Question

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

• A. $\overrightarrow{AB}+\overrightarrow{AC}$
• B. $\overrightarrow{0}$
• C. $2(\overrightarrow{AB}+\overrightarrow{BC})$
• D. $\overrightarrow{AC}+\overrightarrow{BD}$

Answer 1

The correct option is B $\overrightarrow{0}$

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, $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}$