# 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.

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.
• The second rotation is about the new position of $y$-axis after the first rotation, and so on. Jun 24, 2020 at 15:44