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प्रश्न
Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3x − y + z = 1.
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उत्तर
\[\text{ It is given that the line passes through } A \left( 3, - 4, - 2 \right) \text{ and } B \left( 12, 2, 0 \right) . \]
\[\text{ So} , \vec{b} = \vec{AB} = \vec{OB} - \vec{OA} = 12 \hat{i} + 2 \hat{j} + 0 \hat{k} - \left( 3 \hat{i} - 4 \hat{j} - 2 \hat{k} \right) = 9 \hat{i} + 6 \hat{j} + 2 \hat{k} \]
\[\text{ The given line is parallel to the vector } \vec{b} = 9 \hat{i} + 6 \hat{j} + 2 \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = 3 \hat{i} - \hat{j} + \hat{k} . \]
\[ \text{ We know that the angle } \theta \text{ between the line and the plane is given by } \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( 9 \hat{i} + 6 \hat{j} + 2 \hat{k} \right) . \left( 3 \hat{i} - \hat{j} + \hat{k} \right)}{\left| 9 \hat{i} + 6 \hat{j} + 2 \hat{k} \right| \left| 3 \hat{i} - \hat{j} + \hat{k} \right|} = \frac{27 - 6 + 2}{\sqrt{81 + 36 + 4} \sqrt{9 + 1 + 1}} = \frac{23}{11 \sqrt{11}}\]
\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{23}{11 \sqrt{11}} \right)\]
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