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Question
Find the angle between the following pair of line:
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
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Solution
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
The equations of the given lines can be re-written as
\[\frac{x - 5}{1} = \frac{y + 3}{- 1} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
Let
\[\overrightarrow { b_1}\] and \[\overrightarrow { b_2}\] be vectors parallel to the given lines.
Now,
\[\overrightarrow{b_1} = \hat{i} - \hat{j} + \hat{k} \]
\[ \overrightarrow{b_2} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( \hat{i} - \hat{j} + \hat{k} \right) . \left( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2} \sqrt{3^2 + 4^2 + 5^2}}\]
\[ = \frac{3 - 4 + 5}{\sqrt{3} \sqrt{50}}\]
\[ = \frac{4}{5\sqrt{6}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{4}{5\sqrt{6}} \right)\]
Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.
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