Three vectors a, b and c satisfy the condition a+b+c=0. Evaluate … - Vidyadip

Question

Three vectors $\vec{a}, \vec{b}$ and $\vec c$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Evaluate the quantity $\mu=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$, if $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{c}|=2$.

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Answer

Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, we have
$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}=0$
or $\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0$
Therefore $\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-|\vec{a}|^{2}=-9$ .....(i)
Again $\vec{b} \cdot(\vec{a}+\vec{b}+\vec{c})=0$
or $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}=-|\vec{b}|^{2}=-16$ .....(ii)
Similarly $\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}=-4$ ......(iii)
Adding (i), (ii) and (iii), we have
$2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})=-29$
or $2 \mu=-29$ i.e. $\mu=\frac{-29}{2}$

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