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(C) Let the angle between vectors $\vec{P}$ and $\vec{Q}$ be $	heta$.The magnitude of the resultant $A$ is given by the law of parallelogram of vecto

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$2(P^2 + Q^2)$

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(C) Let the angle between vectors $\vec{P}$ and $\vec{Q}$ be $	heta$.The magnitude of the resultant $A$ is given by the law of parallelogram of vectors:$A = |\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$Squaring both sides, we get:$A^2 = P^2 + Q^2 + 2PQ \cos 	heta$  --- $(1)$When vector $\vec{Q}$ is reversed, it becomes $-\vec{Q}$. The angle between $\vec{P}$ and $-\vec{Q}$ is $(180^\circ - 	heta)$.The magnitude of the new resultant $B$ is:$B = |\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ \cos(180^\circ - 	heta)}$Since $\cos(180^\circ - 	heta) = -\cos 	heta$, we have:$B = \sqrt{P^2 + Q^2 - 2PQ \cos 	heta}$Squaring both sides, we get:$B^2 = P^2 + Q^2 - 2PQ \cos 	heta$  --- $(2)$Adding equations $(1)$ and $(2)$:$A^2 + B^2 = (P^2 + Q^2 + 2PQ \cos 	heta) + (P^2 + Q^2 - 2PQ \cos 	heta)$$A^2 + B^2 = 2P^2 + 2Q^2 = 2(P^2 + Q^2)$

Resultant of two vectors $\vec{P}$ and $\vec{Q}$ is of magnitude $A$. If $\vec{Q}$ is reversed, then the resultant is of magnitude $B$. The value of $A^2 + B^2$ is

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