Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then $\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}$ equals

$\vec{OP}$ $2\vec{OP}$ $3\vec{OP}$ $4\vec{OP}$

$4\vec{OP}$

Since the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both. $∴\vec{OA}+\vec{OC}=2\vec{OP}$ and $\vec{OB}+\vec{OD}=2\vec{OP}$ $⇒\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}=4\vec{OP}$