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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 |
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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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