Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement?
Oximes can undergo the Beckmann rearrangement.  Oximes also exist as stable syn and anti isomers.  In the figure below, $\ce{R_1}$ is anti to the hydroxyl group; another isomer exists where it is syn the the hydroxyl group.

In the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group.  As the hydroxyl group begins to depart, the $\ce{R}$ group which is anti to the leaving group begins to migrate and form a bond with nitrogen.



(Wikipedia)

This step is said to be concerted (rather than stepwise) with the $\ce{N-O^{+}H2}$ bond being broken and the new $\ce{R-N}$ bond being formed at (more or less) the same time.  Whether you think of the $\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\ce{R}$ group as starting to bond to the available $\sigma^{*}$ orbital of the elongating $\ce{N-O}$ bond (equivalent descriptions), you see that having the $\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\ce{N-O}$ bond would be preferred. Hence, the $\ce{R}$ group anti to the $\ce{N-O}$ bond migrates preferentially.

There are many reactions showing this geometric preference, the Baeyer-Villiger oxidation being an example.