Question

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

Answer 1

Given line

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

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

So comparing with general cartesian form of eq

$a=3,b=-5,c=6$

another line passing through $P(-2,4,-5)$ we can assume it as position vector and parallel to given line so normal vector will be same

eq of line

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