Question

Find the vector and Cartesian equations of the line passing through the point $(2,1,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}.$

Answer 1

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

The position vector of the point $(2,1,3)$ and parallel to vector $\overrightarrow{b}$ is

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

$\Rightarrow \overrightarrow{r}(2\overrightarrow{i}+\overrightarrow{j}+3\overrightarrow{k})+\lambda (b_1\overrightarrow{i}+b_2\overrightarrow{j}+b_3\overrightarrow{k})$--- (1)

The equation of the lines are

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$--- (2)

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

Line (1) and (2) are $\perp$ to each other.

$\therefore b_1+2b_2+3b_3=0$--- (4)

Also, Line (1) and (3) are $\perp$ to each other.

$\therefore -3b_1+2b_2+5b_3=0$--- (5)

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

$\frac{b_1}{2\left(5\right)-3\left(2\right)}=\frac{b_2}{5-3(-3)}=\frac{b_3}{2-2(-3)}$

$\frac{b_1}{4}=-\frac{b_2}{14}=\frac{b_3}{8}$

$\Rightarrow \frac{b_1}{2}=-\frac{b_2}{7}=\frac{b_3}{4}$

Therefore direction ratios of $\overrightarrow{b}$ are $2,-7,4.$

$\overrightarrow{b}=2\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}$

Substituting $\overrightarrow{b}=2\overrightarrow{i}+7\overrightarrow{j}+4\overrightarrow{k}$ in equation (1), we obtain

$\overrightarrow{r}(2\overrightarrow{i}+\overrightarrow{j}+3\overrightarrow{k})+\lambda (2\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k})$

This is the equation of the required line.