Slashing an Ellipsoid

The equation of the given ellipsoid is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

where

$a = 10,\:b = 15,\:c=30$

Define

$r = [x, y, z]^T$

Then the equation of the ellipsoid can be written as

$r^T Q r = 1$

where

$Q = \begin{bmatrix} \frac{1}{a^2} & 0 & 0 \\ 0 & \frac{1}{b^2} & 0 \\ 0 & 0 & \frac{1}{c^2} \end{bmatrix}$

On the other hand, a point in the cutting plane can be written in parametric form as

$r = r_0 + V u$

where $r_0$ is the position vector of any point on the plane, and matrix $V$ is composed of two columns of two vectors $v_1$ and $v_2$ that are perpendicular to the normal vector to the plane. Further we will take $v_1$ and $v_2$ to be perpendicular to each other and to be of unit length.

The normal to the given plane is

$n = [1,1,1]^T$

So I chose

$v_1 = \frac{1}{\sqrt{2}} [-1, 1, 0]^T$

and

$v_2 = \hat{n} \times v_1$

where $\hat{n}$ is the normalized version of n.

Now we can substitute the equation of the plane into the equation of the ellipsoid

$(r_0 + V u)^T Q (r_0 + V u) = 1$

Expanding

$r_0^T Q r_0 + u^T V^T Q V u + 2 u^T V^T Q r_0 = 1$

Rearranging

$u^T V^T Q V u + 2 u^T V^T Q r_0 = 1 -r_0^T Q r_0$

Now the left hand side can be written as

$(u + u_0)^T V^T Q V (u + u_0) - u_0^T V^T Q V u_0$

where

$u_0 = (V^T Q V)^{-1} V^T Q r_0$

Finally, set

$u = R w - u_0$

where $R$ is the orthonormal matrix of the normalized eigenvectors of $V^T Q V$ . Using this change of variables, we arrive at

$w^T D w = 1 - r_0^T Q r_0 + u_0^T V^T Q V u_0$

where $D$ is the diagonal matrix resulting from diagonalization of $V^T Q V$ through $R$ .

Dividing by the right hand side gives the standard equation of the ellipse

$w^T E w = 1$

where

$E = \dfrac{D} { 1 - r_0^T B r_0 + u_0^T V^T B V u_0}$

Now the diagonal entries of $E$ are the reciprocals of the squares of the lengths of the semi-minor and semi-major axes.

Doing the necessary calculations, gives the following values for semi-axes

$a = 9.3053,\;b = 16.1173$

Therefore, the answer is $9.3053 + 16.1173 = \boxed{25.423}$

I will post the outline that I used for a hybrid analytical/computational solution.

1) Find two unit vectors $(\vec{u}_1 \text{and} \, \vec{u}_2 )$ that are perpendicular to each other, and to the normal vector $\vec{N}$ . This is easily done using dot product and cross product formulas.

2) Sweep an angular parameter $\theta$ from $0$ to $2 \pi$

3) Let $\vec{P} = (0,0,20)$ . Let $\vec{I}$ be the intersection point with the ellipsoid. For each $\theta$ :

$\vec{v} = \cos \theta \, \vec{u}_1 + \sin \theta \, \vec{u}_2 \\ \vec{I} = \vec{P} + \alpha \vec{v} \\ \frac{(P_x + \alpha v_x)^2}{a^2} + \frac{(P_y + \alpha v_y)^2}{b^2} + \frac{(P_z + \alpha v_z)^2}{c^2} = 1$

4) For each $\theta$ , solve the quadratic for the positive value of $\alpha$ , and calculate the coordinates of $\vec{I}$ . Append these coordinates to an array.

5) Let $\vec{I}_k$ and $\vec{I}_{k+1}$ be two consecutive points in the array. Find the area of the intersection ellipse by summing the infinitesimal areas $(dA)$ calculated using the cross product formula:

$\vec{v}_k = \vec{I}_k - \vec{P} \\ \vec{v}_{k+1} = \vec{I}_{k+1} - \vec{P} \\ dA_k = \frac{1}{2} |\vec{v}_k \times \vec{v}_{k+1} | \\ A = \Sigma_{\text{All k}} \, dA_k$

5) Search the buffer to find the max distance between any two points. This value is twice the semi-major axis length.

$D_{max} = |\vec{I}_j -\vec{I}_{k}|_{max} \\ \text{semi-major axis} = \frac{D_{max}}{2}$

6) Calculate the semi-minor axis length based on the ellipse area and the semi-major axis length.

$\text{semi-minor} = \frac{A}{\pi (\text{semi-major })}$

7) Results:

$\text{semi-major } \approx 16.1 \\ \text{semi-minor } \approx 9.3$