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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)` equals :

A

0

B

2AB

C

3AB

D

4AB

Text Solution

Verified by Experts

The correct Answer is:
D
Since, ABCDEF is a regular hexagon. We know rom the hexagon that AD||BC
`impliesAD=2BC`
similarly, EB=2FA
and FC=2AB
thus, AD+EB+FC
`=2(BC+FA+AB)`
`=2(FC)=4AB`
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