Physics - Vectors Question with Solution | TestHub

Let the angle between vectors $\vec{P}$ and $\vec{Q}$ be $\theta$.

The magnitude of the resultant of $\vec{P}$ and $\vec{Q}$ is $A$.

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

When $\vec{Q}$ is reversed, the angle between $\vec{P}$ and $-\vec{Q}$ becomes $180^{\circ }-\theta$.

The magnitude of the new resultant is $B$.

$B^2=P^2+Q^2+2PQ\cos (180^{\circ }-\theta )$

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

Adding the two equations, we get:

$A^2+B^2=(P^2+Q^2+2PQ\cos \theta )+(P^2+Q^2-2PQ\cos \theta )$

$A^2+B^2=2(P^2+Q^2)$