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If $a+b+c=0,|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is
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The correct answer is:
$\frac{\pi}{3}$
$\frac{\pi}{3}$
Let $a+b+c=0 \Rightarrow(a+b)=-c$ $\Rightarrow(a+b)^2=c^2$
$$
\begin{aligned}
& \Rightarrow a^2+b^2+2 a \cdot b=c^2 \\
& \Rightarrow 9+25+2.3 .5 \cos \theta=49 \\
& (\because|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7) \\
& \therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow a^2+b^2+2 a \cdot b=c^2 \\
& \Rightarrow 9+25+2.3 .5 \cos \theta=49 \\
& (\because|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7) \\
& \therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
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