The sum of two forces overrightarrow{P} and o | vedclass

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(B) Given that $|\overrightarrow{P} + \overrightarrow{Q}| = |\overrightarrow{P}|$.Squaring both sides,we get $P^2 + Q^2 + 2PQ \cos 	heta = P^2$,where

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(B) Given that $|\overrightarrow{P} + \overrightarrow{Q}| = |\overrightarrow{P}|$.Squaring both sides,we get $P^2 + Q^2 + 2PQ \cos 	heta = P^2$,where $	heta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$.This simplifies to $Q^2 + 2PQ \cos 	heta = 0$,or $Q(Q + 2P \cos 	heta) = 0$.Since $Q 
eq 0$,we have $Q + 2P \cos 	heta = 0$.Now,let $\overrightarrow{R'} = 2\overrightarrow{P} + \overrightarrow{Q}$. The angle $\alpha$ that $\overrightarrow{R'}$ makes with $\overrightarrow{Q}$ is given by $	an \alpha = \frac{|2\overrightarrow{P}| \sin 	heta}{|\overrightarrow{Q}| + |2\overrightarrow{P}| \cos 	heta} = \frac{2P \sin 	heta}{Q + 2P \cos 	heta}$.Substituting $Q + 2P \cos 	heta = 0$ into the expression,we get $	an \alpha = \frac{2P \sin 	heta}{0} = \infty$.Therefore,$\alpha = 90^{\circ}$.

The sum of two forces $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$ such that $|\overrightarrow{R}| = |\overrightarrow{P}|$. The angle $\alpha$ (in degrees) that the resultant of $2\overrightarrow{P}$ and $\overrightarrow{Q}$ will make with $\overrightarrow{Q}$ is

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