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Question

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

A
(A2+B2+AB3)12
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B
A+B
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C
(A2+B2+3AB)12
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D
(A2+B2+AB)12
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Solution

The correct option is D (A2+B2+AB)12
|A×B|=3 A.B
ABsinθ=3ABcosθ
tanθ=3θ=60
Now,|R|=|A+B|=A2+B2+2ABcosθ
=A2+B2+2AB(12)=(A2+B2+AB)1/2

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