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Questions  

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

a
(A2+B2+AB3)12
b
A+B
c
(A2+B2+3AB)12
d
(A2+B2+AB)12

detailed solution

Correct option is D

|A→×B→| =3A→.B→ABsinθ = 3    ABcosθ ⇒tanθ = 3    ∴θ = 600Now |R→| = |A→+B→| = A2+B2+2ABcosθ= A2+B2+2AB12 = A2+B2+AB12

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