Let |a→|=2,|b→|=3 and the angle between the vectors a→ and b→ be π4. Then |(a→+2b→)×(2a→-3b→)|2 is equal to [2023]
(1)
We have,
|(a→+2b→)×(2a→-3b→)|2=|-3(a→×b→)+4(b→×a→)|2
=|-7(a→×b→)|2=49 |a→|2|b→|2sin2θ=49×4×9×12=882