Resolving $\mathbf r=a(\cos f\,\mathbf e_1+\sin f\,\mathbf e_2)$ into the sky coordinates gives
$$
\boxed{x=a(\cos\Omega\cos f-\sin\Omega\cos I\sin f)},
$$
$$
\boxed{y=a(\sin\Omega\cos f+\cos\Omega\cos I\sin f)},
$$
$$
\boxed{z=a\sin I\sin f}.
$$
In particular, $z=0$ at the <orbital nodes>, and $dz/df=a\sin I\gt0$ at $f=0$, so that node is indeed the <ascending node>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 1 / ii / Solution

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An orbital node is an intersection of an orbital plane with its reference plane.


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