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Question
Find the angle between the line \[\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}\] and the plane
3x + 4y + z + 5 = 0.
Solution
\[\text{ The given line is parallel to the vector } \vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = 3 \hat{i} + 4 \hat{j} + \hat{k} . \]
\[\text{ We know that the angle } \theta \text{ between the line and the plane is given by} \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( 3 \hat{i} - \hat{j} + 2 \hat{k} \right) . \left( 3 \hat{i} + 4 \hat{j} + \hat{k} \right)}{\left| 3 \hat{i} - \hat{j} + 2 \hat{k} \right| \left| 3 \hat{i} + 4 \hat{j} + \hat{k} \right|} = \frac{9 - 4 + 2}{\sqrt{9 + 1 + 4} \sqrt{9 + 16 + 1}} = \frac{7}{\sqrt{14} \sqrt{26}} = \frac{7}{\sqrt{2} \sqrt{7} \sqrt{2} \sqrt{13}} = \frac{\sqrt{7}}{\sqrt{52}} = \sqrt{\frac{7}{52}}\]
\[ \Rightarrow \theta = \sin^{- 1} \left( \sqrt{\frac{7}{52}} \right)\]
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