If |vec{P}+vec{Q}|=|vec{P}|=|vec{Q}|,then the | vedclass

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(B) Given that $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$. Let $|\vec{P}|=|\vec{Q}|=P$.Using the vector addition formula,$|\vec{P}+\vec{Q}| = \sqrt{P^2+Q

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Given that $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$. Let $|\vec{P}|=|\vec{Q}|=P$.Using the vector addition formula,$|\vec{P}+\vec{Q}| = \sqrt{P^2+Q^2+2PQ \cos 	heta}$.Since $|\vec{P}+\vec{Q}|=P$,we have $P = \sqrt{P^2+P^2+2P^2 \cos 	heta}$.Squaring both sides,$P^2 = 2P^2 + 2P^2 \cos 	heta$.Dividing by $P^2$,$1 = 2 + 2 \cos 	heta$.$2 \cos 	heta = -1 \Rightarrow \cos 	heta = -\frac{1}{2}$.Therefore,$	heta = 120^{\circ}$.

If $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$,then the angle between $\vec{P}$ and $\vec{Q}$ is: (in $^{\circ}$)

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