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

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(B) Let $	heta$ be the angle between $\vec{P}$ and $\vec{Q}$. The resultant $\vec{R} = \vec{P} + \vec{Q}$.When $\vec{Q}$ is doubled,the new resultant

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$Q$

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(B) Let $	heta$ be the angle between $\vec{P}$ and $\vec{Q}$. The resultant $\vec{R} = \vec{P} + \vec{Q}$.When $\vec{Q}$ is doubled,the new resultant is $\vec{R}' = \vec{P} + 2\vec{Q}$.It is given that $\vec{R}'$ is perpendicular to $\vec{P}$,so their dot product is zero:$(\vec{P} + 2\vec{Q}) \cdot \vec{P} = 0$$\vec{P} \cdot \vec{P} + 2(\vec{Q} \cdot \vec{P}) = 0$$P^2 + 2PQ \cos 	heta = 0$$2PQ \cos 	heta = -P^2$Now,the magnitude of the original resultant $\vec{R}$ is:$R = |\vec{P} + \vec{Q}| = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$Substituting $2PQ \cos 	heta = -P^2$ into the equation:$R = \sqrt{P^2 + Q^2 - P^2}$$R = \sqrt{Q^2} = Q$

The resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If $\vec{Q}$ is doubled,the new resultant vector is perpendicular to $\vec{P}$. What is the magnitude of $\vec{R}$?

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