Question

When two vectors of magnitudes $P$ and $Q$ are inclined at an angle $\theta$, the magnitude of their resultant is $2P$. When the inclination is changed to ${180}^o-\theta$, the magnitude of the resultant is halved. Find the ratio of $P$ to $Q$.

Answer 1

When the angle is $\theta$, the magnitude of resultant is

$2P=\sqrt{P^2+Q^2+2PQcos\theta }$ (1)

When the angle is $180-\theta$, the magnitude of resultant is

$P=\sqrt{P^2+Q^2+2PQcos\left(180-\theta \right)}$ (2)

Squaring and adding the two equations, we get

$2P^2+2Q^2=5P^2$

$2Q^2=3P^2$

$\frac{P}{Q}=\sqrt{\frac{2}{3}}$