Physics - Vectors Question with Solution | TestHub

Given $\vec{R}=\vec{P}+\vec{Q}$ and $\vec{R}$ is perpendicular to $\vec{P}$.

$\vec{R}\cdot \vec{P}=0$

$(\vec{P}+\vec{Q})\cdot \vec{P}=0$

$P^2+PQ\cos \theta =0$

$Q\cos \theta =-P$

Also given $|\vec{R}|=|\vec{P}|$, which implies $R^2=P^2$.

$P^2+Q^2+2PQ\cos \theta =P^2$

$Q^2+2PQ\cos \theta =0$

Substituting $Q\cos \theta =-P$ into the above equation:

$Q^2+2P(-P)=0$

$Q^2=2P^2\Rightarrow Q=\sqrt{2}P$

Substituting $Q=\sqrt{2}P$ in $Q\cos \theta =-P$:

$\sqrt{2}P\cos \theta =-P$

$\cos \theta =-\frac{1}{\sqrt{2}}$

$\theta =\frac{3\pi }{4}$

Answer: $\frac{3\pi }{4}$