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If `sqrt(A^(2)+B^(2))` represents the magnitude of resultant of two vectors (A+ B) and (A - B),then the angle between two vectors isA. `cos^(-1)[-(2(A^(2)-B^(2)))/((A^(2)+B^(2)))]`B. `cos^(-1)[-(2(A^(2)-B^(2))/(A^(2)B^(2))]`C. `cos^(-1)[-((A^(2)-B^(2)))/2((A^(2)-B^(2)))]`D. `cos^(-1)[-((A^(2)-B^(2)))/(A^(2)+B^(2))]` |
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Answer» Correct Answer - C As we know that the magnitude of the resultant of two vectors X and Y, `R^(2)=X^(2)+Y^(2)+2XYcostheta` …(i) where,`theta` is the angle between X and Y. Putting, X=(A+B) Y=(A-B) and `R=sqrt(A^(2)+B^(2))` in Eq. (i), we get `A^(2)+B^(2)=(A+B)^(2)+(A-B)^(2)+2(A+B)(A-B)costheta` `!A^(2)+B^(2)=A^(2)+B^(2)+2AB+A^(2)+B^(2)-2AB+2(A^(2)-B^(2))costheta` `!(-(A^(2)+B^(2)))/(2(A^(2)-B^(2)))=costheta` we get, `theta=cos^(-1)[-((A^(2)+B^(2)))/(2(A^(2)-B^(2)))]` |
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