<s>[INST] 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?
 [/INST] 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.
</s>