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The Cartesian Equations of a Line Ab Are 2 X − 1 √ 3 = Y + 2 2 = Z − 3 3 . Find the Direction Cosines of a Line Parallel to Ab.

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प्रश्न

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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उत्तर

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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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 15 | पृष्ठ ४१

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