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.
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.