How to get the axial tilt vector(x,y,z) relative to ecliptic

According to the orbital plane XYZ, I understand that the axial tilt is the angle between the planet axis rotation and any normal of the plane (actually Z in right handed system). BUT, this is not sufficient to define a vector direction, I need another angle into the XY plane. Where do I find this information? Actually I believe it's relative to the direction of the periapsis. Am I right?

Following all the nice conversation on the answers, I maybe need to ask

How to get the axial tilt vector(x,y,z) relative to ecliptic or orbital plane, given ONLY ONE angle, i.e. 23.439281 deg for the Earth, 25.19 for Mars, etc.

• Does this post answer your question? Sep 1 '16 at 16:48
• nope, still discussion from specialist and no one give simple definition.. one is speaking about the velocity vector, the other to another referential.. nothing clear for a newbiz like me. BTW many thank to taking the time to answer. Sep 1 '16 at 17:19
• This is the answer: astronomy.stackexchange.com/a/8800/6 Unfortunately I'm not sure how to simply it from there. Sep 1 '16 at 17:20
• When you say "pole" do you mean "How do they define north and south?" Sep 1 '16 at 17:45
• no, i just want to obtain the vector of the axial tile relative to the ecliptic plane (z to the eccliptic pole) or relativ to the planet orbital plane, which must have the form of (alpha, theta) or (x,y,z). Actually, for the earth, it's seem a rotation X is doing the job. but for others planet with orbital plane different than the eccliptic i do not find any clear explanation anywhere.. Sep 1 '16 at 17:58

Earth is a special case since the equatorial and ecliptic coordinate systems are defined in terms of its own rotation and orbit. Earth's north pole vector in equatorial coordinates is $$\vec N_{\oplus,eq} = (0, 0, 1)$$ To transform this to ecliptic coordinates, we rotate about the $x$ axis by the obliquity $\varepsilon$ = 23.44$^\circ$ and get $$\vec N_{\oplus,ecl} = (0, \sin \varepsilon, \cos \varepsilon) = (0, 0.3978, 0.9175)$$ In a spherical coordinate system, two angles define a unique direction. In equatorial coordinates these are the right ascension $\alpha$ and declination $\delta$. This IAU report, table 1, gives $\alpha$ and $\delta$ values for each major planet's north pole as of 2000-01-01 and formulas to compute them for other years. To convert these to rectangular form $$(x_{eq}, y_{eq}, z_{eq}) = (\cos \alpha \cos \delta, \sin \alpha \cos \delta, \sin \delta)$$ Then if you want J2000 ecliptic coordinates $$(x_{ecl}, y_{ecl}, z_{ecl}) = (x_{eq}, y_{eq} \cos \varepsilon + z_{eq} \sin \varepsilon, z_{eq} \cos \varepsilon - y_{eq} \sin \varepsilon)$$ But if you want another planet's orbital plane to be the $xy$-plane, then you also need the orbit's inclination and longitude of ascending node; I leave that transformation as an exercise for the motivated reader.
• No, that is only valid for Earth. In general there is no such shortcut; a single angle is insufficient to define a unique direction. Rotating the ecliptic pole about another planet's ascending node (not the x axis) by that planet's orbital inclination gives you the pole of the other planet's orbit. The obliquity of the planet's rotational axis gives you an angle away from that orbital axis but does not determine direction. You really need both $\alpha$ and $\delta$ as above. Sep 3 '16 at 14:13
• @uhoh Table 1 of the IAU WGCCRE report gives most planets' polar $\alpha_0$ and $\delta_0$ [ICRF] as functions of $T$ [Julian centuries; T(J2000.0) = 0]. Dec 9 '19 at 3:30