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Write the Direction Cosines of the Line Whose Cartesian Equations Are 6x − 2 = 3y + 1 = 2z − 4.

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प्रश्न

Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.

 
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उत्तर

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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अध्याय 27: Straight Line in Space - Very Short Answers [पृष्ठ ४१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Very Short Answers | Q 6 | पृष्ठ ४१

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