I am not sure that I understand the question, but unbounded convex polyhedra in $R^3$ are determined, up to a translation, by their vertices and *recession cone*; see Section 1.4 in Alexandrov's book on Convex polyhedra. Roughly speaking, the recession cone encodes the behavior of the polyhedron at infinity. See also Theorem 3 on page 391 of that book, where it is discussed how to determine an unbounded convex polyhedron from the projection of its vertices into some plane, the curvatures at these vertices, and a given convex polyhedral cone.

The cone, in turn, may be determined by a convex polygon in the plane mentioned above and the apex of the cone. In short there is a way to parametrize the space of unbounded convex polyhedra in terms of a finite number of points, which prescribe its vertices and its recession cone. In this conception, the unbounded polyhedron is just the convex hull of the union of its vertices together with the recession cone, once it is properly positioned relative to the vertices.

By the way, the space of unbounded convex hypersurfaces of $R^n$ which are homeomorphic to $R^{n-1}$ and the recession cones of the corresponding convex bodies are extensively studied in

Deformations of unbounded convex bodies and hypersurfaces,
Amer. J. Math.,134 (2012),1585-1611

where it is shown for instance that the space of these convex hypesurfaces admits a strong deformation retraction into the Grassmannian space of hyperplanes. This is true in the polyhedral category as well.