planet - North Pole Right Ascension/Declination to axial tilt conversion

This is wrong, but may help someone find the right answer:

You can convert right ascension and declination to a 3 dimensional
unit vector in the J2000.0 ICRF reference frame using the standard
formula for converting spherical coordinates to rectangular
coordinates and using a radius of 1:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}

The ra and dec of the north ecliptic pole is ra,dec of
{270, 66.5607083333} per Wikipedia, so the unit vector representing
the north ecliptic pole is:

{0, -0.3977771648286046, 0.9174820582119942}

As it turns out, there is conflicting data for Ceres, perhaps because
of the semi-recent DAWN flyby.

Instead, I'll use Vesta, where wikipedia explicitly states (https://en.wikipedia.org/wiki/4_Vesta#Rotation):

north pole pointing in the direction of right ascension 20 h 32 min,
declination +48 [... which] gives an axial tilt of 29 [degrees]

Converting to degrees, Vesta's north pole's ra,dec is
{308,48}. Using the formula above to find the unit vector, we get:

{0.411957936296447, -0.527282113378167, 0.743144825477394}

If we take the dot product of two vectors and divide by the product of
their lengths, we get the cosine of the angle between them. In this
case, both vectors have length 1, and the dot product is 0.891563
whose arc-cosine is right around 27 degrees.

So, if Vesta's orbit were the same plane as Earth's orbit, the axial
tilt would be 27 degrees.

However, since Vesta's orbit is inclined 7.14043 degrees to the
ecliptic, this answer is incorrect.

I don't think you can find the correct answer without using Vesta's
inclination and the longitude of its ascending node.

Both of these values are known, but I can't figure out how to use them
to get the correct asnwer.