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. `2vec(OP)`
B. `4vec(OP)`
C. `6vec(OP)`
D. `8vec(OP)`

Answer

B

Explanation

From the figure, observe that
`vec(OA)+vec(AP)=vec(OP)`

Similarly, `vec(OB)+vec(BP)=vec(OP)`
`vec(OC)+vec(CP)=vec(OP)`
`vec(OD)+vec(DP)=vec(OP)`
`rArr vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(AP)`
`+vec(BP)+vec(CP)+vec(DP)=4vec(OP) rarr (i)`
In a parallelogram, diagonals bisect each other.
`:. vec(AP)=vec(PC) ` (`:. ` Since AP=CP)
`rArr vec(AP)=-vec(PC)`
`rArr vec(AP)+vec(PC)=O`
Similarly `vec(BP)+vec(DP)=O`
`:. (i) rArr vec(OA)+vec(OB)+vec(OC)+vec(OD)=4vec(OP)`