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Question
Write the angle between two vectors \[\vec{a} \text{ and } \vec{b}\] with magnitudes \[\sqrt{3}\] and 2 respectively if \[\vec{a} \cdot \vec{b} = \sqrt{6} .\]
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Solution
\[\text{ Let } \theta \text{ be the angle between } \vec{a} \text{ and } \vec{b} .\]
\[\text{ Given },\]
\[\left| \vec{a} \right| = \sqrt{3}; \left| \vec{b} \right| = 2; \vec{a} . \vec{b} = \sqrt{6}\]
\[\text{ We know that }\]
\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta\]
\[ \Rightarrow \sqrt{6} = \left( \sqrt{3} \right)\left( 2 \right) \cos \theta\]
\[ \Rightarrow \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]
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