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Could someone draw the Poincaré model for $\mathbb{H}^3$ with this $P_t$? I can not find $P_t$ orthogonal to $\partial B^3$ and perpendicular to the line that contains $\overrightarrow{v}$.

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OK, $P_t$ is essentially a sphere with center $c\vec v$ for some $c\in\mathbb R,\lvert c\rvert>1$ (because of “perpendicular to the line through $\vec v$) which is orthogonal to the unit sphere and contains the point $t\vec v$. Now if two spheres intersect orthogonally, the inversion of one in the other leaves the sphere as a whole invariant. Which means that you can invert $t\vec v$ in the unit sphere to obtain $t^{-1}\vec v$ and know that this is another point on the sphere you want. Furthermore, since the center is on the line between them, these two points form a diameter. The center has to be in the middle between them, and the radius chosen so that both these points are on the sphere:

$$c=\frac{t+t^{-1}}2\qquad r=\frac{\left\lvert t-t^{-1}\right\rvert}2$$

Of course $t=0$ is a special case: In that situation your sphere will have “infinite radius” and will in fact be the plane through the origin orthogonal to $\vec v$.

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