If `A B C D` is a rhombus whose diagonals cut at the origin `O ,` then proved that ` vec O A+ vec O B+ vec O C+ vec O D` =0

Options

A. `A vec(B) + A vec(C )`
B. `vec(0)`
C. `2(vec(AB)+vec(BC))`
D. `A vec(C ) +vec(BD)`

Answer

B

Explanation

Since the diagonals of a rhombus bisect each other.
`therefore O vec(A) = -O vec(C ) and O vec(B)=-vec(OD)`
`rArr O vec(A)+O vec(B)+O vec(C ) +vec(OD) = vec(0)`