Consider two closed convex cones $A$ and $B$ in $\mathbb{R}^3$. Assume that they are convex even without zero vector, i.e. $A \setminus \{0\}$ and $B \setminus \{0\}$ are also convex (it helps to avoid weird cases like a plane being convex cone).
Suppose that they do not have common directions, i.e. $A \cap B = \{0\}$. It seems that in this case there must exist a strict separating plane, i.e. there exists some unit vector $d$ such that $d \cdot a > 0$ for $a \in A \setminus \{0\}$ and $d \cdot b < 0$ for $b \in B \setminus \{0\}$.


I have tried to prove it myself, but to no avail. If we take a unit sphere $S$, then we may consider convex sets $\mathcal{H}(A \cap S)$ and $\mathcal{H}(B \cap S)$. They are compact and disjoint, so perpendicular bisector between the pair of closest points should separate these sets. If these two closest points are both on unit sphere $S$, then the bisector passes through origin and everything is OK. But in other cases the bisector is not helpful.


After some searching, I have found several topics which may be related to the question:


1. [Hahn–Banach separation theorem](https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#Hahn.E2.80.93Banach_separation_theorem) gives separating plane for convex sets. However, the plane provided may touch both cones.
2. [Farkas's lemma](https://en.wikipedia.org/wiki/Farkas%27_lemma) also provides a separating plane for cones. But in classical formulation one cone must be a ray, and the other one must be cone hull of finite number of vectors. Also, the separating plane constructed is nonstrict for one of the cones.
3. $A \cap S$ and $B \cap S$ should be [geodesically convex](https://en.wikipedia.org/wiki/Geodesic_convexity) on the unit sphere $S$. Since they do not intersect, there might be a strict separating big circle between them. I'm not even sure how to treat this sort of objects though.


I would be glad to see either a reference to literature or a simple proof.



Convex cones: strict separation
I think you may be misreading the Hahn-Banach theorem in this case, as it should give you a strict separating plane here. Anyway, you can avoid this problem by enlarging the cones a bit before applying Hahn-Banach. This is no problem in the finite dimensional case.


Here's what I believe is a direct proof: Find a codimension-1 hyperplane $H$ such that the cross-section $A\cap H$ is compact and convex (such and $H$ should exist by your hypotheses). The intersection $B \cap H$ will be closed and convex, though perhaps not compact. Treat $H$ itself as a linear space and apply Hahn-Banach to *strictly* separate $A\cap H$ from $B\cap H$ by a codimension-1 (in $H$) hyperplane $G$. Complete $G$ to a codimension-1 (in the ambient space) hyperplane passing through the origin, which will separate the cones $A\setminus \{0\}$ and $B\setminus \{0\}$ as you wanted.


In the infinite dimensional case, it is of course more difficult and requires stronger hypotheses (one of the cones has to be closed and the other have a compact cross-section) and there is some literature on this. For example:



> 
> *Separation Properties of Convex Cones*, by V. L. Klee, Jr.
>  [PAMS **6** 313-318 (1955)](http://www.jstor.org/stable/2032366)
> 
> 
> 


