Question

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

Answer 1

It is given that the line passes through the point (-2,4, -5) and is parallel to
$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
The direction ratios of the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$, are (3, 5, 6)
The required line is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
It is known that the equation of the line passes through the point $(x_1,y_1,z_1)$ and with direction ratios a, b, c, is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
Also, the line passes through (-2, 4, - 5) and having direction ratios (3, 5, 6) therefore, the equation of the line is $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$