 Solution

A Shockley partial dislocation in an FCC slip plane has Burgers vector of the family $a⟨112⟩/6$. Its vector must lie in the specified plane, and the two vectors must sum to the perfect dislocation vector.

For the first plane normal ${\mathbf{n}}_1=[\bar{1}11]$, one allowed dissociation isThe vectors add to $a[303]/6$, and ${\mathbf{n}}_1\cdot [211]=-2+1+1=0$, ${\mathbf{n}}_1\cdot [1\bar{1}2]=-1-1+2=0$.

For ${\mathbf{n}}_2=[111]$,Their sum is $a[\bar{3}30]/6$, and both have zero dot product with $[111]$. Interchanging leading and trailing labels does not change these vector identities.

Each perfect vector has squared magnitude $a^2/2$, whereas the sum of the two partial squared magnitudes is $a^2/3$. The line-energy estimate proportional to $G{b}^2$ therefore favors separation, opposed by the stacking-fault energy of the ribbon between the partials.