I'm using this resource (will download a .pdf) to calculate the positions of all the Lagrange points in a 2-body system, but it's only concerned with the x and y components of their position vectors (2D treatment of the problem, in other words), but reality is made up of three dimensions and I would like to capture that. In the case of most of the planets I guess Z is pretty close to the ecliptic plane, but for bodies with inclined orbits like Eris, I guess that is not a good approximation.

Could somebody tell me how to calculate the z component of the vector that describes the positions of all the Lagrange points?

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    $\begingroup$ Long time no see! I'd even forgot about you, shame on me! Looking for that cool astronomical and spaceflight simulation website that's been discussed here Lagrange points only exist in-plane. The two massive bodies have an orbital plane and the zeros of the forces (in the rotating frame) must also be in the same plane. Lagrange points are mathematical concepts and are only defined for a two-body orbit in a 2D plane. $\endgroup$
    – uhoh
    Sep 4 '21 at 13:16
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    $\begingroup$ Howdy! How's life treating you? Well, that's reassuring to know that we put spacecraft into orbit around mathematical concepts. So in the case of Earth, the Lagrange points are simply chilling on the ecliptic plane? And in the case of Pluto, the, ehrm, Plutonian plane? I'll get to it... $\endgroup$ Sep 4 '21 at 13:30
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    $\begingroup$ Thanks for the kind words (read the question you referred to)! You can actually create your own scenarios now, and add rings to any celestial body: gravitysimulator.org/misc/create-new-gravity-simulation. $\endgroup$ Sep 4 '21 at 13:46
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    $\begingroup$ The z component is identically zero. $\endgroup$ Sep 4 '21 at 13:53
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    $\begingroup$ @HappyKoala In the case of Sun-Earth system with a circular orbit, yes. As soon as you add some reality (elliptical orbit, or other planets, or...) then the conditions under which Lagrange derived his points no longer exist. There are still halo-like orbits around where you'd expect them to be but the points themselves only exist in theory and only in the 2 dimensional orbital plane, and only for circular orbits. $\endgroup$
    – uhoh
    Sep 4 '21 at 14:36

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