Finding the center and axes of an ellipse in 3D I am studying ellipses and need to prove the intersection of a plane and an elliptical cylinder is still an ellipse in 3D. My approach is to find its center and axes and have been struggling.
I have a parametric equation, $\left(x,\, y,\, z\right) = \left(\frac{1}{4}-\frac{3}{2}\cos t-\frac{1}{4}\sin t,\, 2\cos t,\, \sin t \right)$ . This is an intersection between a plane $4x+3y+z=1$ and an elliptical cylinder $y^2+4z^2=4$.
How do I find its center and two axes? Equivalently, how can I express it in the vector form:
$$\mathbf x (t)=\mathbf c+(\cos t)\mathbf u+(\sin t)\mathbf v$$
where $\mathbf u$ and $\mathbf v$ are orthogonal vectors from the center $\mathbf c$ whose norms represent the lengths of the axes and whose directions represent the directions of the axes.

$$\left(x,\, y,\, z\right) = \left(\frac{1}{4}-\frac{3}{2}\cos t-\frac{1}{4}\sin t,\, 2\cos t,\, \sin t \right)=(\frac{1}{4},0,0)+(\frac{-3}{2},2,0)\cos t+(\frac{-1}{4},0,1)\sin t$$
So $c=(\frac{1}{4},0,0)$, $u=(\frac{-3}{2},2,0)$ and $v=(\frac{-1}{4},0,1)$
