मराठी

Find the Direction Cosines of the Line X + 2 2 = 2 Y − 7 6 = 5 − Z 6 Also, Find the Vector Equation of the Line Through the Point A(−1, 2, 3) and Parallel to the Given Line.

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प्रश्न

Find the direction cosines of the line 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]  Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  

बेरीज
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उत्तर

The equation of the given line is 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]

The given equation can be re-written as

\[\frac{x + 2}{2} = \frac{y - \frac{7}{2}}{3} = \frac{z - 5}{- 6}\]

This line passes through the point \[\left( - 2, \frac{7}{2}, 5 \right)\] and has direction ratios proportional to 2, 3, −6. So, its direction cosines are  \[\frac{2}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{3}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{- 6}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}\]  Or \[\frac{2}{7}, \frac{3}{7}, \frac{- 6}{7}\] The required line passes through the point having position vector  \[\overrightarrow{a} = - \hat{i} + 2 \hat{j} + 3 \hat{k}\] and is parallel to the vector \[\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}\]  So, its vector equation is  \[\overrightarrow{r} = \left( - \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right)\]

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पाठ 27: Straight Line in Space - Exercise 28.2 [पृष्ठ १७]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
Exercise 28.2 | Q 24 | पृष्ठ १७

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