1.

If |AxB|=√3(A.B), then the valtue of |A+B| is(1) (A^2+B^2 + AB)^1/2 (2) (A^2+B^2 +AB/√3)^1/2(3) A+B(4) (A^2+B^2 +√3AB)1/2

Answer»

|A×B| = √3(A•B)

=> ABsin∅ = √3ABcos∅=> sin∅/cos∅ = √3 => tan∅ = √3 .. so, angle between them = 60°

now, |A+B| = √A²+B²+2ABcos60 = √A²+B²+AB

option 1



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