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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are unit vectors such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, then $\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}$ is equal to
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1517 Upvotes
Verified Answer
The correct answer is:
$-\frac{3}{2}$
Given that, $a, b$ and $c$ are unit vectors.
i.e., $\quad|a|=|b|=|c|=1...(i)$
Now, we have
$\begin{gathered}
(a+b+c)=0 \\
\Rightarrow \quad|a+b+c|^{2}=|0|^{2} \\
\Rightarrow \quad|a|^{2}+|b|^{2}+|c|^{2}+2(a \cdot b+b \cdot c+c \cdot a)=0 \\
\Rightarrow \quad 1+1+1+2(a \cdot b+b \cdot c+c \cdot a)=0 [from Eq. (i)] \\
\Rightarrow \quad(a \cdot b+b \cdot c+c \cdot a)=-\frac{3}{2}
\end{gathered}$
i.e., $\quad|a|=|b|=|c|=1...(i)$
Now, we have
$\begin{gathered}
(a+b+c)=0 \\
\Rightarrow \quad|a+b+c|^{2}=|0|^{2} \\
\Rightarrow \quad|a|^{2}+|b|^{2}+|c|^{2}+2(a \cdot b+b \cdot c+c \cdot a)=0 \\
\Rightarrow \quad 1+1+1+2(a \cdot b+b \cdot c+c \cdot a)=0 [from Eq. (i)] \\
\Rightarrow \quad(a \cdot b+b \cdot c+c \cdot a)=-\frac{3}{2}
\end{gathered}$
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