The resultant of two vectors vec{P} and vec{Q | vedclass

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(D) Let $	heta$ be the angle between vectors $\vec{P}$ and $\vec{Q}$.By the law of vector addition,the resultant $\vec{R} = \vec{P} + \vec{Q}$.The ma

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$R^2 + S^2 = 2(P^2 + Q^2)$

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(D) Let $	heta$ be the angle between vectors $\vec{P}$ and $\vec{Q}$.By the law of vector addition,the resultant $\vec{R} = \vec{P} + \vec{Q}$.The magnitude squared is $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$ ... $(1)$When the direction of $\vec{Q}$ is reversed,the new vector is $-\vec{Q}$. The new resultant is $\vec{S} = \vec{P} - \vec{Q}$.The magnitude squared is $S^2 = P^2 + Q^2 - 2PQ \cos 	heta$ ... $(2)$Adding equations $(1)$ and $(2)$:$R^2 + S^2 = (P^2 + Q^2 + 2PQ \cos 	heta) + (P^2 + Q^2 - 2PQ \cos 	heta)$$R^2 + S^2 = 2(P^2 + Q^2)$Therefore,option $D$ is correct.

The resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. When the direction of $\vec{Q}$ is reversed,the resultant is given by $\vec{S}$. Which one of the following is true for vectors $\vec{R}$ and $\vec{S}$?

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