Find the angle between two vectors

It is given that,

$|\vec{a}|=\sqrt{3},|\vec{b}|=2$ and, $\vec{a} \cdot \vec{b}=\sqrt{6}$

Now, we know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.

$\therefore \sqrt{6}=\sqrt{3} \times 2 \times \cos \theta$

$\Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \times 2}$

$\Rightarrow \cos \theta=\frac{1}{\sqrt{2}}$

$\Rightarrow \theta=\frac{\pi}{4}$

Hence, the angle between the given vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$