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)

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