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Question
Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 3\hat{i} - 2\hat{j} - 6\hat{k} \text{ and } \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}\]
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Solution
\[\text { Let }\theta \text{ be }\ \text{ the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\left| \vec{a} \right| = \sqrt{\left( 3 \right)^2 + \left( - 2 \right)^2 + \left( - 6 \right)^2} = \sqrt{49} = 7\]
\[\left| \vec{b} \right| = \sqrt{\left( 4 \right)^2 + \left( - 1 \right)^2 + \left( 8 \right)^2} = \sqrt{81} = 9\]
\[ \vec{a} . \vec{b} = 12 + 2 - 48 = -34 \]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{-34}{\left( 7 \right)\left( 9 \right)} = \frac{-34}{63}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{-34}{63} \right)\]
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