what is the lowest point of a tilted elliptical plate? I'd like to know the lowest point $z_\min$ of an ellipse with radius $r_x, r_y$ in (Euclidian) XY that's tilted in XYZ - first rotated around X axis by $\gamma$, then rotated around Y axis by $\alpha$.
If it's only tilted around one axis, it's easy.
$$-z_\min = r_x|\sin(\alpha)|$$ (tx for pointing out Achille)
For the case $\alpha \neq 0$ and $\gamma \neq 0$, I thought about identifying the new direction and corresponding radius that lead to lowest point, but not sure exactly how to go about it.
What's the way to find $z_\min$ with both $\alpha \neq 0$ and $\gamma \neq 0$?
 A: Consider a parametric version of an ellipse laying in the $xy$ plane is centered at the origin. The following vector will describe the ellipse:
$$\begin{bmatrix}
r_x\cos(\theta)\\
r_y\sin(\theta)\\0
\end{bmatrix}$$
and $0\le \theta <2\pi.$
A rotation by angle $\gamma$ around the $x$ axis can be performed by the following matrix:
$$\begin{bmatrix}
1&0&0\\
0&\cos(\gamma)&-\sin(\gamma)\\
0&\cos(\gamma)&\sin(\gamma)
\end{bmatrix}$$
With this the vector describing the ellipse canges
$$\begin{bmatrix}
1&0&0\\
0&\cos(\gamma)&-\sin(\gamma)\\
0&\cos(\gamma)&\sin(\gamma)
\end{bmatrix}\begin{bmatrix}
r_x\cos(\theta)\\
r_y\sin(\theta)\\0
\end{bmatrix}=\begin{bmatrix}
r_x\cos(\theta)\\
r_y\sin(\theta)\cos(\gamma)\\
r_y\sin(\theta)\cos(\gamma)
\end{bmatrix}.$$
A rotation by angle $\alpha$ around the $y$ axis can be performed by the following matrix:
$$\begin{bmatrix}
\cos(\alpha)&0&\sin(\alpha)\\
0&1&0\\
-\sin(\alpha)&0&\cos(\alpha)
\end{bmatrix}.$$
Multiplying the vector by this matrix we get
$$\begin{bmatrix}
\cos(\alpha)&0&\sin(\alpha)\\
0&1&0\\
-\sin(\alpha)&0&\cos(\alpha)
\end{bmatrix}\begin{bmatrix}
r_x\cos(\theta)\\
r_y\sin(\theta)\cos(\gamma)\\
r_y\sin(\theta)\cos(\gamma)
\end{bmatrix}=$$ 
$$=\begin{bmatrix}
r_x\cos(\theta)\cos(\alpha)+r_y\sin(\theta)\cos(\gamma)\sin(\alpha)\\
r_y\sin(\theta)\cos(\gamma)\\
-r_x\cos(\theta)\sin(\alpha)+r_y\sin(\theta)\cos(\gamma)\cos(\alpha)
\end{bmatrix}.$$
We are interested only in the $z$ coordinate of the twice rotated ellipse:
$$z(\theta)=-r_x\cos(\theta)\sin(\alpha)+r_y\sin(\theta)\cos(\gamma)\cos(\alpha)$$
and we are interested in the lowest point. So, we need to find the absolute minimum of $z(\theta)$ over the interval $0\le \theta <2\pi$. 
We have to solve the following equation:
$$\frac{dz}{d\theta}=r_x\sin(\theta)\sin(\alpha)+r_y\cos(\theta)\cos(\gamma)\cos(\alpha)=0.$$
Let $A=r_x\sin(\alpha), B=r_y\cos(\gamma)\cos(\alpha), \text{ and }x=\sin(\theta)$. With this we have two equations
$$\begin{cases} i.&Ax+B\sqrt{1-x^2}=0, \text{ if } 0\le \theta <\frac{\pi}{2} \text{ or }\frac{3\pi}{2}\le \theta <2\pi\\
ii.&Ax-B\sqrt{1-x^2}=0, \text{ if } \frac{\pi}{2}\le \theta <\frac{3\pi}{2}\end{cases}.$$
These equations are easy to solve. However, the final result gets complicated...
A: Instead of planar ellipse, let's us recall some results we know about ellipsoid.
Given any ellipsoid with semi-major axis $a,b,c$ and the corresponding symmetry axis pointing along direction $\hat{p},\hat{q},\hat{r}$ (represented as unit vectors). The equation
of ellipsoid is given by
$$\frac{(\hat{p}\cdot\vec{x})^2}{a^2} + \frac{(\hat{q}\cdot\vec{x})^2}{b^2} + \frac{(\hat{r}\cdot\vec{x})^2}{c^2} = 1\tag{*1}$$
If $\hat{n}$ is any unit vector, the maximum extent of ellipsoid $(*1)$ 
along direction $\hat{n}$ is equal to
$$\max\big\{\; \hat{n}\cdot\vec{x} : \vec{x} \text{ satisfies } (*1) \;\big\} = 
\sqrt{ 
a^2 (\hat{p}\cdot\hat{n})^2 +
b^2 (\hat{q}\cdot\hat{n})^2 +
c^2 (\hat{r}\cdot\hat{n})^2
}\tag{*2}$$
If we treat the planar ellipse at hand as the limiting case of a flat ellipsoid
and let $\hat{n}$ be the unit vector pointing towards the $-ve$ $z$-direction.
We can read off the answer using $(*2)$ once we know the unit vectors $\hat{p}, \hat{q}, \hat{r}$ corresponds to the semi-major axis $a = r_x, b = r_y, c = 0$.
Start from same information as other answer, the transformation matrices for rotation about the $x$-axis for angle $\gamma$ and rotation about the $y$-axis for angle $\alpha$ are:
$$\mathcal{R}_{x,\gamma} = \begin{bmatrix}
1&0&0\\
0&\cos(\gamma)&-\sin(\gamma)\\
0&\sin(\gamma)&\cos(\gamma)
\end{bmatrix}
\quad\text{ and }\quad
\mathcal{R}_{y,\alpha} =
\begin{bmatrix}
\cos(\alpha)&0&\sin(\alpha)\\
0&1&0\\
-\sin(\alpha)&0&\cos(\alpha)
\end{bmatrix}.
$$
Multiply $\mathcal{R}_{x,\gamma}$ by $\mathcal{R}_{y,\alpha}$ from the left, the $3$ column vectors of the resulting matrix
$$\mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma} = \begin{bmatrix}
\cos\alpha  & \sin\alpha\sin\gamma & \sin\alpha\cos\gamma\\
0 & \cos\gamma & -\sin\gamma\\
-\sin\alpha & \cos\alpha\sin\gamma & \cos\alpha\cos\gamma
\end{bmatrix}
$$
will be the $3$ unit vectors $\hat{p}$, $\hat{q}$, $\hat{r}$ we seek. i.e
$$
\hat{p} = 
\mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma}
\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} = 
\begin{bmatrix}
\cos\alpha \\ 0 \\ -\sin\alpha 
\end{bmatrix}
\quad\text{ and }\quad
\hat{q} = 
\mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma}
\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} = 
\begin{bmatrix}
\sin\alpha\sin\gamma\\
\cos\gamma\\
\cos\alpha\sin\gamma
\end{bmatrix}
$$
Throw this to $(*2)$ and recall $\hat{n} = (0,0,-1)$, we will obtain
$$
\begin{cases}
\hat{p} \cdot \hat{n} &= \sin\alpha,\\
\hat{q} \cdot \hat{n} &= -\cos\alpha\sin\gamma
\end{cases}
\quad\implies\quad
z_{min} = -\sqrt{ (r_x \sin\alpha)^2 + (r_y\cos\alpha\sin\gamma)^2}
$$
Please note that if you tilt along one axis (say $\gamma = 0$) without any "shearing", then $z_{min} = -|r_x \sin\alpha|$ instead of $ -r_x\tan\alpha$.
