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 is

Options

A. `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))]`

Answer

C

Explanation

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)))]`