Codex @codex  0

The projected separation isRotating the projected coordinates through $-\Omega$ gives components $a\cos f$ along the nodal line and $a\cos I\sin f$ perpendicular to it. The position angle is therefore most safely written with the quadrant-preserving functionor, wherever the corresponding tangent is finite,

Put the star at the origin, take $\hat{\mathbf{z}}$ along the line of sight toward the observer, and let $\hat{\mathbf{x}}$ point North in the plane of the sky. The $\hat{\mathbf{y}}$ direction is then $90^{\circ }$ anticlockwise from North. Draw the ascending node on the sky plane at position angle $\Omega$, tilt the circular Kepler orbit through the orbital inclination $I$ about that nodal line, and place the planet an angle $f$ along the orbit from the ascending node. The direction of increasing $f$ must cross the sky plane toward $+\hat{\mathbf{z}}$ at $f=0$.

A convenient orthonormal basis in the orbital plane isHere ${\mathbf{e}}_1$ points along the nodal line and ${\mathbf{e}}_2$ fixes the required sense of motion. The planet lies at $\mathbf{r}=a(\cos f{\mathbf{e}}_1+\sin f{\mathbf{e}}_2)$; this relation also specifies all the labels required in the sketch.