How to project the position vector of an ellipse onto a plane? 
An ellipse E is arbitrarily oriented in 3-D space with the origin at one of its foci. Use the standard elements provided in the figure. Assume the reference direction $\gamma$ is the positive Y-axis direction.
Find the projected position vector of the ellipse onto the plane of reference as a function of the angle $\nu$.
After some calculation, I obtained the position vector of the celestial body to be $$r=\frac{a(1-e^2)}{1+e\cos(\nu)}$$ Where $a$ is the semi-major axis and $e$ is the eccentricity.
Now, I begin with the $(r,\nu)$ coordinates in the plane of the ellipse and project it onto the reference plane using the $\cos(i)$ factor. How, then, can I find the orthogonal components $(x,y)$ of this projected position vector on the reference plane?
A parameterized solution would be appreciated as well.
 A: You just need to find a unit vector $\hat n$ giving the direction of the planet, and then multiply it by $r(\nu)$ given by your formula.
Suppose then your planet is at $(0,1,0)$: to carry it to its actual position $\hat n$ you need to perform first a rotation about $z$-axis by an angle $\omega+\nu$, then a rotation about $y$-axis by an angle $i$ and finally a rotation about $z$-axis by an angle $\Omega$:
$$
\hat n=
\pmatrix{\cos\Omega &-\sin\Omega & 0\\ \sin\Omega &\cos\Omega & 0\\ 0 & 0 & 1}\cdot
\pmatrix{\cos i & 0 &\sin i\\ 0 & 1 & 0 \\ -\sin i & 0 &\cos i}\cdot
\pmatrix{\cos(\omega+\nu) &-\sin(\omega+\nu) & 0\\ \sin(\omega+\nu) &\cos(\omega+\nu) & 0\\ 0 & 0 & 1}\cdot
\pmatrix{0\\ 1\\0},
$$
giving the result:
$$
\hat n=
\pmatrix{-\cos i \cos\Omega \sin(\omega+\nu)-\sin\Omega \cos(\omega+\nu)\\ 
\cos\Omega \cos(\omega+\nu)-\cos i \sin \Omega\sin(\omega+\nu)\\ 
\sin i \sin(\omega+\nu)}.
$$
Multiply that by
$$
r(\nu)=\frac{a(1-e^2)}{1+e\cos(\nu)}
$$
and you'll have the coordinates of the planet.
