Question

Find the Cartesian equations 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

The equation of the given line is $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$

It can be re-written as

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

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, $-$5, 6.

Hence, the cartesian equations of the line passing through the point ($-$2, 4, $-$5) and parallel to a vector having direction ratios proportional to 3, $-$5, 6 is $\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$.