The resultant vec{R} of vec{P} and vec{Q} is | vedclass

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(D) Let $	heta$ be the angle between $\vec{P}$ and $\vec{Q}$.Since the resultant $\vec{R} = \vec{P} + \vec{Q}$ is perpendicular to $\vec{P}$,the angl

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$\frac{3 \pi}{4}$

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(D) Let $	heta$ be the angle between $\vec{P}$ and $\vec{Q}$.Since the resultant $\vec{R} = \vec{P} + \vec{Q}$ is perpendicular to $\vec{P}$,the angle between $\vec{R}$ and $\vec{P}$ is $90^{\circ}$.The formula for the direction of the resultant is $	an \alpha = \frac{Q \sin 	heta}{P + Q \cos 	heta}$,where $\alpha$ is the angle between $\vec{R}$ and $\vec{P}$.Since $\alpha = 90^{\circ}$,$	an 90^{\circ}$ is undefined,which implies the denominator must be zero:$P + Q \cos 	heta = 0 \Rightarrow \cos 	heta = -\frac{P}{Q} \dots (1)$Given $|\vec{P}| = |\vec{R}|$,we use the magnitude formula $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$.Substituting $R = P$:$P^2 = P^2 + Q^2 + 2PQ \cos 	heta$$0 = Q^2 + 2PQ \cos 	heta$$Q^2 = -2PQ \cos 	heta$$\cos 	heta = -\frac{Q}{2P} \dots (2)$From equations $(1)$ and $(2)$:$-\frac{P}{Q} = -\frac{Q}{2P} \Rightarrow Q^2 = 2P^2 \Rightarrow Q = \sqrt{2}P$.Substituting $Q = \sqrt{2}P$ into equation $(1)$:$\cos 	heta = -\frac{P}{\sqrt{2}P} = -\frac{1}{\sqrt{2}}$.Therefore,$	heta = 135^{\circ} = \frac{3 \pi}{4}$ radians.

The resultant $\vec{R}$ of $\vec{P}$ and $\vec{Q}$ is perpendicular to $\vec{P}$. Also,$|\vec{P}| = |\vec{R}|$. Find the angle between $\vec{P}$ and $\vec{Q}$.

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