Physics - Vectors Question with Solution | TestHub

Given $\vec{P}\cdot \vec{Q}=|\vec{P}\times \vec{Q}|$

$PQ\cos \theta =PQ\sin \theta$

$\tan \theta =1\Rightarrow \theta =45^{\circ }$

The magnitude of the resultant vector $\vec{R}=\vec{P}+\vec{Q}$ is given by:

$|\vec{R}|=\sqrt{P^2+Q^2+2PQ\cos \theta }$

Substituting $\theta =45^{\circ }$:

$|\vec{R}|=\sqrt{P^2+Q^2+2PQ\cos 45^{\circ }}$

$|\vec{R}|=\sqrt{P^2+Q^2+2PQ\left(\frac{1}{\sqrt{2}}\right)}$

$|\vec{R}|=\sqrt{P^2+Q^2+\sqrt{2}PQ}$

Answer: $\sqrt{P^2+Q^2+\sqrt{2}PQ}$