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Question
Find the angle between the following pair of line:
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
Solution
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3}\text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
The equation of the given line can be re-written as
\[\frac{x - 5}{2} = \frac{y + 3}{1} = \frac{z - 1}{- 3} \text{ and } \frac{x}{3} = \frac{y - 1}{2} = \frac{z + 5}{- 1}\]
Let
\[\overrightarrow{b_1}\] and \[\overrightarrow{b_2}\] be vectors parallel to the given lines.
Now,
\[\overrightarrow{b_1} = 2 \hat{i} + \hat{j} - 3 \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + 2 \hat{j} - \hat{k}\]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 2 \hat{i} + \hat{j} - 3 \hat{k} \right) . \left( 3 \hat{i} + 2 \hat{j} - \hat{k} \right)}{\sqrt{2^2 + 1^2 + \left( - 3 \right)^2} \sqrt{3^2 + 2^2 + \left( - 1 \right)^2}}\]
\[ = \frac{6 + 2 + 3}{\sqrt{14} \sqrt{14}}\]
\[ = \frac{11}{14}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{11}{14} \right)\]
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