Question

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

• A. $\overrightarrow{OA}$
• B. $2\overrightarrow{OP}$
• C. $3\overrightarrow{OP}$
• D. $4\overrightarrow{OP}$

Answer 1

The correct option is D $4\overrightarrow{OP}$

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