Finding the extrema of $E(\vec{r})=\frac1a x^2+\frac1b y^2+ \frac1c z^2$ with respect to constraints geometrically

Question

I have a function, $$E(\vec{r})=\frac1a x^2+\frac1b y^2+ \frac1c z^2.$$ Where $\vec{r}=(x,y,z)$ and $a>b>c>0$. I wish to find the maximum and minimum of this function with respect to the constraints, $$\vec{r}\cdot\vec{r}=1,\qquad\vec{k}\cdot\vec{r}=0,$$ where $\vec{r}=(k_1,k_2,k_3)$, is an arbitrary constant vector. I do not wish to solve this using Lagrange multipliers, but instead via a geometrical argument.

The constraints are clearly the unit sphere intersecting with a plane normal to $\vec{k}$ through the origin. So this is a unit circle rotated in 3 dimensions through the origin, dependent on the direction of the vector $\vec{k}$.

So to analyse to the maxima and minima of $E$ w.r.t. to the constraints above we can consider the intersections of the rotated circle through the origin with the family of ellipsoids

$$E(\vec{r})=\text{constant}$$

My question is, is this the right way to look at this problem from a geometrical perspective? If we let $E=constant=\lambda$, imagine this $\lambda>>1$, so that the ellipsoid is very large and outside of the unit sphere. Then as we lower the value of $E$ and hence $\lambda$ eventually as some point on the intersection of the plane $\vec{k}\cdot\vec{r}=0$ with the unit sphere and the ellipsoid will touch. This point will be the point that maximises $E$ and similarly the last point to touch the ellipsoid as it is shrunk further will be the minimum, is this correct? If so how could these points be found?

Here is a diagram to help explain what i'm talking about.

EDIT#1: Here is the tilted circle (formed from the intersection on a sphere and a plane), in parametric form

$$\vec{r}(\theta)=\left(\begin{array}{c} \mathrm{k_1}\, \mathrm{k_2}\, \left(\sqrt{1 - {\mathrm{k_3}}^2} - 1\right)\\ {\mathrm{k_2}}^2\, \sqrt{1 - {\mathrm{k_3}}^2} + {\mathrm{k_3}}^2\, \sqrt{1 - {\mathrm{k_3}}^2} - {\mathrm{k_2}}^2 - {\mathrm{k_3}}^2 + 1\\ - \mathrm{k_2}\, \mathrm{k_3} \end{array}\right)\sin(\theta)+\left(\begin{array}{c} {\mathrm{k_2}}^2 - {\mathrm{k_2}}^2\, \sqrt{1 - {\mathrm{k_3}}^2} + \sqrt{1 - {\mathrm{k_3}}^2}\\ \mathrm{k_1}\, \mathrm{k_2}\, \left(\sqrt{1 - {\mathrm{k_3}}^2} - 1\right)\\ - \mathrm{k_1}\, \mathrm{k_3} \end{array}\right)\cos(\theta)$$

EDIT#2: Or it can be described more neatly by defining $\vec{k_\perp}=(-k_2,k_1,0)$, then

$$\vec{r}(\theta)=\vec{k_\perp}\sin(\theta)+(\vec{k_\perp}\times\vec{k})\cos(\theta)$$

Answer 1

Before answering the question, I want to make a few modifications.

• Ellipsoid equation: $E(r)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

• Constant normal vector $\vec k=(k_1,k_2,k_3)$ is a unit vector.

Now, let's consider the geometric meaning of the question. The question requires extremes of $E(r)$ whose corresponding ellipsoid intersecting with a plane normal to $\vec k$, which is a circle or an ellipse, has the boundary of intersection tangent to the unit circle, as what you understand.

Thus, the solution, if we rotate all the stuffs so that $\vec k$ becomes $z$ direction and of course the ellipsoid inclines. Then the intersection is in the x-y plane so we can easily find the extreme.

According to the above, the hardest thing is to find a rotation $R$ so that $$R(\vec k)=Rk=z=(0,0,1)$$ $$R^TR=RR^T=I$$ Here're 9 equations to determine $3^2$ unknowns of $R$.

With this rotation, new coordinate system establishes, viz $$r'=Rr$$ where $r'=(x',y',z')$.

In the new coordinate system, we can rewrite the equation of ellipsoid in terms of $r'$. Then set $z'=0$ and we obtain the equation of ellipse of the intersection. The form of equation shall be $$0=Dx'^2+2Ex'y'+Fy'^2=(x',y')\begin{pmatrix}D&E\\E&F\end{pmatrix}(x',y')^T=:(x',y')Q(x',y')^T$$ The eigenvalues, say $\lambda_{min},\lambda_{max}$, of $Q$ corresponds the length of semi-major axis of ellipsoid or that of radius of circle. And corresponding eigenvectors, say $v'_{max},v'_{min}$, represents the direction of axes.

Set $\lambda=1$ will, conversely, obtain $v'$ and also $v=R^{-1}v'=R^Tv'$, so do, of course, $$E(r)_{extreme}=E(v)$$