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Question
Find the angle between the lines
\[\vec{r} = \left( 2 \hat{i} - 5 \hat{j} + \hat{k} \right) + \lambda\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = 7 \hat{i} - 6 \hat{k} + \mu\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)\]
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Solution
Let
θ be the angle between the given lines. The given lines are parallel to the vectors \[\overrightarrow{b_1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \] and
\[\overrightarrow{b_2} = \hat{i} + 2 \hat{j} + 2 \hat{k}\] respectively.
So, the angle θ between the given lines is given by
\[\cos\theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right|\left| \overrightarrow{b_2} \right|}\]
\[ = \frac{\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)}{\sqrt{3^2 + 2^2 + 6^2}\sqrt{1^2 + 2^2 + 2^2}}\]
\[ = \frac{3 \times 1 + 2 \times 2 + 6 \times 2}{\sqrt{49}\sqrt{9}}\]
\[ = \frac{19}{21}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{19}{21} \right)\]
Thus, the angle between the given lines is
\[\cos^{- 1} \left( \frac{19}{21} \right)\]
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