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Question
Find the angle between the line
Solution
\[\text{ The given line is parallel to the vector } \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = 10 \hat{i} + 2 \hat{j} - 11 \hat{k} . \]
\[\text{ We know that the angle } \theta \text{ between the line and the plane is given by} \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) . \left( 10 \hat{i} + 2 \hat{j} - 11 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right| \left| 10 \hat{i} + 2 \hat{j} - 11 \hat{k} \right|} = \frac{20 + 6 - 66}{\sqrt{4 + 9 + 36} \sqrt{100 + 4 + 121}} = \frac{- 40}{\left( 7 \right) \left( 15 \right)} = \frac{- 8}{21}\]
\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{- 8}{21} \right)\]
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