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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Solution

$It is given that the line passes through A (3, - 4, - 2) and B (12, 2, 0) . So, \vec{b} = \vec{A B} = \vec{O B} - \vec{O A} = 12 \hat{i} + 2 \hat{j} + 0 \hat{k} - (3 \hat{i} - 4 \hat{j} - 2 \hat{k}) = 9 \hat{i} + 6 \hat{j} + 2 \hat{k} The given line is parallel to the vector \vec{b} = 9 \hat{i} + 6 \hat{j} + 2 \hat{k} and the given plane is normal to the vector \vec{n} = 3 \hat{i} - \hat{j} + \hat{k} . We know that the angle θ between the line and the plane is given by \sin θ = \frac{\vec{b} . \vec{n}}{|\vec{b}| |\vec{n}|} = \frac{(9 \hat{i} + 6 \hat{j} + 2 \hat{k}) . (3 \hat{i} - \hat{j} + \hat{k})}{|9 \hat{i} + 6 \hat{j} + 2 \hat{k}| |3 \hat{i} - \hat{j} + \hat{k}|} = \frac{27 - 6 + 2}{\sqrt{81 + 36 + 4} \sqrt{9 + 1 + 1}} = \frac{23}{11 \sqrt{11}} \Rightarrow θ = \sin^{- 1} (\frac{23}{11 \sqrt{11}})$

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