Question

Find the vector and cartersian equation 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

$\text{Equation of the line through P(2,1,3) is}$
$\frac{x-2}{a}=\frac{y-1}{b}=\frac{z-3}{c}...\left(i\right)$

$\text{As (i) is}$ $\perp$ to lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\text{and}\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$

$\therefore a+2b+3c=0\text{and}$
$-3a+2b+5c=0$

$\text{Solving for a,b,c, we have}$

$\frac{a}{10-6}=\frac{b}{-9-5}=\frac{c}{2+6}$

$\Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}$

$\Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}$

$\therefore$ $\text{from (i), we get required cartesian form i.e.}$
$\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$

$\text{Its vector form is}$ $\overrightarrow{r}=2\hat{i}+\hat{j}+3\hat{k}+\lambda (2\hat{i}-7\hat{j}+4\hat{k})$.