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Question
The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\] Write the direction cosines of a line parallel to this line.
Solution
We have
\[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6}\]
The equation of the given line can be re-written as
\[\frac{x - 4}{- 3} = \frac{y + 3}{3} = \frac{z + 2}{6}\]
The direction ratios of the line parallel to the given line are proportional to -3, 3 , 6 .
Hence, the direction cosines of the line parallel to the given line are proportional to
\[\frac{- 3}{\sqrt{\left( - 3 \right)^2 + 3^2 + 6^2}}, \frac{3}{\sqrt{\left( - 3 \right)^2 + 3^2 + 6^2}}, \frac{6}{\sqrt{\left( - 3 \right)^2 + 3^2 + 6^2}} \]
\[ = \frac{- 1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\]
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