InterviewSolution
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InterviewSolution
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If `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`, then the value of `|vec(A)+vec(B)|` isA. `(A^(2)+B^(2)+AB)^(1//2)`B. `(A^(2)+B^(2)+(AB)/(sqrt(3)))^(1//2)`C. `A+B`D. `(A^(2)+B^(2)+sqrt(3)AB)^(1//2)` |
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Answer» Correct Answer - A Given `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`….(i) but `|vec(A)xxvec(B)|=|vec(A)||vec(B)| sin theta= AB sin theta` and `vec(A).vec(B)=|vec(A)||vec(B)|cos theta= AB cos theta` Make these substitution in Eq.(i), we get `AB sin theta= sqrt(3) AB cos theta` or `tan theta =sqrt(3)implies theta= 60^(@)` The resultant of vector `vec(A)` and `vec(B)` can be given by the law of parallelogram. `:. |vec(A)+vec(B)|= sqrt(A^(2)+B^(2)+2AB cos 60^(@))` `=sqrt(A^(2)+B^(2)+2ABxx1/2)` `=(A^(2)+B^(2)+AB)^(1//2)` |
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