An ellipsoid centered at the origin, with semi-axes of lengths $10, 15, 30$ along the $x, y, z$ directions, is cut by a plane whose equation is $x + 3 y + 2 z = 40 $
The intersection of the ellipsoid and the plane is an ellipse. Find the coordinates of its center and its semi-axes (as $3D$ vectors).
My attempt:
The given plane can be parameterized and points on it can be expressed explicity in terms of two spanning vectors. Then this vector is used in the equation of the ellipsoid to generate an implicit equation in the two coordinates of the planar points. However, I was not able to proceed with that idea. Are there simpler ways to deal with this problem?
Edit:
I followed the suggestion of Intelligenti pauca, outlined in his comment below.
First, I transformed the given ellipsoid into the unit sphere. The given ellipsoid equation is
$ r^T Q r = 1 $
where $Q = \begin{bmatrix} \dfrac{1}{10^2} && 0 && 0 \\ 0 && \dfrac{1}{15^2} && 0 \\ 0 && 0 && \dfrac{1}{30^2} \end{bmatrix} $
So I defined the transformation as follows:
$ Q^{1/2} r = r' $
Then I would have
$ r'^T r' = 1 $
From the above,
$ r = Q^{-1/2} r' $
i.e. $ x = 10 x', y = 15 y' , z = 30 z' $
Substitute that into the plane equation, then
$10 x' + 45 y' + 60 z' = 40 $
Let $\hat{n}$ be the unit normal vector to this plane, then
$ \hat{n} = \dfrac{ (2, 9, 12) }{\sqrt{229} } $
Next, I want to find the equation of the circle of intersection of this plane with the unit sphere.
The center of the circle of intersection is away from the origin a distance $d$ where
$ d= \dfrac{40}{\sqrt{ 10^2 + 45^2 + 60^2 } } = \dfrac{8}{\sqrt{229} } $
And this means the center of the circle of intersection is at
$ C = d \hat{n} = \dfrac{ ( 16, 72, 96 ) }{ 229 } $
The radius of the circle is $r$ which is given by
$ r = \sqrt{1 - d^2} = \dfrac{\sqrt{165}}{\sqrt{229}} $
Two orthogonal unit vector which are orthogonal to $ n = (10, 45, 60) $ are
$ u_1 = \dfrac{(0, 4,-3)}{5} $
and
$ u_2 = \hat{n} \times u_1 = \dfrac{ (-75, 6, 8) }{5 \sqrt{229}} $
Now the equation of the circle is
$P(t) = C + r \ u_1 \ \cos t + r \ u_2 \ \sin t $
Now, we're ready to find the equation of the intersection ellipse by scaling back,
$ E(t) = Q^{-1/2} P(t) = Q^{-1/2} C + r \ Q^{-1/2} u_1 \ \cos t + r \ Q^{-1} u_2 \ \sin t $
And this evaluates to,
$ E(t) = \dfrac{ (160, 1080, 2880) }{229} + \sqrt{\dfrac{165}{229}} (0, 12, -18) \ \cos t + \dfrac{\sqrt{165}}{229} ( -150, 18, 48 ) \ \sin t $
This is the equation of the ellipse in terms of conjugate semi-axes. And now I need to find the semi-axes of the ellipse. For that, let
$ C_e = \dfrac{ (160, 1080, 2880) }{229} $
$ v_1 = \sqrt{\dfrac{165}{229}} (0, 12, -18) $
$ v_2 = \dfrac{\sqrt{165}}{229} ( -150, 18, 48 ) $
So the equation of the ellipse is
$ E(t) = C_e + v_1 \ \cos t + v_2 \ \sin t $
Note that $v_1$ is not orthogonal to $v_2$.
This is how I proceeded:
Set $t = t' + t_0$, and substitute that into the above expression.
$E(t') = C_e + v_1 \cos(t' + t_0) + v_2 \sin(t' + t_0) $
Expand, and collect the $\cos(t')$ terms and the $\sin(t')$ terms
$E(t') = C_e + ( \cos(t_0) v_1 + \sin(t_0) v_2 ) \cos(t') + ( -\sin(t_0) v_1 + \cos(t_0) v_2) \sin(t') $
Now, we want to find the value of $t_0$ such that the vector multiplying $\cos(t')$ is perpendicular to the vector multiplying $\sin(t') $, therefore,
$ ( \cos(t_0) v_1 + \sin(t_0) v_2 ) \cdot (-\sin(t_0) v_1 + \cos(t_0) v_2) = 0 $
Expanding,
$ (v_1 \cdot v_2) ( \cos^2(t_0) - \sin^2(t_0) ) + (v_2 \cdot v_2 - v_1 \cdot v_1 ) \sin(t_0) \cos(t_0) = 0 $
And this simplifies to:
$ (v_1 \cdot v_2) \cos(2 t_0) + \dfrac{1}{2} (v_2 \cdot v_2 - v_1 \cdot v_1 ) \sin(2 t_0) = 0 $
Therefore,
$ t_0 = \dfrac{1}{2} \tan^{-1} \bigg(\dfrac{ 2 v_1 \cdot v_2 }{ v_1 \cdot v_1 - v_2 \cdot v_2 } \bigg) $
Using the numerical values of $v_1$ and $v_2$, I found that
$ t_0 = -0.117319923 $
And the semi-axes vectors are:
$w_1 = \cos(t_0) v_1 + \sin(t_0) v_2 = (0.984856129, 9.997839809, -15.48918778) $
$w_2 = - \sin(t_0) v_1 + \cos(t_0) v_2 = (-8.356069825, 2.195014587, 0.885513032) $
Hence the semi-axes lengths are
$ \| w_1 \| = 18.46189807 $ and $ \| w_2 \| = 8.684821546 $
Question: I would appreciate if someone would verify my solution, and also if someone would present an alternative (probably simpler) method of solution. Thank you all.
From the equation of the plane, we have that $z=20-\frac{x}{2}-\frac{3y}{2}$. Putting this into the ellipsoid equation $$\frac{x^2}{10^2}+\frac{y^2}{15^2}+\frac{z^2}{30^2}=1$$ we find the projection of the ellipse $E$ onto the $xy$-plane: $$37x^2+6xy+25y^2-80x-240y-2000=0.$$ Now it suffices to find the center and the vectors representing the semi-axes of the projected ellipse $E_p$ and then lift them back to the given plane. For example, we find that the center of $E_p$ is $(\frac{160}{229},\frac{1080}{229},0)$. Therefore the center of $E$ is $$\mathbf{c}=\left(\frac{160}{229},\frac{1080}{229},20-\frac{80}{229}-3\cdot \frac{540}{229}\right)=\frac{1}{229}(160,1080,2880).$$