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For two groups carrying angular momenta $L_a,L_b$ with mutual inclination $\Delta I$, resolving total angular momentum parallel and perpendicular to the second group's normal gives its orbital inclination relative to the invariable plane:For a positive denominator, $\tan I_b=\sin \Delta I/(L_b/L_a+\cos \Delta I)$. The larger-angular-momentum group lies closer to the invariable plane.

Initially the two inner orbital normals coincide, so their combined angular momentum is $L_{12}=L_1+L_2$ along one direction. Use the outer orbital normal as the vertical axis. The inner pair's vector then has perpendicular component $L_{12}\sin \Delta I$ and parallel component $L_{12}\cos \Delta I$. Adding the outer angular momentum givesThe invariable plane is perpendicular to this total vector. Thus the outer orbit's orbital inclination to that plane equals the angle between its own normal and the total normal. The inclination to the invariable plane of two orbital planes isThis uses the small, prograde inclinations for which the denominator is positive; in general use the corresponding two-argument arctangent to choose the quadrant. If $A\gg 1$, $I_3\simeq \Delta I/(A+1)$ for small $\Delta I$: the distant planet carries most of the angular momentum and lies close to the invariable plane.