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If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\vec{a}|=7,|\vec{b}|=5,|\vec{c}|=3$ then angle between vector $\vec{b}$ and $\vec{c}$ is
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The correct answer is:
$60^{\circ}$
$60^{\circ}$
$\vec{a}+\vec{b}+\vec{c}=0 \Rightarrow \vec{b}+\vec{c}=-\vec{a} \Rightarrow(\vec{b}+\vec{c})^2=(\vec{a})^2=5^2+3^2+2 \vec{b} \vec{c}=7^2$
$\Rightarrow 2|\vec{b}||\vec{c}| \cos \theta=49-34=15 \Rightarrow 2 \times 5 \times 3 \cos \theta=15 \Rightarrow \cos \theta=1 / 2 \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$
$\Rightarrow 2|\vec{b}||\vec{c}| \cos \theta=49-34=15 \Rightarrow 2 \times 5 \times 3 \cos \theta=15 \Rightarrow \cos \theta=1 / 2 \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$
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