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Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{b}}|=6$ and $|\overrightarrow{\mathrm{c}}|=14$
What is a. $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}$. $\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}$. a. equal to?
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What is a. $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}$. $\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}$. a. equal to?
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Verified Answer
The correct answer is:
$-166$
We have $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
So $|\vec{a}+\vec{b}+\vec{c}|=0$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=(10)^{2}+(6)^{2}+(14)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=100+36+196+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow-\frac{332}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
$\Rightarrow-166=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
So $|\vec{a}+\vec{b}+\vec{c}|=0$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=(10)^{2}+(6)^{2}+(14)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow 0=100+36+196+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$
$\Rightarrow-\frac{332}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
$\Rightarrow-166=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
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