Find 3 vectors in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$ Find $3$ vectors $\vec{a},\vec{b},\vec{c}$ in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$
I know that the dot-product of two vectors in $\mathbb{R}^3$ is represented like this $(\vec{a},\vec{b})=|\vec{a}||\vec{b}|\cos(\angle\vec{a},\vec{b})$ so I've used random vector $\vec{a}(1,0,0)$ and added the following three conditions:
1.$(\vec{a},\vec{b})=|\vec{a}||\vec{b}|\cos(\frac{\pi}{3}) \Rightarrow 2b_1=b_1^2+b_2^2+b_3^2$
2.$(\vec{a},\vec{c})=|\vec{a}||\vec{c}|\cos(\frac{\pi}{3}) \Rightarrow 2c_1=c_1^2+c_2^2+c_3^2$
3.$(\vec{b},\vec{c})=|\vec{b}||\vec{c}|\cos(\frac{\pi}{3}) \Rightarrow 2(c_1b_1+c_2b_2+c_3b_3)=(b_1^2+b_2^2+b_3^2)(c_1^2+c_2^2+c_3^2) \Rightarrow b_1c_1=c_2b_2+c_3b_3$
And this is the point where I stuck... Any ideas how to solve this problem?
 A: Consider a regular tetrahedron.
A regular tetrahedron can be inscribed inside a cube.
$(0,0,0), (0,1,1),(1,1,0),(0,1,1)$ could be the vertexes of a regular tetrahedron.
Alternative.  Pick an arbitrary vector $a=(1,0,0)$ will do.
Find a vector that meets $a$ at the correct angle.
$b = (\cos \frac {\pi}{3},\sin\frac {\pi}{3}, 0)$
solve for $c = (x_1,x_2,x_3)$ such that
$c\cdot a = x_1 = \frac 12\\
c\cdot b = \frac 12 x_1 + \frac {\sqrt 3}{2}x_2 = \frac 12\\
\|c\| = 1 \implies x_1^2 + x_2^2 + x_3^2 = 1$
A: In more general case, I would approach this problem like this:


*

*You have the first vector $\vec{a}(1,0,0)=(a_1,0,0)$. I use $a_1=|\vec{a}|=1$.

*Create a vector $\vec{b}$ in the plane of $\vec{a}$, and a direction perpendicular to $\vec{a}$. In this case I can choose $(0,1,0)$. You can then write $\vec{b}=(b_1,b_2,0)$. The condition to find out the components are $$b_1^2+b_2^2=1\\a_1b_1=\cos(\pi/3)$$ Note that you can choose any angle with this approach. In this case, $\vec{b}=(\cos(\pi/3),\sin(\pi/3),0)$ An equally good solution is to have the second component with a minus sign.

*For the third vector, you have three components, one along $\vec{a}$, one along the vector perpendicular to $\vec{a}$ in the $ab$ plane, and one component in the direction perpendicular to $ab$ plane. In this case $\vec{c}=(c_1,c_2,c_3)$. You apply the conditions to find out the components:$$\vec{a}\cdot\vec{c}=a_1c_1=\cos(\pi/3)\\\vec{c}\cdot\vec{c}=b_1c_1+b_2c_2=\cos(\pi/3)\\c_1^2+c_2^2+c_3^2=1$$ Note once again that I could choose any other angles with this approach. From the first equation $c_1=\cos(\pi/3)$. In the second equation $\cos(\pi/3)^2+c_2\sin(\pi/3)=\cos(\pi/3)$, so $c_2=\frac{1}{2\sqrt{3}}$, and $c_3$ you get from the normalization equation: $c_3=\sqrt{2/3}$

A: One can use the idea of forming tetrahedron in the following way. First, consider equilateral triangle in $xy$-plane centered at origin. We know that the vertices correspond to the third roots of unity, and thus are given by $(1,0,0)$, $(-\frac 12,\frac{\sqrt3} 2,0)$ and $(-\frac 12,-\frac{\sqrt3} 2,0)$.
Now, we want to raise this triangle in $z$ direction so the given vectors with origin correspond to vertices of tetrahedron. Write them as $$(1,0,h), (-\frac 12,\frac{\sqrt3} 2,h), (-\frac 12,-\frac{\sqrt3} 2,h).$$
We can calculate $h$ in different ways, for example, you can use only elementary geometry (Pythagoras theorem), but probably the simplest is just to calculate angles between them. By symmetry, it is enough to calculate angle between two of them, for example:
$$ \cos\varphi = \frac{\langle (1,0,h), (-1/2,\sqrt3/2,h) \rangle}{\|(1,0,h)\| \|(-1/2,\sqrt3/2,h)\|} = \frac{h^2-\frac 12}{h^2 + 1}$$
(Feel free to check that any choice would give the same answer depending on $h$.)
Now set $\varphi = \pi/3$ to get equation 
$$ \frac{h^2-\frac 12}{h^2 + 1} = \frac 12$$
and thus $h = \pm \sqrt 2$.
A: Let see if we can find the most symmetric triple of vectors all in the positive octant. Let the $3$ unit vectors be 
\begin{eqnarray*}
( \alpha, \alpha , \beta) , ( \alpha,\beta ,\alpha , )( \beta,\alpha, \alpha )
\end{eqnarray*}
We require that they are of unit length & their dot products are $1/2$. So we have
\begin{eqnarray*}
2\alpha^2+\beta^2=1 \\
\alpha^2+2 \alpha \beta= \frac{1}{2}.
\end{eqnarray*}
These are easily solved to give $ \alpha=\frac{1}{3 \sqrt{2}}$,$\beta=\frac{2 \sqrt{2}}{3}$.
So the three vectors $\color{red}{(\frac{1}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}},\frac{2 \sqrt{2}}{3}),(\frac{1}{3 \sqrt{2}},,\frac{2 \sqrt{2}}{3}\frac{1}{3 \sqrt{2}}),(\frac{2 \sqrt{2}}{3},\frac{1}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}})}$ will do.
