Given three vectors $a,b,c$ which are linearly independent, and since there's a dot product $<,>$, I gotta prove that the orthocenter $h$ exists and it's unique.
I really have no idea how to do it. The professor gave us a hint about drawing in $\mathbb{R}^{3}$, what I did, but for me, every triangle I draw with these vectors will make them linearly dependent, which goes against the hypothesis.
Three points determine a plane, so it suffices to prove this for $\mathbb{R}^2$.
The set of vectors $x$ such that $(a-x)\cdot(b-c)=0$ determines a line. Likewise, $(b-x)\cdot(c-a)=0$ determines another line. If $a$, $b$ and $c$ are not collinear, these lines will not be parallel, and determine a unique intersection, $h$. To prove our original claim, it thus suffices to prove that $(c-h)\cdot(a-b)=0$. And we can do this by noticing the following chain of equalities: $$(c-h)\cdot(a-b)=$$ $$a\cdot c-b\cdot c-a\cdot h+b\cdot h=$$ $$-((a\cdot b-a\cdot c-b\cdot h+c\cdot h)+(b\cdot c-a\cdot b-c\cdot h+a\cdot h))=$$ $$(a-h)\cdot(b-c)+(b-h)\cdot(c-a)=0.$$