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

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

\[ = \frac{x - \frac{1}{2}}{\sqrt{3}} = \frac{y + 2}{4} = \frac{z - 3}{6}\]

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

\[\frac{\sqrt{3}}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}}, \frac{4}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}}, \frac{6}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}} \]

\[ = \frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}\]