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Question
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
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Solution
We have ,
6x − 2 = 3y + 1 = 2z − 4
The equation of given line can be re-written as ,
\[\frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y + \frac{1}{3}}{\frac{1}{3}} = \frac{z - 2}{\frac{1}{2}}\]
\[ \Rightarrow \frac{x - \frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 2}{3}\]
The direction ratios of the line parallel to AB are proportional to 1, 2, 3.
Hence, the direction cosines of the line parallel to AB are proportional to
\[\frac{1}{\sqrt{1^2 + 2^2 + 3^2}}, \frac{2}{\sqrt{1^2 + 2^2 + 3^2}}, \frac{3}{\sqrt{1^2 + 2^2 + 3^2}}\]
\[ = \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\]
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