If ABCD is a rhombus whose diagonals cut at the origin O, then $\vec{OA}+\vec{OB}+\vec{OC} +\vec{OD}$ equals
Answer & explanation
Correct answer: option 2
Since the diagonals of a rhombus bisect each other.
$∴\vec{OA}=-\vec{OC}$ and $\vec{OB}=-\vec{OD}$
$⇒\vec{OA}+\vec{OB}+\vec{OC} +\vec{OD}=\vec 0$