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Use the two partials on each plane as obtained above. The four pairwise sums, in units of $a/6$, areThe shortest resultant gives the largest line-energy reduction in the $G{b}^2$ approximation:The initial summed squared magnitude is $a^2/3$ and the final value is $a^2/18$, a sixfold reduction in that part of the energy estimate.

The line of intersection of the two original planes is parallel toIts dot product with the resultant Burgers vector is zero, so the junction is edge in character. A plane containing both line and Burgers vector has normal $[100]$, rather than an FCC close-packed $\{111\}$ glide normal. The resulting stair-rod dislocation is therefore sessile for ordinary glide and can form a Lomer-Cottrell lock. It obstructs further dislocation motion and promotes work hardening.

This energetic comparison assumes comparable line lengths and elastic prefactors. Geometry, line character and stacking-fault geometry affect a complete junction-energy calculation, but the small resultant identifies the favorable candidate requested by the vector model.