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Question
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]
Solution
\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]
Let `vec b_1 and vec b_2 ` be vectors parallel to the given lines .
Now,
\[\overrightarrow{b_1} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]
\[ \overrightarrow{b_2} = 2 \hat{i} + 4 \hat{j} - 4 \hat{k}\]
If θ is the angle between the given lines, then
\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) . \left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)}{\sqrt{1^2 + 2^2 + \left( - 2 \right)^2} \sqrt{2^2 + 4^2 + \left( - 4 \right)^2}}\]
\[ = \frac{2 + 8 + 8}{3 \times 6}\]
\[ = 1\]
\[ \Rightarrow \theta = 0° \]
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