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Question
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
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Solution
The equation of the line 5x − 25 = 14 − 7y = 35z can be re-written as
\[\frac{x - 5}{\frac{1}{5}} = \frac{y - 2}{\frac{- 1}{7}} = \frac{z}{\frac{1}{35}}\]
\[ \Rightarrow \frac{x - 5}{7} = \frac{y - 2}{- 5} = \frac{z}{1}\]
Since the required line is parallel to the given line, so the direction ratios of the required line are proportional to 7, -5 , 1 .
The vector equation of the required line passing through the point (1, 2,-1,) and having direction ratios proportional to 7,-5,1 is
\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} - \hat{k} \right) + \lambda\left( 7 \hat{i} - 5 \hat{j} + \hat{k} \right)\]
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