मराठी

Write the Direction Cosines of the Line Whose Cartesian Equations Are 2x = 3y = −Z.

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प्रश्न

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

 
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उत्तर

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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पाठ 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
पाठ 27 Straight Line in Space
Very Short Answers | Q 10 | पृष्ठ ४१

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