Question

Vectors $\overline{A}$ and $\overline{B}$ are equal in magnitude. The magnitude of $\overline{A}+\overline{B}$ is larger than the magnitude of $\overline{A}-\overline{B}$ by a factor of $n$, then the angle between them is

• A. 2 ${\tan }^{-1}\left(1/n\right)$
• B. ${\tan }^{-1}\left(1/n\right)$
• C. ${\tan }^{-1}\left(1/2n\right)$
• D. 2 ${\tan }^{-1}\left(1/2n\right)$

Answer 1

The correct option is A 2 ${\tan }^{-1}\left(1/n\right)$

We have $|\overline{A}+\overline{B}|=n|\overline{A}-\overline{B}|$
Thus we get: $A^2+B^2+2ABcos\theta =n^2(A^2+B^2-2ABcos\theta )$
Since the magnitudes of the two vectors are equal we get
$2A^2(1+cos\theta )=n^{2}2A^2(1-cos\theta )$
$\frac{1+cos\theta }{1-cos\theta }=n^2$

Now as $cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}$

We get : $tan\frac{\theta }{2}=\frac{1}{n}$
$\theta =2ta{n}^{-1}\frac{1}{n}$