How to show that you can always move away from a convex object?

Question

Assume I have two convex objects in 3D, $A$ and $B$, and the two are in contact. How can I show that there always exists a vector $v$, such that if I move $A$ along $v$ the contact between $A$ and $B$ is broken?

Answer 1

The Hyperplane separation theorem states that two disjoint convex sets in 3D, $A$ and $B$, can be separated by a plane. Let $A$ and $B$ touch at one (or more) points. These must lie on a plane. Now move set $A$ along the normal (perpendicular) to the separating plane. Calculate the new distance to any point on $B$... it has become larger.

Answer 2

Not a rigorous proof …

If A and B are in contact at a point P, then they have a common surface normal, N, at that point. Consider the common tangent plane $\pi$ — the plane containing P normal to N. It seems obvious that A must lie one side of $\pi$ and B must lie on the other side, otherwise convexity would be violated. Moving A along the vector N would work, it seems to me.