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Question
Find the angle between the line \[\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{1}\] and the plane 2x + y − z = 4.
Solution
\[\text{ The given line is parallel to the vector } \vec{b} = \hat{i} - \hat{j} + \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = 2 \hat{i} + \hat{j} - \hat{k} . \]
\[\text{ We know that the angle θ between the line and the plane is given by } \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( \hat{i} - \hat{j} + \hat{k} \right) . \left( 2 \hat{i} + \hat{j} - \hat{k} \right)}{\left| \hat{i} - \hat{j} + \hat{k} \right| \left| 2 \hat{i} + \hat{j} - \hat{k} \right|} = \frac{2 - 1 - 1}{\sqrt{1 + 1 + 1} \sqrt{4 + 1 + 1}} = 0\]
\[ \Rightarrow \theta = \sin^{- 1} \left( 0 \right) = 0\]
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