Question

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

Answer 1

Let the direction ratios of the required line be a, b, c. Since the required line is perpendicular to the given lines, therefore,

$a+2b+3c=0$--- (1)

$-3a+2b+5c=0$---(2)

Solving (1) and (2), by cross multiplication, we get

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

$\Rightarrow a=4k,b=-14k,c=8k$

Thus, the required line passing through P(-1, 3, -2) and

having the direction ratios $a=4k,b=-14k,c=8k$ is

$\frac{x+1}{4}=\frac{y-3}{-14}=\frac{z+2}{8}$ or $\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$

which is the required equation of the line.