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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

2 `vec(AB)`

B

3 `vec(AB)`

C

4` A vec(AB)`

D

none of these

Text Solution

Verified by Experts

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

`vec(AD) + vec(EB) + vec(FC) = 2 vec(AO) + 2 vec(OB) + 2 vec(OC)`
`" " = 2 ( vec(AO) + vec(OB) ) + 2 vec(OC)`
`" " = 2 vec(AB) + 2 vec(AB) `
`" " ( because vec(OC) = vec(AB))`
`" " = 4 vec(AB)`
`vecR = vec(AB) + vec(AC) + vec(AD) + vec(AE) + vec(AF)`
`= vec(ED) + vec(AC) + vec(AD) + vec(AE) +vec(CD)`
`( because vec(AB) = vec(ED) and vec(AF)= vec(CD))`
`= ( vec(AC) + vec(CD)) + ( vec(AE) + vec(ED)) + vec(AD)`
`= vec(AD) + vec(AD) + vec(AD) = 3 vec(AD) = 6 vec(AO)`
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