Question

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

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

Answer 1

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

Given, $|A|=|B|$
or $A=B$.......(i)
Let magnitude of $(A+B)$ is $R$ and for $(A-B)$ is $R^{\prime }$
Now, $R=A+B$
and $R^2=A^2+B^2+2AB\cos \theta$
$R^2=2A^2+2A^2\cos \theta$ .........(ii)
[$\because$ using Eq. (i)]
$R^{\prime }=A-B$
$R^{\prime 2}=A^2+B^2-2AB\cos \theta$
$R^{\prime 2}=2A^2-2A^2\cos \theta$.......(iii)
$[\because$ using Eq. (i)]
Given, $R=n{R}^{\prime }$ or ${\left(\frac{R}{R^{\prime }}\right)}^2=n^2$
Dividing Eq. (ii) by Eq. (iii), we get
$\frac{n^2}{1}=\frac{1+\cos \theta }{1-\cos \theta }$
or $\frac{n^2+1}{n^2-1}=\frac{(1+\cos \theta )+(1-\cos \theta )}{(1+\cos \theta )-(1-\cos \theta )}$ [Apply Componendo- Dividendo rule]
or $\frac{n^2-1}{n^2+1}=\frac{(1+\cos \theta )-(1-\cos \theta )}{(1+\cos \theta )+(1-\cos \theta )}$ [Take reciprocal on both sides]
$\Rightarrow \frac{n^2-1}{n^2+1}=\frac{2\cos \theta }{2}=\cos \theta$
or $\theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$