#### Question

Find the angle between two vectors \[\vec{a} \text{ and } \vec{b}\] if

\[\left| \vec{a} \right| = \sqrt{3}, \left| \vec{b} \right| = 2 \text{ and } \vec{a} \cdot \vec{b} = \sqrt{6}\]

#### Solution

Let θ be the angle between \[\vec{a} \text{ and }\vec{b} .\]

Given that

\[\left| \vec{a} \right| = \sqrt{3}, \left| \vec{b} \right| = \text{ 2 and }\vec{a} . \vec{b} = \sqrt{6} . . . \left( 1 \right)\]

We know that

\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta\]

\[ \Rightarrow \sqrt{6} = \left( \sqrt{3} \right)\left( 2 \right) \cos \theta .....................\left[ \text{ Using }\left( 1 \right) \right]\]

\[ \Rightarrow \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]