Question

If $ABCD$ is a rhombus whose diagonals cut at the origin $O$, then $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}$ equals to

• A. $AB+BC$
• B. Zero
• C. $2(AB+BC)$
• D. $AC+BD$

Answer 1

The correct option is B Zero

Given that $ABCD$ is a rhombus.

Since, the diagonals of rhombus bisect each other.
Thus $\overrightarrow{OA}=-\overrightarrow{OC}$ and $\overrightarrow{OB}=-\overrightarrow{OD}$
Therefore, $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=0$