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Question
Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k} \text{ and } \vec{b} = 4\hat{i} + 4 \hat{j} - 2\hat{k}\]
Solution
\[\text { Let }\theta\text{ be the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\left| \vec{a} \right| = \sqrt{\left( 2 \right)^2 + \left( - 1 \right)^2 + \left( 2 \right)^2} = \sqrt{9} = 3\]
\[\left| \vec{b} \right| = \sqrt{\left( 4 \right)^2 + \left( 4 \right)^2 + \left( - 2 \right)^2} = \sqrt{36} = 6\]
\[ \vec{a} . \vec{b} = 8 - 4 - 4 = 0\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{0}{\left( 3 \right)\left( 6 \right)} = 0\]
\[ \Rightarrow \theta = \cos^{- 1} \left( 0 \right) = \frac{\pi}{2}\]
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