Question.
A set of $n$ distinct vectors $\vec a_1,\vec a_2,...,\vec a_n$, where $n\ge2$, is called regular if it satisfies the following two conditions:
$\qquad$(i) there are constants $\alpha$ and $\beta$, with $\alpha\gt0$, such that for any $i$ and $j$,
$$\vec a_i\cdot\vec a_j=
\begin{cases}
\alpha^2,&\text{when $i=j$}\\
\beta,&\text{when $i\ne j$}
\end{cases},$$
$\qquad$(ii) the centroid of $\vec a_1,\vec a_2,...,\vec a_n$ is the origin $0$.
$\qquad$[The centroid of vectors $\vec b_1,\vec b_2,...,\vec b_n$ is the vector $\frac 1m(\vec b_1+\vec b_2+\cdots+\vec b_n)$.]
Prove that (i) and (ii) imply that $(n-1)\beta=-\alpha^2.$
If $\vec a_1=\binom{1}{0}$, where $\vec a_1,\vec a_2,...,\vec a_n$ is a regular set of vectors in 2-dimensional space, show that either $n=2$ or $n=3$, and in each case find the other vectors in the set.
Hence, or otherwise, find all regular sets of vectors in 3-dimensional space for which $\vec a_1=\begin{pmatrix}1\\0\\0\\\end{pmatrix}$ and $\vec a_2$ lies in the $xy$ plane.
My Attempt.
I got stuck on the last part of 3D space. Though I could deal with the 2D part, I could not think of a more systematic and rigorous way to prove; what I argued was that all pairs of vectors within a regular set must have the same "between angle" according to the definition, and it would be impossible to have more than three members within the set to satisfy that unless some vectors are identical.
By the way, I proved with ease that $(n-1)\beta=-\alpha^2$ but I had no idea how to implement it for the proofs of last two parts.
That ambiguity for 2D part devastated me for the 3D part, for which I had no idea how to figure out the upper limit of $n$.
Comment.
Please just give me some hint so that I can make further progress. Thank you very much in advance!