My question is basically as posed in the title. Suppose we are given $n + 1$ unit vectors in $\mathbb{R}^n$ so that the angle between any pair of them is the same. What is that angle? I have (unfounded) reason to believe that the angle is arccos$(\frac{-1}{n})$, but I don't know how to prove it. For example, we can find three unit vectors $\begin{pmatrix} 1 \\ 0\end{pmatrix}$, $\begin{pmatrix} \frac{-1}{2} \\ \frac{\sqrt{3}}{2}\end{pmatrix}$, $\begin{pmatrix} \frac{-1}{2} \\ \frac{-\sqrt{3}}{2}\end{pmatrix}$, and the angle between any pair of them is arccos$(\frac{-1}{2})$, since the dot product of any pair of them is $\frac{-1}{2}$. Can anyone supply the reasoning for general $\mathbb{R}^n$, or correct me if I'm incorrect?
Angle between $n + 1$ equidistant unit vectors in $\mathbb{R}^n$
geometrylinear algebra
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Best Answer
Since they're spaced symmetrically,
$$u_1+u_2+u_3+\cdots+u_n+u_{n+1}=0.$$
Now take the dot product with $u_1$:
$$1+u_1\cdot u_2+u_1\cdot u_3+\cdots+u_1\cdot u_n+u_1\cdot u_{n+1}=0.$$
Again by symmetry, these last $n$ dot products should be the same:
$$1+u_1\cdot u_2+u_1\cdot u_2+\cdots+u_1\cdot u_2+u_1\cdot u_2=0$$
$$1+nu_1\cdot u_2=0$$
$$u_1\cdot u_2=-1/n.$$