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If ABCDEF is a regular hexagon then vec ...

If ABCDEF is a regular hexagon then `vec AD + vec EB + vec FC = `

A

(a)2AB

B

(b)3AB

C

(c)4AB

D

(d)none of these

Text Solution

Verified by Experts

The correct Answer is:
C
Consider the regular hexagon ABCDEF with centre at O (origin).

`AD+EB+FC=2AO+2OB+2OC`
`=2(AO+OB)+2OC`
`=2AB+2AB" "[becauseOC=AB]`
`=4AB`
`R=AB+AC+AD+AE+AF`
`=ED+AC+AD+AE+CD" "[becauseAB=ED and AF=CD]`
`=(AC+CD)+(AE+ED)+AD`
`=AD+AD+AD=3AD=6AO`
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