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Find the Vector Equation of the Line Passing Through the Point (2, −1, −1) Which is Parallel to the Line 6x − 2 = 3y + 1 = 2z − 2. - Mathematics

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Question

Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 

Sum

Solution

The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as

\[\frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y + \frac{1}{3}}{\frac{1}{3}} = \frac{z - 1}{\frac{1}{2}}\]

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

Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 1,2,3.The vector equation of the required line passing through the point (2,-1,-1) and having direction ratios proportional to 1,2,3 is \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \lambda\left( \hat{i}  + 2 \hat{j} + 3 \hat{k}  \right)\]

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Chapter 28: Straight Line in Space - Exercise 28.2 [Page 17]

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RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.2 | Q 20 | Page 17

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