Question

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ have equal magnitudes. If magnitudes of $\overrightarrow{A}+\overrightarrow{B}$ is equal to $n$ times the magnitude of $\overrightarrow{A}-\overrightarrow{B},$ then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is:

• A. ${\cos }^{-1}\left(\frac{n+n^2}{n-n^2}\right)$
• B. ${\cos }^{-1}\left(\frac{n^2+1}{n^2-1}\right)$
• C. ${\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$
• D. ${\cos }^{-1}\left(\frac{2+n}{2-n}\right)$

Answer 1

Answer: (C)

According to the question
$|\overrightarrow{A}+\overrightarrow{B}|=n|\overrightarrow{A}-\overrightarrow{B}|$
Squaring both side
$A^2+B^2+2AB\cos \theta =n^2\left(A^2+B^2-2AB\cos \theta \right)$
$A^2+B^2+2AB\cos \theta =n^2{A}^2+n^2{B}^2-n^2\times 2AB\cos \theta$
$2A^2+2A^2\cos \theta =2n^2{A}^2-2n^2{A}^2\cos \theta$
$2A^2(1+\cos \theta )=2n^2{A}^2(1-\cos \theta )$
$1+\cos \theta =n^2-n^2\cos \theta |\vec{A}+\vec{B}|=\sqrt{A^2+B^2+2AB\cos \theta }$
$1+\cos \theta +n^2\cos \theta =n^2|\vec{A}-\vec{B}|=\sqrt{A^2{B}^2-2AB\cos \theta }$
$\cos \theta =\frac{n^2-1}{n^2+1}|\vec{A}|=|\vec{B}|$
$\theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$