Question

The equations of a line are given by $\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}.$ Write the direction cosines of a line parallel to this line.

Answer 1

We have

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

The equation of the given line can be re-written as

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

The direction ratios of the line parallel to the given line are proportional to $-$3, 3, 6.


Hence, the direction cosines of the line parallel to the given line are proportional to

$$\begin{gathered} \frac{-3}{\sqrt{{\left(-3\right)}^2+3^2+6^2}},\frac{3}{\sqrt{{\left(-3\right)}^2+3^2+6^2}},\frac{6}{\sqrt{{\left(-3\right)}^2+3^2+6^2}} \\ =\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}} \end{gathered}$$