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Question
Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.
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Solution
We have ,
2x = 3y = −z
The equation of the given line can be re-written as
\[\frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{- 1}\]
\[\frac{x}{3} = \frac{y}{2} = \frac{z}{- 6}\]
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
\[\frac{3}{\sqrt{3^2 + 2^2 + \left( - 6 \right)^2}}, \frac{2}{\sqrt{3^2 + 2^2 + \left( - 6 \right)^2}}, \frac{- 6}{\sqrt{3^2 + 2^2 + \left( - 6 \right)^2}}\]
\[ = \frac{3}{7}, \frac{2}{7}, - \frac{6}{7}\]
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