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Three vectors $\bar{a}, \bar{b}, \bar{c}$ satisfy the condition $\bar{a}+\bar{b}+\bar{c}=\overline{0}$.
If $|\bar{a}|=1,|\bar{b}|=3,|\bar{c}|=4$ then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$
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If $|\bar{a}|=1,|\bar{b}|=3,|\bar{c}|=4$ then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$
Solution:
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Verified Answer
The correct answer is:
$-13$
$(\vec{a}+\vec{b}+\vec{c})^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\begin{aligned} & 0=(1)^2+(3)^2+(4)^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ & 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-26 \\ & \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-13\end{aligned}$
$\begin{aligned} & 0=(1)^2+(3)^2+(4)^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ & 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-26 \\ & \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-13\end{aligned}$
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