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Question
Find the angle between the following pair of line:
\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{ and } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]
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Solution
\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{ and }\frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]
The equations of the given lines can be re-written as
\[\frac{x - 2}{2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{ and } \frac{x + 2}{- 1} = \frac{y - 4}{2} = \frac{z - 5}{4}\]
Let
\[\overrightarrow{b_1}\] and \[\overrightarrow{b_2}\] be vectors parallel to the given lines.
Now,
\[\overrightarrow{b_1} = 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \]
\[ \overrightarrow{b_2} = - 1 \hat{i} + 2 \hat{j} + 4 \hat{k}\]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \right) . \left( - 1 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 7^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 2^2 + 4^2}}\]
\[ = \frac{- 2 + 14 - 12}{\sqrt{62} \sqrt{21}}\]
\[ = 0\]
\[ \Rightarrow \theta = \frac{\pi}{2}\]
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