Find the angle between two vectors with the help of scalar product.

If the $\theta$ is the angle between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$, then vector product,

$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta$

$\therefore \quad \cos \theta=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}|}$

$\therefore \quad \cos \theta=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}$

$\therefore \quad \theta=\cos ^{-1}\left(\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\right)$

In Cartesian co-ordinate system,

$\cos \theta=\frac{\left(\mathrm{A}_{x} \mathrm{~B}_{x}+\mathrm{A}_{y} \mathrm{~B}_{y}+\mathrm{A}_{z} \mathrm{~B}_{z}\right)}{\left(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}+\mathrm{A}_{z}^{2}}\right)\left(\sqrt{\mathrm{B}_{x}^{2}+\mathrm{B}_{y}^{2}+\mathrm{B}_{z}^{2}}\right)}$