Question

Two vectors $\overrightarrow{A}\text{and}\overrightarrow{B}$ 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 $\overrightarrow{A}\text{and}\overrightarrow{B}$ is

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

Answer 1

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

Let magnitude of two vectors $\overrightarrow{A}\text{and}\overrightarrow{B}=a$
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{a^2+a^2+2a^2\cos \theta }\text{and}$
$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{a^2+a^2+2a^2\left[\cos \right({180}^{\circ }-\theta }\left)\right]$
$|\overrightarrow{A}+\overrightarrow{B}|=n|\overrightarrow{A}+\overrightarrow{B}|$

$\Rightarrow \frac{a^2+a^2+2a^2\cos \theta }{a^2+a^2-2a^2\cos \theta }=n^2$
$\Rightarrow \frac{a^2(1+1+2\cos \theta )}{a^2(1+1-2\cos \theta )}=n^2$
$\Rightarrow \frac{(1+\cos \theta )}{(1-\cos \theta )}=n^2$

Using componendo and dividendo throrem, we get
$\theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+!}\right)$.