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Question
ABCD is a rhombus whose diagonals intersect at E . Then `vec(EA) + vec(EB) + vec(EC) + vec(ED)` equals to ______.
Options
`vec(0)`
`vec(AD)`
`2vec(BD)`
`2vec(AD)`
Solution
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)`.
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