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Question
Write the angle between the line \[\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{- 2}\] and the plane x + y + 4 = 0.
Solution
\[\text{ The given line is parallel to the vector } \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = \hat{i} + \hat{j} + 0 \hat{k} . \]
\[\text{ We know that the angle } \theta \text{ between the line and the plane is given by } \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) . \left( \hat{i} + \hat{j} + 0 \hat{k} \right)}{\left| \hat{i}+ 2 \hat{j} + 2 \hat{k} \right| \left| \hat{i} + \hat{j} + 0 \hat{k} \right|} = \frac{1 + 2 + 0}{\sqrt{1 + 4 + 4} \sqrt{1 + 1 + 0}} = \frac{3}{3 \sqrt{2}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{1}{\sqrt{2}} \right) = {45}^o \]
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