For two vectors vec{P} and vec{Q}, vec{P} cdo | vedclass

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(C) Given that $\vec{P} \cdot \vec{Q} = |\vec{P} 	imes \vec{Q}|$.Using the definitions of dot and cross products: $PQ \cos 	heta = PQ \sin 	heta$.D

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$\sqrt{P^2 + Q^2 + \sqrt{2}PQ}$

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(C) Given that $\vec{P} \cdot \vec{Q} = |\vec{P} 	imes \vec{Q}|$.Using the definitions of dot and cross products: $PQ \cos 	heta = PQ \sin 	heta$.Dividing both sides by $PQ$, we get $\cos 	heta = \sin 	heta$, which implies $	an 	heta = 1$, so $	heta = 45^\circ$.The magnitude of the resultant vector $\vec{R} = \vec{P} + \vec{Q}$ is given by $R = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$.Substituting $	heta = 45^\circ$ and $\cos 45^\circ = 1/\sqrt{2}$:$R = \sqrt{P^2 + Q^2 + 2PQ(1/\sqrt{2})}$$R = \sqrt{P^2 + Q^2 + \sqrt{2}PQ}$.

For two vectors $\vec{P}$ and $\vec{Q}$, $\vec{P} \cdot \vec{Q} = |\vec{P} \times \vec{Q}|$. The magnitude of $\vec{R} = \vec{P} + \vec{Q}$ is:

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