambda+3-3)^2}=5$

[Squaring both sides]

$\Rightarrow (3\lambda-3)^2+(2\lambda-4)^2+(2\lambda)^2=25$

$\Rightarrow 17\lambda^2-34\lambda=0$

$\Rightarrow \lambda(\lambda-2)=0$

$\Rightarrow \lambda=0~~or~~2$

Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as $(-2,~~-1,~~3)$ and $(4,~~3,~~7)$

Question 15

Show that the points whose position vectors are $2\hat{i}+3\hat{j}$ , $\hat{i}+2\hat{j}+3\hat{k}$ [latex]7\hat{i}-\hat{k}[/latex] collinear.

Sol :

Let the given points be P, Q and R and let their position vector be $\vec{a}$ , $\vec{b}$ and $\vec{c}$ respectively.

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

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

$\vec{c}=7\hat{i}+9\hat{k}$

Vector equation of line passing through P and Q is

$\Rightarrow \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$

$\Rightarrow \vec{r}=(-2\hat{i}+3\hat{j})+\lambda{(\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j})}$ …1

If points P,Q and R are collinear , the R must satisfy (1).

Replacing $\vec{r}$ by $\vec{c}=7\hat{i}+9\hat{k}$ in (1) , we get

$\Rightarrow 7\hat{i}+9\hat{k}=(-2\hat{i}+3\hat{j})+\lambda{(\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j})}$ …1

Comparing the coefficients $\hat{i}$ , $\hat{j}$ and $\hat{k}$ , we get

$7=-2+3\lambda$ , $0=3-\lambda$ , $9=-3\lambda$

$\lambda=3$

These three equations are consistent, i.e. they give the same value of $\lambda$ . Hence , the given three points are collinear

Question 16

Find the cartesian and vector equations of a line which passes through the point $(1,~2,~3)$ and is parallel to the line $\dfrac{-x-2}{1}=\dfrac{y+3}{7}=\dfrac{2z-6}{3}$

Sol :

we have

$\dfrac{-x-2}{1}=\dfrac{y+3}{7}=\dfrac{2z-6}{3}$

It can be re-written as

$\dfrac{x+2}{-1}=\dfrac{y+3}{7}=\dfrac{z-3}{\dfrac{3}{2}}$

$\dfrac{x+2}{-2}=\dfrac{y+3}{14}=\dfrac{z-3}{3}$

This shows that the given line passes through the point $(-2,~-3,~3)$ and its direction ratios are proportional to -2 , 14 , 3 .

Thus , the parallel vector is $\vec{b}=-2\hat{i}+14{j}+3\hat{k}$

We know that the vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to the vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda\vec{b}$

Here,

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

$\vec{b}=-2\hat{i}+14{j}+3\hat{k}$

Vector equation of the required line is

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(-2\hat{i}+14{j}+3\hat{k})$ …..1

Here, $\lambda$ is a parameter

Reducing (1) to cartesian form, we get

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(-2\hat{i}+14{j}+3\hat{k})$

$\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]$

$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1-2\lambda)\hat{i}+(2+14\lambda)\hat{j}+(3+3\lambda)\hat{k}$

Comparing the coefficients of $\hat{i},~~\hat{j}$ and $\hat{k}$ ,we get

$\Rightarrow x=1-2\lambda,~~y=2+14\lambda~~z=3+3\lambda$

$\Rightarrow \dfrac{x-1}{-2}=\lambda,~\dfrac{y-2}{14}=\lambda,~\dfrac{z-3}{3}=\lambda$

Hence, the cartesian form of (1) is

$\Rightarrow \dfrac{x-1}{-2}=\dfrac{y-2}{14}=\dfrac{z-3}{3}=\lambda$

Question 17

The cartesian equation of a line are $3x+1 = 6y-2 = 1-z$ . Find the fixed point through which it passes , its direction ratios and also its vector equation.

Sol :

The cartesian equation of the given line is

$3x+1 = 6y-2 = 1-z$

It can be re-written as

$\dfrac{x+\dfrac{1}{3}}{\dfrac{1}{3}}=\dfrac{y-\dfrac{1}{3}}{\dfrac{1}{6}}=\dfrac{z-1}{-6}$

Thus , the given line passes through the point $(-\dfrac{1}{3},\dfrac{1}{3},1)$ and its direction ratios are proportional to $(2,~1,-6)$ . It is parallel to the vector $\vec{b}=2\hat{i}+\hat{j}-6\hat{k}$