Ellipsoid in the corner

The parametric equation (vector equation) of points on the ellipsoid surface is

$\mathbf{r}(\theta, \phi) = \mathbf{r}_C + V \mathbf{u}(\theta,\phi)$

where $\mathbf{r}_C = (x_C , y_C , z_C)$ is the position vector of the center of the ellipsoid. And the $3 \times 3$ matrix $V$ is given by

$V = [ v_1, v_2, v_3 ] = \mathbf{R D}$ , with $\mathbf{R}$ being an orthogonal matrix whose columns are the unit vectors of three axes of the ellipsoid, and $\mathbf{D}$ is a diagonal matrix of the semi-axes lengths. Finally, the unit vector $\mathbf{u}(\theta,\phi)$ is specified as

$\mathbf{u}(\theta,\phi) = ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta )$

At the three tangency points, one of the coordinates vanishes, and the same coordinate of the derivatives of $\mathbf{r}(\theta, \phi)$

with respect to $\theta$ and to $\phi$ vanishes as well.

Let's take the tangent point with the $xy$ plane. We know that at that point we'll have,

$z_C + v_{1z} \sin \theta_1 \cos \phi_1 + v_{2z} \sin \theta_1 \sin \phi_1 + v_{3z} \cos \theta_1 = 0$

In addition, by differentiating $\vec{r}$ with respect to $\theta$ and $\phi$ and setting the $z$ -component to zero,

we obtain two more equations,

$v_{1z} \cos \theta_1 \cos \phi_1 + v_{2z} \cos \theta_1 \sin \phi_1 - v_{3z} \sin \theta_1 = 0$

and

$-v_{1z} \sin \theta_1 \sin \phi_1 + v_{2z} \sin \theta_1 \cos \phi_1 = 0$

Now in the last equation, $\sin \theta_1 = 0$ is extraneous and can be ignored, and therefore we can divide by it, resulting in

$-v_{1z} \sin \phi_1 + v_{2z} \cos \phi_1 = 0$

And this can be solved for $\phi_1$ , resulting in,

$\cos \phi_1 = - \dfrac{v_{1z}} { \sqrt{{v_{1z}}^2 + {v_{2z}}^2 }}$ and

$\sin \phi_1 = - \dfrac{v_{2z}} { \sqrt{{v_{1z}}^2 + {v_{2z}}^2 }}$

Plugging in the angle $\phi_1$ in the previous equation, we can now solve for $\theta_1$ , in a similar fashion, and

the result is that,

$\cos \theta_1 = -\dfrac{v_{3z}}{ \sqrt{{v_{1z}}^2 + {v_{2z}}^2 + {v_{3z}}^2}}$

$\sin \theta_1 = \dfrac{ \sqrt{{v_{1z}}^2 + {v_{2z}}^2 }}{ \sqrt{{v_{1z}}^2 + {v_{2z}}^2 + {v_{3z}}^2}}$

The choice of signs for $\cos \phi_1$ , $\sin \phi_1$ , $\cos \theta_1$ and $\sin \theta_1$ guarantees that $\theta_1 \in [0, \pi ]$ on the one hand, and also that the $z$ -coordinate of the tangency point (which is 0) is as far as possible from $z_C$ (which is positive) for any values of $v_{1z}$ , $v_{2z}$ and $v_{3z}$ .

Plugging both $\theta_1$ and $\phi_1$ into the first equation, results in

$z_C = \sqrt{{v_{1z}}^2 + {v_{2z}}^2 + {v_{3z}}^2}$

Similarly, we have,

$x_C = \sqrt{{v_{1x}}^2 + {v_{2x}}^2 + {v_{3x}}^2}$

and

$y_C = \sqrt{{v_{1y}}^2 + {v_{2y}}^2 + {v_{3y}}^2}$

So far we have found the center of the ellipsoid, and expressions for the angles $\theta$ and $\phi$ at the tangency point.

Now for the tangency point at the $yz$ plane, we have the following,

$\cos \phi_3 = - \dfrac{v_{1x}} { \sqrt{{v_{1x}}^2 + {v_{2x}}^2 }}$ and

$\sin \phi_3 = -\dfrac{v_{2x}} { \sqrt{{v_{1x}}^2 + {v_{2x}}^2 }}$

and

$\cos \theta_3 = - \dfrac{v_{3x}}{ \sqrt{{v_{1x}}^2 + {v_{2x}}^2 + {v_{3x}}^2}}$

$\sin \theta_3 = \dfrac{ \sqrt{{v_{1x}}^2 + {v_{2x}}^2 }}{ \sqrt{{v_{1x}}^2 + {v_{2x}}^2 + {v_{3x}}^2}}$

Using these values, we can find $\mathbf{r}(\theta_3, \phi_3 )$ using the parametric representation of the ellipsoid. And the result is $\mathbf{r}^* = ( 0, 2.855166 , 6.531751 )$ , making the answer $(2.855166+6.531751) = \boxed{9.386917}$

Another approach to this problem is to consider the algebraic equation of the ellipsoid instead of the vector equation.

The algebraic equation of ellipsoid is given by,

$(\mathbf{r} - \mathbf{r}_C)^T \mathbf{R D R}^T (\mathbf{r} - \mathbf{r}_C) = 1$

where the matrix $\mathbf{R}$ is as before, but the diagonal matrix $\mathbf{D} = \text{diag} \{ \dfrac{1}{a^2}, \dfrac{1}{b^2},\dfrac{1}{c^2} \}$ , where $a, b$ and $c$ are the semi-axes lengths. The gradient of the above expression (the normal to the surface) is given by,

$\nabla = 2 \mathbf{R D R}^T (\mathbf{r} - \mathbf{r}_C)$

Let $\mathbf{r}_1$ be the tangency point with the $xy$ plane, then at that point the gradient is pointing in the opposite direction of $\mathbf{k}$ (the unit vector along the positive $z$ axis ). Hence, there exist a positive constant $\alpha$ , such that,

$\mathbf{R D R}^T (\mathbf{r}_1 - \mathbf{r}_C)= - \alpha \mathbf{k}$

and this implies that,

$\mathbf{r}_1 - \mathbf{r}_C= - \alpha \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} \hspace{24pt}(*)$

Substituting this expression into the algebraic equation of the ellipsoid, and solving for $\alpha$ , we get,

$\alpha = \dfrac{1}{\sqrt{ \mathbf{k}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} }}$

Since the $z$ coordinate of $\mathbf{r}_1 = 0$ , it follows that, by pre-multiplying both sides of equation $(*)$ by $\mathbf{k}^T$ ,

$z_C = \alpha \mathbf{k}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} = \sqrt { \mathbf{k}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} }$

and similarly,

$x_C = \sqrt{ \mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} }$

$y_C = \sqrt{ \mathbf{j}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{j} }$

where $\mathbf{i}$ and $\mathbf{j}$ are the unit vector along the positive $x$ and $y$ axes. Thus we have determined $\mathbf{r}_C$

And using equation $(*)$ and its variations, we can obtain expressions for the three tangent points $\mathbf{r}_1$ with the $xy$ plane , $\mathbf{r}_2$ , with the $xz$ plane, and $\mathbf{r}_3$ with the $yz$ plane, and they are as follows,

$\mathbf{r}_1 = \mathbf{r}_C - \dfrac{ \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} }{ \sqrt{ \mathbf{k}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{k} } }$

$\mathbf{r}_2 = \mathbf{r}_C - \dfrac{ \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{j} }{ \sqrt{ \mathbf{j}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{j} } }$

$\mathbf{r}_3 = \mathbf{r}_C - \dfrac{ \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} }{ \sqrt{ \mathbf{i}^T \mathbf{R D}^{-1} \mathbf{R}^T \mathbf{i} } }$