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If $\vec{a}, \vec{b}, \vec{c}$ are the unit vector such that $\vec{a}+\vec{b}+\vec{c}=0$, then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$
Solution:
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Verified Answer
- We have $\vec{a}+\vec{b}+\vec{c}=0$
$$
\begin{aligned}
&\Rightarrow \quad \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{0} \\
&\Rightarrow \quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=0 \\
&\Rightarrow \quad|\overrightarrow{\mathrm{a}}|^2+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0 \\
&\Rightarrow \quad 1+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0
\end{aligned}
$$
Similarly, by taking dot product with $\vec{b}$ and $\vec{c}$, we get
$$
\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+1+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0
$$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+1$
Adding (i), (ii) and (iii), we have
$$
\begin{aligned}
&2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})+3=0 \\
&\therefore \quad \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{3}{2}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \quad \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{0} \\
&\Rightarrow \quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=0 \\
&\Rightarrow \quad|\overrightarrow{\mathrm{a}}|^2+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0 \\
&\Rightarrow \quad 1+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0
\end{aligned}
$$
Similarly, by taking dot product with $\vec{b}$ and $\vec{c}$, we get
$$
\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+1+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0
$$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+1$
Adding (i), (ii) and (iii), we have
$$
\begin{aligned}
&2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})+3=0 \\
&\therefore \quad \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{3}{2}
\end{aligned}
$$
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