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Use the upper-half-plane model of the hyperbolic plane, whose geodesics are vertical lines and semicircles perpendicular to the real boundary. Hyperbolic and Euclidean angles agree. An isometry takes one of the ultraparallel hyperbolic lines to the imaginary axis. The other then has two finite endpoints on the same side; reflecting if necessary, write them as $0<a<b$. Its Euclidean centre and radius are $c=(a+b)/2$ and $r=(b-a)/2$.

A geodesic perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $q$. Two intersecting Euclidean circles meet orthogonally exactly when their squared centre distance equals the sum of their squared radii. Orthogonality to the second line therefore requiresSince $a<\sqrt{ab}<b$, the two semicircles intersect at one point in the upper half-plane. This constructs the common perpendicular of ultraparallel hyperbolic lines and proves uniqueness, because its radius is forced.

For the three-line continuation, the three common perpendiculars can be pairwise disjoint, but need not be. Explicit configurations of three ultraparallel lines make both claims transparent. For the positive example, take the three semicircles of radius $1/2$ centred at $-5/2,0,5/2$. Their endpoint intervals are disjoint. Their common perpendiculars are the semicircles with centres $-5/4,5/4,0$ and radii $\sqrt{21}/4,\sqrt{21}/4,\sqrt{6}$, respectively. The two smaller semicircles are disjoint because $2(\sqrt{21}/4)<5/2$; each is nested inside the largest because $5/4+\sqrt{21}/4<\sqrt{6}$. Hence their full geodesics are disjoint. For the negative example, take three concentric semicircles of radii $1,2,3$. They are pairwise ultraparallel, but every common perpendicular is the imaginary axis, so the perpendiculars coincide.