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If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$, then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is
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Verified Answer
The correct answer is:
3
$\begin{array}{l}
\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\
|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\
\Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\
\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}
\end{array}$
Also,
$\begin{array}{l}
|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\
=1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\
\Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\
\therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3
\end{array}$
\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\
|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\
\Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\
\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}
\end{array}$
Also,
$\begin{array}{l}
|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\
=1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\
\Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\
\therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3
\end{array}$
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