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$\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $|\overline{\mathrm{a}}|=3,|\overline{\mathrm{b}}|=5,|\overline{\mathrm{c}}|=7$, then the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$
$\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=0 \Rightarrow \overline{\mathrm{c}}=-(\overline{\mathrm{a}}+\overline{\mathrm{b}})$ and let angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ be $\theta$
$$
\begin{aligned}
& \therefore|\overline{\mathrm{c}}|^2=(\overline{\mathrm{a}}+\overline{\mathrm{b}})^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \\
& =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| \cdot \cos \theta \\
& \therefore(7)^2=(3)^2+(5)^2+2(3)(5) \cos \theta \\
& \therefore 49=9+25+30 \cos \theta \Rightarrow \cos \theta=\frac{15}{30}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore|\overline{\mathrm{c}}|^2=(\overline{\mathrm{a}}+\overline{\mathrm{b}})^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \\
& =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+2|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| \cdot \cos \theta \\
& \therefore(7)^2=(3)^2+(5)^2+2(3)(5) \cos \theta \\
& \therefore 49=9+25+30 \cos \theta \Rightarrow \cos \theta=\frac{15}{30}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
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