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Question
Find the angle between the following pair of line:
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]
Solution
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and }\frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]
Let
\[\overrightarrow{b_1}\] and\[\overrightarrow{b_2}\] be vectors parallel to the given lines.
Now,
\[\overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} - 3 \hat{k} \]
\[ \overrightarrow{b_2} = - \hat{i} + 8 \hat{j} + 4 \hat{k} \]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 2 \hat{i} + 3 \hat{j} - 3 \hat{k} \right) . \left( - \hat{i} + 8 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 3^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 8^2 + 4^2}}\]
\[ = \frac{- 2 + 24 - 12}{9\sqrt{22}}\]
\[ = \frac{10}{9\sqrt{22}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{10}{9\sqrt{22}} \right)\]
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