Let `A B C D` be a p[arallelogram whose diagonals intersect at `P` and let `O` be the origin. Then prove that ` vec O A+ vec O B+ vec O C+ vec O D=4 vec O Pdot`

Options

A. `vec(OP)`
B. 2`vec(OP)`
C. `3vec(OP)`
D. `4vec(OP)`

Answer

D

Explanation

Diagonals of a parallelogram bisect each other. Therefore, P is the mid point of AC and BD both.

So, in `DeltaOAC, vec(OA)+vec(OC)=2vec(OP)`
and in `DeltaODB, vec(OB)+vec(OD)=2vec(OP)`
`rArr vec(OA)+vec(OC)=2 vec(OP) and vec(OB)+vec(OD)=2vec(OP)`
`rArr vec(OA) + vec(OB)+vec(OC)+vec(OD)=4vec(OP)`