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Question
Find the angle between the vectors \[\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}\]
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Solution
\[\text{ Let }\theta \text{ be the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\left| \vec{a} \right| = \sqrt{\left( 2 \right)^2 + \left( - 3 \right)^2 + \left( 1 \right)^2} = \sqrt{14}\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( - 2 \right)^2} = \sqrt{6}\]
\[ \vec{a} . \vec{b} = 2 - 3 - 2 = - 3\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 3}{\sqrt{14}\sqrt{6}} = \frac{- 3}{\sqrt{84}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 3}{\sqrt{84}} \right)\]
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