At most how many distinct unit vectors can be selected in $\mathbb R^n$ such that from any three of them, at least two are orthogonal?

Question

Let $n$ be a positive integer. At most how many distinct unit vectors can be selected in $\mathbb R^n$ such that from any three of them, at least two are orthogonal?

Clearly we can achive value $2n$ if we take vectors $\pm e_i$, for $i=1,2,...,n$, where $\{e_1,e_2,...e_n\}$ is the standard basis for $\mathbb R^n$.

Call this set $S$. Let's prove that it can have no more than $2n$ elements with strong induction. Suppose $|S|\geq 2n+1$. Then clearly exists two vectors $a,b$ in $S$ that are not colinear nor orthogonal. Let $A=\{v\in S;\;v\cdot a=0 \}$ and $B=\{v\in S;\;v\cdot b=0 \}$. Then each vector $v$ is in set $A$ or in set $B$ (or both). Let $\bar A$ and $\bar B$ be span of $A$ respectively $B$. If $|A|=p$ and $|B|=q$ then $\dim \bar A\leq p/2$ and $\dim\bar B\leq q/2$ since $A$ and $B$ are the same type of set as $S$ is. $$p/2+q/2 \geq \dim \bar A + \dim \bar B \geq \dim (\bar A+\bar B) $$ Since $\bar A+\bar B \leq \mathbb{R}^{n-1}$ we have $p/2+q/2 \leq n-1$ so $p+q+2\leq 2n$ and we are done.

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Well I have a feeling that this is wrong. Some better idea how to solve this?

Answer 1

Cannot delete this answer because it is accepted. But it's wrong

Answer 2

Can someone explain to me how can I assure $x$ has to be orthogonal to both $a$ and $b$? I know by property of $S$ that $x$ has to be orthogonal to at least one of them, but not necessarily to both.