मराठी

Find the Equations of the Line Passing Through the Point (−1, 2, 1) and Parallel to the Line 2 X − 1 4 = 3 Y + 5 2 = 2 − Z 3 .

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प्रश्न

Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]

बेरीज
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उत्तर

The equation of line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3}\]  can be re-written as \[\frac{x - \frac{1}{2}}{2} = \frac{y + \frac{5}{3}}{\frac{2}{3}} = \frac{z - 2}{- 3}\] 

The direction ratios of the line parallel to line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3}\] are proportional to 2, \[\frac{2}{3}\]   \[-3\] . Equation of the required line passing through the point ( -1, 2, 1) having direction ratios proportional to 2, \[\frac{2}{3}\]   \[-3\]  is

 \[\frac{x - \left( - 1 \right)}{2} = \frac{y - 2}{\frac{2}{3}} = \frac{z - 1}{- 3}\]

\[ = \frac{x + 1}{2} = \frac{y - 2}{\frac{2}{3}} = \frac{z - 1}{- 3}\]

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पाठ 27: Straight Line in Space - Exercise 28.2 [पृष्ठ १६]

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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 13 | पृष्ठ १६

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