A Planar Cut Through 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

$r^T B r = 1$

where

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

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

$n = [1,3,2]^T$

So I chose

$v_1 = \frac{1}{\sqrt{5}} [-2, 0, 1]^T$

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

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

Expanding

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

Rearranging

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

Now the left hand side can be written as

$(u + u_0)^T V^T B V (u + u_0) - u_0^T V^TBVu_0$

where

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

Finally, set

$u = R w - u_0$

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

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

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

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

$w^T E w = 1$

where

$E = D / (1 - r_0^T B r_0 + u_0^T V^T B V u_0)$

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

$a = 8.6848,\;b = 18.4619$

The area of the ellipse is given by

$A = \pi a b = 503.717$

So now my previously posted answer works. Man, this was a really unpleasant job.

I can't hardly wait to see someone explain how this area was figured out, in the space provided here.

One way to do this (not the way I actually did it) is to do a rotation transform to move both the cutting plane and the ellipsoid so that the plane is coincident with the $x,y$ plane. Then find the equation of the ellipse on the plane (by letting $z=0$ ), and from there figure out the major and minor axes. There are different ways of doing that, but given the coefficients of the implicit 2D equation of the ellipse on the plane, the semi-axes can be computed directly.