Question

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

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

Answer 1

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

Let the position vectors of $A,B,C,D,P$ be $\overline{a},\overline{b},\overline{c},\overline{d},\overline{p}$ respectively.
We need $\overline{a}+\overline{b}+\overline{c}+\overline{d}$
Also, $\overline{p}=\frac{\overline{a}+\overline{b}}{2}=\frac{\overline{c}+\overline{d}}{2}$.........$(\because$ diagonals bisect each other$)$
So, the question needed becomes $4\overline{p}=4\overline{OP}$