Question

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ have equal magnitudes. The magnitude of $(\overrightarrow{A}+\overrightarrow{B})$ is ${}^{\prime }{n}^{\prime }$ times the magnitude of $(\overrightarrow{A}-\overrightarrow{B})$. The angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

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

Answer 1

The correct option is C ${\cos }^{-1}\left[\frac{n^2-1}{n^2+1}\right]$

Given, $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+B^2+2AB\cos \theta }$
As the magnitude of $\overrightarrow{A}$ is equal to the magnitude of $\overrightarrow{B}$ . Hence,
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^2+A^2+2A^2\cos \theta }$
Similarly, $\overrightarrow{R}=\overrightarrow{A}-\overrightarrow{B}$
$|\overrightarrow{A}-\overrightarrow{B}|=\sqrt{A^2+B^2+2A^2\cos \theta }$
Given,
$|\overrightarrow{A}+\overrightarrow{B}|=n|\overrightarrow{A}-\overrightarrow{B}|$
$|\overrightarrow{A}+\overrightarrow{B}{|}^2=n^2|\overrightarrow{A}-\overrightarrow{B}{|}^2$
$A^2+A^2+2A^2\cos \theta =n^2(A^2+A^2-2A^2\cos \theta )$
$2A^2(1+\cos \theta )=2A^2{n}^2(1-\cos \theta )$
$1+\cos \theta =n^2-n^2\cos \theta$
$\cos \theta (1+n^2)=n^2-1$
$\cos \theta =\frac{n^2-1}{n^2+1}$
$\theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$
Final Answer: $\left(c\right)$