Question

ABCD is a rhombus whose diagonals intersect at E . Then `vec(EA) + vec(EB) + vec(EC) + vec(ED)` equals to ______.

पर्याय

• `vec(0)`

• `vec(AD)`

• `2vec(BD)`

• `2vec(AD)`

Answer 1

ABCD is a rhombus whose diagonals intersect at E . Then `vec(EA) + vec(EB) + vec(EC) + vec(ED)` equals to `underlinebb(vec(0))`.

Explanation:

Given, ABCD is a rhombus whose diagonals bisect each other.

`|vec(EA)| = |vec(EC)|` and `|vec(EB)| = |vec(ED)|` but since they are opposite to each other so they are of opposite signs

`\implies vec(EA) = -vec(EC)` and `vec(EB) = -vec(ED)`

`\implies vec(EA) + vec(EC) = vec(0)`  ...(i)

and `vec(EB) + vec(ED) = vec(0)`  ...(ii)

Adding (i) and (ii), we get

`vec(EA) + vec(EB) + vec(EC) + vec(ED) = vec(0)`.