$\vec a$, $\vec b$, $\vec c$ are unit vectors so that $\vec a+\vec b+\vec c=\vec 0$. What is $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$? 
Let $\vec a$, $\vec b$, $\vec c$ be unit vectors so that $\vec a+\vec b+\vec c=\vec 0$. How do I find $\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$?

 A: Hint: The three vectors form an equilateral triangle of unit side length, so we know all lengths and angles. Now use the fact that $\vec u \cdot \vec v = |\vec u||\vec v| \cos \theta$.
A: Hint
$$\vec a+\vec b+\vec c=\vec 0\implies (\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)=0.$$ That is
$$3+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$$
A: $$|\vec a|^2+\vec a\cdot\vec b+\vec a\cdot\vec c=0$$
$$\vec a\cdot\vec b+|\vec b|^2+\vec b\cdot\vec c=0$$
$$ a\cdot\vec c+\vec b\cdot\vec c+|\vec c|^2=0$$
A: $\vec a+\vec b+\vec c=\vec 0 \Rightarrow |\vec a+\vec b+\vec c|^2=0$,$$|a|^2+|b|^2+|c|^2+2(\vec a.\vec b +\vec b.\vec c+\vec a.\vec c)=0$$And since, $|a|^2+|b|^2+|c|^2=3$,therefore $$\vec a.\vec b+\vec b.\vec c+\vec a.\vec c= -\frac{3}{2}$$
A: Hint: if $\vec a+\vec b+\vec c=0$, then:
$$\begin{align}
\| \vec a+\vec b+\vec c \|^2 = \left(\vec a+\vec b+\vec c \right) \cdot \left(\vec a+\vec b+\vec c \right)  
 &=0 \\\
\| \vec a \|^2 + \| \vec b \|^2 + \| \vec c \|^2 + 2\left( \color{blue}{\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a} \right) &=0
\end{align}$$
Since the vectors are unit vectors, the norms are... You're looking for the blue expression.
A: Starting  with $$
(\vec{a}+\vec{b}+\vec{c})\cdot (\vec{a}+\vec{b}+\vec{c})=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot \vec{b} +\vec{b}\cdot \vec{c}+\vec{a} \cdot \vec{c})
$$
$$
\implies 0=1+1+1+2(\vec{a}\cdot \vec{b} +\vec{b}\cdot \vec{c}+\vec{a} \cdot \vec{c})
$$
$$
\implies \vec{a}\cdot \vec{b} +\vec{b}\cdot \vec{c}+\vec{a} \cdot \vec{c}=\frac{-3}{2}
$$
A: We have 3 unit vectors, sum of these vectors is 0. This means that these 3 vectors form equilateral triangle. We have an equilateral triangle ABC and need to calculate $\vec{AB} \cdot \vec{BC} + \vec{BC}\cdot\vec{CA} + \vec{CA}\cdot\vec{CA}$.
$\vec{AB}\cdot\vec{BC} = -1/2$ because these are unit vectors, and the angle between these vectors is 120 degrees (the triangle is equilateral). situation is the same with other two products, and the final answer is -3/2.
No need for any magic tricks in this problem.
