ABCDEF is a regular hexagon in the x-y plance with vertices in the anticlockwise direction. If `vecAB=2hati`, then `vecCD` is

Options

A. `hati+sqrt(3)hatj`
B. `hati-sqrt(3)hatj`
C. `-hati+sqrt(3)hatj`
D. `sqrt(3)hati-hatj`

Answer

C

Explanation

`vecAB` is along the x-axis and BD is along the y-axis
`vec(AB) = 2veci rArr AB=BC=CD=………………=2`
From the figure, BM=BC`sin60^(@)`
`=2sin60^(@)`
`=sqrt(3)`
`therefore vec(BD) = 2sqrt(3)vecj`
`vec(BC) = BC cos60^(@)veci+BCsin60^(@)vecj`
`=veci+sqrt(3)vecj`
`therefore vec(CD) = vec(BD) = 2sqrt(3)vecj-(veci+sqrt(3)vecj)`
`=-veci+sqrt(3)vecj`