Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane? E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane.

I can't think of a reason that they would need to be in the same plane, so what is the reason for this strange phenomenon?
When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond.  For these two p orbitals to form a pi bond they must be in the same plane (or very close).  

This is illustrated in the following figure.  This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.



image source

Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.



This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.