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Question
Find the angle between the following pair of line:
\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]
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Solution
\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]
The equations of the given lines can be re-written as
\[\frac{x - 2}{3} = \frac{y + 3}{- 2} = \frac{z - 5}{0} \text{ and }\frac{x + 1}{1} = \frac{y - \frac{3}{2}}{\frac{3}{2}} = \frac{z - 5}{2}\]
Let
\[\overrightarrow{ b_1}\] and \[\overrightarrow{ b_2}\] be vectors parallel to the given lines.
Now,
\[\overrightarrow{b_1} = 3 \hat{i} - 2 \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} + \frac{3}{2} \hat{j} + 2 \hat{k} \]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 3 \hat{i} - 2 \hat{j} + 0 \hat{k} \right) . \left( \hat{i} + \frac{3}{2} \hat{j} + 2 \hat{k} \right)}{\sqrt{3^2 + \left( - 2 \right)^2 + 0^2} \sqrt{1^2 + \left( \frac{3}{2} \right)^2 + 2^2}}\]
\[ = \frac{3 - 3 + 0}{\sqrt{13} \sqrt{\frac{29}{4}}}\]
\[ = 0\]
\[ \Rightarrow \theta = \frac{\pi}{2}\]
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