Question

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P}+\vec{Q}$ is $n$ times the magnitude of $\vec{P}-\vec{Q}$, then angle between $\vec{P}$ and $\vec{Q}$ is :

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

Answer 1

Answer: (D)

$|\vec{P}|=|\vec{Q}|=x \ldots \ldots$ (i)
$|\vec{P}+\vec{Q}|=n|\vec{P}-\vec{Q}| $
$P^{2}+Q^{2}+2 P Q \cos \theta=n^{2}\left(P^{2}+Q^{2}-2 P Q \cos \theta\right)$
Using (i) in above equation
$\cos \theta=\frac{n^{2}-1}{1+n^{2}}$
$\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$