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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
A regular hexagon ABCDEF.

We know from the hexagon that AD is parallel to BC orr `AD=2BC` is parallel to FA or EB=2FA and FC is parallel to AB or FC=2AB.
Thus, `AD+EB+FC=2BC+2FA+2AB`
`=2(FA+AB+BC)=2(FC)=2(2AB)=4AB`.
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