I have the following ellipsoid before me:
$(x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1$.
There are three points $P$, $Q$ and $R$ on the ellipsoid which when joined with the origin form a set of three mutually perpendicular semi diameters of ellipsoid. I have to prove that plane formed by $P$ ,$Q$ and $R$ touches a fixed sphere.
I started off by assuming the equation of the plane $PQR$ as
$lx+my+nz=p$.
Three points $P$,$Q$ and $R$ were taken as $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$.
Thereafter I substituted the coordinates of three points in the equation of the plane to get three equations as below:
$lx_1+my_1+nz_1=p$;
$lx_2+my_2+nz_2=p$;
$lx_3+my_3+nz_3=p$
Multiplying first equation by $x_1$, second by $x_2$ and third by $x_3$, we get:
$l=p(x_1+x_2+x_3)/(x_1^2+x_2^2+x_3^2)$.
We can get $m$ and $n$ wherein $x$ is replaced with $y$ and $z$ respectively.
After substituting these values of $l$ ,$m$ and $n$ in the equation of the plane, we get
$x(x_1+x_2+x_3)/(x_1^2+x_2^2+x_3^2)+y(y_1+y_2+y_3)/(y_1^2+y_2^2+y_3^2)+z(z_1+z_2+z_3)/(z_1^2+z_2^2+z_3^2)= 1$.
Next I intend to find out the perpendicular distance of this plane from origin and see whether that comes out to be a constant.
But the equation of the plane seems quite daunting to me. And I am not able to simplify the expression of perpendicular distance to show that it is a constant quantity. Please suggest me how can I do that from here? Or is there any easier way out there? Any help would be highly appreciated.
Best Answer
For any $p = (x,y,z) \in \mathbb{R}^3$, we will abuse notation and use same symbol to denote corresponding $3\times 1$ column vector.
Let $\Lambda$ be the $3\times 3$ diagonal matrix with diagonal entries $\frac1{a^2},\frac1{b^2},\frac1{c^2}$. In terms of column vectors, the condition for a point $p$ to lie on the ellipse becomes $$p^T \Lambda p = 1$$
Let $n_1,n_2,n_3$ be any three unit vectors orthogonal to each other. For each $k = 1,2,3$, construct a ray start at origin pointing along direction $n_k$. Let it intersect the ellipse at $\lambda_k n_k$. We have
$$(\lambda_k n_k)^T \Lambda (\lambda_k n_k) = 1 \quad\implies\quad \frac1{\lambda_k^2} = n_k^T \Lambda n_k$$
Since the three unit vectors are orthogonal to each other, we can choose a Cartesian coordinate system having them as coordinate axes. In this new coordinate system (let's call the coordinates $(u, v, w)$), the equation of the plane becomes
$$\frac{u}{\lambda_1} + \frac{v}{\lambda_2} + \frac{w}{\lambda_3} = 1$$ and the distance $d$ between the plane and origin is given by formula:
$$d = \frac{1}{\sqrt{\frac1{\lambda_1^2} + \frac1{\lambda_2^2} + \frac1{\lambda_3^2}}} \quad\iff\quad \frac{1}{d^2} = \frac1{\lambda_1^2} + \frac1{\lambda_2^2} + \frac1{\lambda_3^2}$$ In terms of $\Lambda$ and $n_k$, we have
$$\frac{1}{d^2} = \sum_{k=1}^3 n_k^T \Lambda n_k = {\rm Tr}\left(\Lambda \sum_{k=1}^3 n_kn_k^T \right)$$
Notice $\sum\limits_{k=1}^3 n_k n_k^T = I_3$, the identity matrix. We have
$$\frac{1}{d^2} = {\rm Tr}(\Lambda) = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} $$
From this, we can conclude the distance $d$ is independent of explicit choices of $n_1,n_2,n_3$. As a result, all the planes so constructed are touching a fixed sphere centered at origin with radius $\frac{abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}$.