Question

The cartesian equations of a line AB are $\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}.$ Find the direction cosines of a line parallel to AB.

Answer 1

We have

$\frac{2x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$

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

$$\begin{gathered} \frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\frac{y+2}{2}=\frac{z-3}{3} \\ =\frac{x-{\displaystyle \frac{1}{2}}}{{\displaystyle \sqrt{3}}}=\frac{y+2}{4}=\frac{z-3}{6} \end{gathered}$$

Thus, the direction ratios of the line parallel to AB are proportional to $\sqrt{3}$, 4, 6.

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

$$\begin{gathered} \frac{\sqrt{3}}{\sqrt{{\left(\sqrt{3}\right)}^2+4^2+6^2}},\frac{4}{\sqrt{{\left(\sqrt{3}\right)}^2+4^2+6^2}},\frac{6}{\sqrt{{\left(\sqrt{3}\right)}^2+4^2+6^2}} \\ =\frac{\sqrt{3}}{\sqrt{55}},\frac{4}{\sqrt{55}},\frac{6}{\sqrt{55}} \end{gathered}$$