 Solution

Take the outward surface normal as $[111]$ toward the observer and retain the supplied downward vector $\mathbf{a}=[\bar{1}\bar{1}2]$. The opposite vector is immediately $\mathbf{b}=[11\bar{2}]$. With the diagram's orientation, the remaining arrows areAll have zero dot product with $[111]$. The last three are in-plane close-packed $⟨110⟩$ directions; $\mathbf{a},\mathbf{b}$ are opposite $⟨112⟩$ directions perpendicular to the corresponding straight close-packed step edge.