Let $v$ be non-zero with $B(v,v)=0$. I would like to show that an equation $B(u,v)=1$ defines a horocycle in the hyperboloid model.
A horocycle in the Poincare disk model has a touching point $v$ on the boundary of the unit circle. And if we see this $v$ in the hyperboloid model, $v$ will be a point at infinity on the asymptotic cone. In the hyperboloid model, a horocycle with $v$ is the intersection of the hyperboloid ($-x^2-y^2+z^2=1$) and some plane that is parallel to the vector $v$. Plus, as the asymptotic cone is defined by the equation $x^2+y^2=z^2$, we have $B(v,v)=0$ and $v \neq 0$.
If I have understood the equation correctly, for any $u$ in a horocycle that is (Euclidean Sense) parallel to the vector $v$, we have $B(u,v)=1$.
However, I'm not sure why we should have $B(u,v)=1$ for any $u$ on a horocycle.
Can you please help me in understanding this equation?
I found that $B(u,v)=1$ gives the equation of the plane. And this plane must have the normal vector $(v_1,v_2,-v_3)$ if $v=(v_1,v_2,v_3)$. Thus, by wikipedia definition of horocycle in the hyperboloid model, the equation $B(u,v)=1$ defines a horocycle if we just let $u \in H$ where $H$ stands for the hyperboloid.
Did I get it right?