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Question

Let vectors a and b make an angle θ=2π3 between them. If a=1 and b=2, then (a+3b)×(3ab)2 is

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Solution

(a+3b)×(3ab)2
=3a×aa×b+9b×a3b×b2
=3×0a×b9a×b3×02
=10a×b2
=100[absinθ]2
=100[2sin2π3]2
=100[2sin(ππ3)]2
=100[2sinπ3]2
=100(2×32)2
=100×3
=300

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