Question:
Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2 , respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$.
Solution:
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 $\frac{\pi}{4}$