= Solution
A <Shockley partial dislocation> in an FCC slip plane has <Burgers vector> of the family $a\langle112\rangle/6$. Its vector must lie in the specified plane, and the two vectors must sum to the perfect <dislocation> vector.
For the first plane normal $\mathbf n_1=[\bar111]$, one allowed dissociation is
$$
\boxed{\frac a2[101]
=\frac a6[211]+\frac a6[1\bar12].}
$$
The vectors add to $a[303]/6$, and $\mathbf n_1\cdot[211]=-2+1+1=0$, $\mathbf n_1\cdot[1\bar12]=-1-1+2=0$.
For $\mathbf n_2=[111]$,
$$
\boxed{\frac a2[\bar110]
=\frac a6[\bar211]+\frac a6[\bar12\bar1].}
$$
Their sum is $a[\bar330]/6$, 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 $Gb^2$ therefore favors separation, opposed by the stacking-fault energy of the ribbon between the partials.
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