Suppose you have a cube of side length $2a$. You want to inscribe an ellipsoid in the cube such that its semi-axes are in the ratio $1:2:4$. Find a possible orientation of this ellipsoid.
This problem is closely related to this recent problem
My attempt
As discussed in the solutions of the above-referenced problem, the following relation holds
$ b^2 + (2 b)^2 + (4 b)^2 = 3 a^2 $
Where $b$ is the smallest semi-axis length. Therefore,
$ b = \dfrac{a} {\sqrt{7}} $
Now we have the three semi-axes lengths. From the results of this problem, it turns out that there are infinite possible orientations for the ellipsoid whose dimensions we now have.
The orientation is determined by three angles, and we have the following constraints on the angles
$ e_1^T R D^{-1} R^T e_1 = a^2 $
$ e_2^T R D^{-1} R^T e_2 = a^2 $
where
$ D^{-1} = \begin{bmatrix} b^2 && 0 && 0 \\ 0 && 4 b^2 && 0 \\ 0 && 0 && 16 b^2 \end{bmatrix}$
Matrix $R$ can be parameterized as follows
$ R = [ u_1 , u_2, u_3 ] R_0 $
where
$ u_3 = \begin{bmatrix} \sin \alpha \cos \beta \\ \sin \alpha \sin \beta \\ \cos \alpha \end{bmatrix} $
$ u_1 = \begin{bmatrix} \cos \alpha \cos \beta \\ \cos \alpha \sin \beta \\ - \sin \alpha \end{bmatrix} $
$ u_2 = \begin{bmatrix} - \sin \beta \\ \cos \beta \\ 0 \end{bmatrix} $
and
$ R_0 = \begin{bmatrix} \cos \gamma && -\sin \gamma && 0 \\ \sin \gamma && \cos \gamma && 0 \\ 0 && 0 && 1 \end{bmatrix} $
To obtain a finite number of possibilities we have to impose a constraint on these angles $\alpha, \beta, \gamma$.
For example, we can set $\alpha$ or $\beta$ or $\gamma$ to a specific value, then search for solutions for the remaining angles.
I still have to experiment with that.
Your comments on the above analysis, and alternative solutions are appreciated.
