Question

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

• A. $co{s}^{-1}\left[-\frac{2\left(A^2-B^2\right)}{\left(A^2+B^2\right)}\right]$
• B. $co{s}^{-1}\left[-\frac{A^2-B^2}{A^2{B}^2}\right]$
• C. $co{s}^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$
• D. $co{s}^{-1}\left[-\frac{\left(A^2-B^2\right)}{A^2+B^2}\right]$

Answer 1

Answer: (C)

As we know that the magnitude of the resultant of two vectors $X$ and $Y$,
$R^2=X^2+Y^2+2XY\cos \theta$...(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
$A^2+B^2=(A+B{)}^2+(A-B{)}^2+2(A+B)(A-B)\cos \theta$
$\Rightarrow A^2+B^2=A^2+B^2+2AB+A^2$
$+B^2-2AB+2\left(A^2-B^2\right)\cos \theta$
$\Rightarrow \frac{-\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}=\cos \theta$
we get, $\theta ={\cos }^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$