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प्रश्न
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.
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उत्तर
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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