Let a vector α i ^ + β j ^ be obtained by rotating the vector 3 i ^ + j ^ by an angle 45 °…

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 (in sq. units) of triangle having vertices α,β,0,β and 0,0 is equal to
  1. 12
  2. 1
  3. 12
  4. 22

Solution

Area of ΔOA'B=12OA'cos15°×OA'sin15°

=12OA'2sin30°2

Now, OA=OA', since rotation will not change the magnitude.

=3+1×18=12 sq. units

Asked in: JEE Main 2021 (16 Mar Shift 1)

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