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