Question

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

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

Answer 1

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

Consider the problem

Since

$P$ which is the intersection of diagonals of parallelogram its bisects the diagonal

Thus

$OP=\frac{(OA+OC)}{2}$

i.e.

$OA+OC=2OP$ ----- (i)

Similarly

$OB+OD=OP$ ----- (ii)

Adding (i) and (ii)

we get

$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}$

Hence the option $D$ is the correct answer.