Question

Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.

Answer 1

We have
2x = 3y = −z

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

$$\begin{gathered} \frac{x}{{\displaystyle \frac{1}{2}}}=\frac{y}{{\displaystyle \frac{1}{3}}}=\frac{z}{-1} \\ \frac{x}{3}=\frac{y}{2}=\frac{z}{-6} \end{gathered}$$

The direction ratios of the line parallel to AB are proportional to 3, 2, $-$6.

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

$$\begin{gathered} \frac{3}{\sqrt{3^2+2^2+{\left(-6\right)}^2}},\frac{2}{\sqrt{3^2+2^2+{\left(-6\right)}^2}},\frac{-6}{\sqrt{3^2+2^2+{\left(-6\right)}^2}} \\ =\frac{3}{7},\frac{2}{7},-\frac{6}{7} \end{gathered}$$