Problem source

A set of $n$ distinct vectors $\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n},$
where $n\geqslant2$, is called \textit{regular }if it satisfies the
following two conditions: 
\begin{questionparts}
\item there are constants $\alpha$ and $\beta$, with $\alpha>0$, such
that for any $i$ and $j$, 
\[
\mathbf{a}_{i}\cdot\mathbf{a}_{j}=\begin{cases}
\alpha^{2} & \mbox{ when }i=j\\
\beta & \mbox{ when }i\neq j,
\end{cases}
\]
\item the centroid of $\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}$
is the origin $\mathbf{0}.$ {[}The centroid of vectors $\mathbf{b}_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{m}$
is the vector $\frac{1}{m}(\mathbf{b}_{1}+\mathbf{b}_{2}+\cdots+\mathbf{b}_{m}).${]} 
\end{questionparts}
Prove that \textbf{(i) }and \textbf{(ii) }imply that $(n-1)\beta=-\alpha^{2}.$
If $\mathbf{a}_{1}=\begin{pmatrix}1\\
0
\end{pmatrix},$ where $\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}$ 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 $\mathbf{a}_{1}=\begin{pmatrix}1\\
0\\
0
\end{pmatrix}$ and $\mathbf{a}_{2}$ lies in the $x$-$y$ plane.

Solution source

\begin{align*}
&& \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\
\Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\
&&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\
&&&= (n-1)\beta + \alpha^2 \tag{i} \\
\Rightarrow && (n-1)\beta &= -\alpha^2
\end{align*}

Suppose we have $\mathbf{a}_j = \binom{x}{y}$, $j \neq 1$ then $x = \beta$. We also must have $\beta^2 + y^2 = 1$, so there are at most two values for $y$, ie two extra vectors. ie $n = 2, 3$.

If $n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1$.

If $n = 3$

\begin{align*}
&& \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\
\Rightarrow && \beta = -1/2 \\
\Rightarrow && y &= \pm \frac{\sqrt{3}}{2}
\end{align*}

Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\
0\\
0
\end{pmatrix}$, $\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\
\sin \theta \\
0
\end{pmatrix}$

(since we need $\mathbf{a}_2 \cdot \mathbf{a}_2 = 1$). $\beta = \cos \theta$).

We can have $\cos \theta = - 1$.

Suppose we have $\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}$, so $x = \cos \theta$, and $y^2 + z^2 = \sin^2 \theta$, so we can write it as:

$\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi  \\ \sin \theta \sin \phi \end{pmatrix}$.  We must also have $\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi  \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta$, so $\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}$. Therefore there is one value for $\cos \phi$, so at most two values for $\sin \phi$, Therefore we can have either $2, 3,4$ or $5$ different values in the set.

$n = 2$, we've already handled.

If $n = 3$, then $\beta = -\frac12$, $\cos \phi = -1$, so we can only have two different values for $\sin \theta$, ie:

$\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix},  \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}  \right \}$

Finally, if $n = 4$, we have $\beta = -\frac13$, $\cos \phi = \frac{-1/3}{2/3} = -\frac12$. $\sin \theta = \pm \frac{\sqrt{3}}{2}$

$\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix},  \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix}    \right \}$

$\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix},  \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix}    \right \}$

If $n = 5$, then $\beta = -\frac14$, $\cos \phi = \frac{-1/4}{3/4} = -\frac13$. $\sin \theta = \frac{\sqrt{15}}{4}$, $\sin \phi = \frac{2\sqrt{2}}{3}$

$\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix},  \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix},  \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix},  \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6}  \end{pmatrix},  \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix},  \right \}$