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Let a vector a hati + beta hatj be obt...

Let a vector `a hati + beta hatj ` be obtained by rotating the vector `sqrt3 hati + hatj` by an angle `45^(@)` about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices `(alpha, beta) , (0, beta) and (0,0)` is equal to:

A

1

B

`1/2`

C

`(1)/(sqrt2)`

D

`2sqrt2`

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