= Solution
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 is
$$
\mathbf e_1=(\cos\Omega,\sin\Omega,0),
\qquad
\mathbf e_2=(-\sin\Omega\cos I,\cos\Omega\cos I,\sin I).
$$
Here $\mathbf e_1$ 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.
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