Find three vectors in ${R^3}$ such that the angle between all of them is pi/3? Is there a simple way to do this? I have found that ${(a.b)/|a||b|}$ must be equal to ${1/2}$ but from there I am stuck how to proceed. Any help?
P.s. This is from the MIT 2016 Linear Algebra course and is not homework. 

 A: Diagonals of cube faces forms a tetrahedron

A: Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1 = \cos \frac{\pi}{6} i + \sin \frac{\pi}{6} j$ and $v_2 = \cos \frac{\pi}{6} i - \sin \frac{\pi}{6} j$. We have by construction $\angle(v_1, v_2) = \frac{\pi}{3}$.
Now let's find $\alpha$ such that $v_3 = \cos \alpha i + \sin \alpha k$ solve the problem.
That will be the case providing that $\angle(v_1,v_3) = \frac{\pi}{3}$, i.e. if $\cos \frac{\pi}{6} \cos \alpha = \cos \frac{\pi}{3}$, i.e. $\cos \alpha = \frac{1}{\sqrt 3}$ and $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \frac{\sqrt2}{\sqrt{3}}$.
Finally $$\begin{cases}
v_1 &= \frac{\sqrt 3}{2} i + \frac{1}{2}j\\
v_2 &= \frac{\sqrt 3}{2} i - \frac{1}{2}j\\
v_3 &= \frac{1}{\sqrt 3}(i + \sqrt 2 k)
\end{cases}$$
is a solution.
A: The position vectors of the vertices of the equilateral triangle $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ with respect to the center of the triangle $(1/3,1/3, 1/3)$. We can also multiply the obtained vectors by $3$. Therefore we get 
$(3,0,0)-(1,1,1) = (2,-1,-1)$,   $(-1,2,-1)$, $(-1,-1,2)$ 
Check: all of the norms are $\|(-1,-1,2)\|=\sqrt{6}$, and the dot products are $(-1,-1,2)\cdot (-1,2,-1)=-3$.
