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Question
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.
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Solution
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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