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If |A×B|=3  A.B then the value of |A+B|  is

a
(A2+B2+AB)1/2
b
(A2+B2+AB3)1/2
c
(A+B)
d
(A2+B2+3AB)1/2

detailed solution

Correct option is A

As |A→×B→|=3A.B  so     AB sinθ=3ABcosθor    tanθ=3=tan600  or θ=600∴     |A→+B→|=A2+B2+2ABcos600=(A2+B2+AB)1/2

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