If $\theta$ is the angle between $\vec{A}$ and $\vec{B}$, then scalar product

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

$=\mathrm{BA} \cos \theta$

$\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}$ $=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$

$[\because \mathrm{AB}=\mathrm{BA}]$