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(C) Let the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ be $\alpha$. The magnitude of the resultant $\overrightarrow{R}$ is gi

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$Q$

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(C) Let the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ be $\alpha$. The magnitude of the resultant $\overrightarrow{R}$ is given by:$R^2 = P^2 + Q^2 + 2PQ \cos \alpha \quad \dots(1)$When $\overrightarrow{Q}$ is doubled to $2\overrightarrow{Q}$,let the new resultant be $\overrightarrow{R_1}$. The magnitude squared is:$R_1^2 = P^2 + (2Q)^2 + 2P(2Q) \cos \alpha = P^2 + 4Q^2 + 4PQ \cos \alpha \quad \dots(2)$Given that the new resultant $\overrightarrow{R_1}$ is perpendicular to $\overrightarrow{P}$,the vector sum $\overrightarrow{R_1} = \overrightarrow{P} + 2\overrightarrow{Q}$ implies that $\overrightarrow{P} \cdot \overrightarrow{R_1} = 0$. Since $\overrightarrow{R_1} = \overrightarrow{P} + 2\overrightarrow{Q}$,we have $\overrightarrow{P} \cdot (\overrightarrow{P} + 2\overrightarrow{Q}) = 0$,which gives $P^2 + 2PQ \cos \alpha = 0$,or $2PQ \cos \alpha = -P^2 \quad \dots(3)$Substitute equation $(3)$ into equation $(1)$:$R^2 = P^2 + Q^2 + (-P^2) = Q^2$Therefore,$R = Q$.

The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$. If $\overrightarrow{Q}$ is doubled,the new resultant is perpendicular to $\overrightarrow{P}$. Then $R$ equals

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