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Question
Find the angle between the following pair of line:
\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]
Solution
\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]
Let \[\overrightarrow {b_1}\] and \[\overrightarrow {b_2}\] be vector parallel to the given line.
Now,
\[\overrightarrow{b_1} = \hat{i} + \hat{j} + 2 \hat{k} \]
\[ \overrightarrow{b_2} = \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k}\]
If θ is the angle between the given line, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( \hat{i} + \hat{j} + 2 \hat{k} \right) . \left( \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right)}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{\left( \sqrt{3} - 1 \right)^2 + \left( \sqrt{3} + 1 \right)^2 + 4^2}}\]
\[ = \frac{\left( \sqrt{3} - 1 \right) - \left( \sqrt{3} + 1 \right) + 8}{\sqrt{6} \sqrt{24}}\]
\[ = \frac{6}{12}\]
\[ = \frac{1}{2}\]
\[ \Rightarrow \theta = \frac{\pi}{3}\]
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