ABCDEF is a regular hexagon with centre at the origin such that $\vec{AD}+\vec{EB}+\vec{FC}$ is equal to $λ\vec{ED}$ then λ, is:

4

As the hexagon is regular,

$\vec{OA}=-\vec{OD}$ … (∵ $|\vec{OA}|=|\vec{OD}|$ and OA and DO are parallel)

Similarly, $\vec{OB}=-\vec{OE}$ and $\vec{OC}=-\vec{OF}$

⇒ $\vec d=-\vec a,\,\vec e=-\vec b$ and $\vec f=-\vec c$

Now, $\vec{AD}+\vec{EB}+\vec{FC}=(\vec d-\vec a)+(\vec b-\vec e)+(\vec c-\vec f)$

$=-\vec a+(2\vec b)+(2\vec c)$ … (i)

Also $\vec{OA}=\vec{CB}⇒\vec a=\vec b-\vec c$

Now (i) becomes $-2\vec a+2\vec b+2(\vec b-\vec a)=4(\vec{AB})=4(\vec{ED})$ $(∵\vec{ED}=\vec{AB})$