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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.
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Solution
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
\[\frac{\sqrt{3}}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}}, \frac{4}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}}, \frac{6}{\sqrt{\left( \sqrt{3} \right)^2 + 4^2 + 6^2}} \]
\[ = \frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}\]
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