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