If ABCDEF is a regular hexagon with $\vec{AB}=\vec a$ and $\vec{BC}=\vec b$, then $\vec{CE}$ equals

$\vec b-2\vec a$

In ΔABC, we have

$\vec{AB}+ \vec{BC}=\vec{AC}⇒\vec{AC} = \vec a + \vec b$   ...(i)

Since AD is parallel to BC and AD = 2 BC. Therefore,

$\vec{AD}=2\vec b$

In ΔACD, we have

$\vec{AC}+\vec{CD}=\vec{AD}$

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

$∴\vec{CE}=\vec{CD}+\vec{DE}=\vec b-\vec a+(-\vec a)=\vec b-2\vec a$