Let a vector αi^+βj^ be obtained by rotating the vector 3i^+j^ by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices (α,β),(0,β) and (0,0) is equal to
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a
12
b
,1
c
22
d
12
answer is D.
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Detailed Solution
The vector 3i^+j^ makes an angle of 30° with x− axis. ∴α=2cos75°=3−12,β=2sin75°=3+12∴Area is 12α⋅β=123−12=12