In the following figure ABCDEF is a regular hexagon. If $\overrightarrow{A B}=\vec{a}$ and $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$ then $\overrightarrow{\mathrm{CD}}$ in terms of a and b is :

\(\vec{a} + \vec{b}\) \(\vec{a} - \vec{b}\) \(\vec{b} - \vec{a}\) \(3\vec{b} - \vec{a}\)

\(\vec{b} - \vec{a}\)

$∴ \vec{CD}=\vec{CA}+\vec{AD}$ $=-(\vec a+\vec b)+2\vec b = -\vec a-\vec b+2\vec b$ $=-\vec a+\vec b⇒\vec b - \vec a$