Orbital inclination ($I$)

Orbital inclination is the angle between an orbital plane and a reference plane.

In a right-handed reference frame, the rotation matrix from an orbital plane frame whose first axis points along the ascending node is $R_z(\Omega )R_x(I)$. Its columns are the orbital frame's orthonormal basis expressed in reference coordinates. The unit vector normal is $(\sin \Omega \sin I,-\cos \Omega \sin I,\cos I)$; its dot product with another such normal gives the cosine of their mutual inclination.

The mutual inclination is the angle between two orbital planes, equivalently the angle between consistently oriented orbital normal vectors.

For two nearly edge-on orbits with small orbital inclination difference $\Delta I$ and small longitude of ascending node difference $\Delta \Omega$, expanding the dot product of their normals gives $I_m^2=(\Delta I{)}^2+(\Delta \Omega {)}^2+O({\delta }^4)$, with angles in radians and all offsets from edge-on of order $\delta$. Thus equality of observed sky-plane inclinations alone does not imply aligned orbital planes.

The invariable plane is perpendicular to a system's total angular momentum. For an isolated planetary system, it is fixed while individual orbital planes precess.

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.

An orbit is edge-on to an observer when its orbital plane contains the line of sight, so its inclination to the plane of the sky is $90^{\circ }$.

Projected observations that do not identify which node is ascending may determine an orbital plane only up to reversal of its normal or addition of $\pi$ to its nodal longitude.