Minimum distance from $2x^2 + 2xy + 4yz + z^2 = 1 $ to the origin

Question

Let Σ be the surface in $R^3$ given by $2x^2 + 2xy + 4yz + z^2 = 1 $

By writing this equation as $x^TAx$ = 1, with A a real symmetric matrix, show that there is an orthonormal basis such that, if we use coordinates $(u, v, w)$ with respect to this new basis, Σ takes the form $λu^2 + µv^2 + νw^2 = 1.$ Find $λ$, $µ$ and $ν$ and hence find the minimum distance between the origin and Σ.

Hint: it is not necessary to find the basis explicitly.

I found matrix $A$ = $\begin{pmatrix} 2& 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1 \end{pmatrix} $ , and figured that $λ$, $µ$ and $ν$ should be its eigenvalues, after changing the basis to the eigenvectors. The eigenvalues are $ 3, \sqrt{3}, -\sqrt{3} $, so I have $3u^2 + \sqrt{3}^2 -\sqrt{3}w^2 = 1.$

I feel like I have made a mistake somewhere, I don't know how to proceed.

Answer 1

Let be $u$, $v$ and $w$ the orthogonal base of eigenvectors of $A$. Then you can write each $x\in\mathbb R^3$ as the linear combination of $u$,$v$ and $w$. So there exist $\alpha,\beta,\gamma\in\mathbb R$ such that $x=\alpha u+\beta v+\gamma w$ and you get $$ x^TAx=1\Leftrightarrow 3\alpha^2 u^2+\sqrt{3}\beta^2 v^2-\sqrt{3}\gamma^2 w^2=1. $$ But we can do better. We can normalize $u$, $v$ and $w$ such that we get $$ 3\alpha^2 u^2+\sqrt{3}\beta^2 v^2-\sqrt{3}\gamma^2 w^2=1\Leftrightarrow 3\alpha^2+\sqrt{3}\beta^2-\sqrt{3}\gamma^2=1. $$ To compute the minimal distance of $\Sigma$ to the orgin, we have to compute the minimum of $\tilde f(x)=\|x\|^2=x_1^2+x_2^2+x_3^2$, which is the square of the distance, within the restriction $x^TAx=1$.

But considering $x=\alpha u+\beta v+\gamma w$, we can also minimize $f(\alpha,\beta,\gamma)=\alpha^2+\beta^2+\gamma^2$ with the restiction $3\alpha^2+\sqrt{3}\beta^2-\sqrt{3}\gamma^2=1$.

Here you see that we don't need the explicit eigenvectors. Should the rest be fine?

Hint:

Define $g(\alpha,\beta,\gamma)=3\alpha^2+\sqrt{3}\beta^2-\sqrt{3}\gamma^2$. You have to find $\lambda\in\mathbb R$ and $(\alpha,\beta,\gamma)\in\mathbb R^3$ such that $\nabla f(\alpha,\beta,\gamma)=\lambda\nabla g(\alpha,\beta,\gamma)$ and $g(\alpha,\beta,\gamma)=1$.

Result:

We compute $$\nabla f(\alpha,\beta,\gamma)=\begin{pmatrix}2\alpha\\2\beta\\2\gamma\end{pmatrix}\text{ and }\nabla g(\alpha,\beta,\gamma)=\begin{pmatrix}3\alpha\\\sqrt{3}\beta\\-\sqrt{3}\gamma\end{pmatrix}.$$ The equation $\nabla f(\alpha,\beta,\gamma)=\lambda\nabla g(\alpha,\beta,\gamma)$ can hold if and only if two values out of $\alpha$,$\beta$ and $\gamma$ are $0$. (Otherwise we get a contradiction). Further, we get the non-zero value from $g(\alpha,\beta,\gamma)=1$. We deduce the critical points $$X_1=\begin{pmatrix}\frac1{\sqrt{3}}\\0\\0\end{pmatrix},~X_2=\begin{pmatrix}-\frac1{\sqrt{3}}\\0\\0\end{pmatrix},~X_3=\begin{pmatrix}0\\\frac1{\sqrt[4]{3}}\\0\end{pmatrix}\text{ and }X4=\begin{pmatrix}0\\\frac1{\sqrt[4]{3}}\\0\end{pmatrix}.$$ Finally, we check $f(X_1)=f(X_2)=\frac13$ and $f(X_3)=f(X_4)=\frac1{\sqrt{3}}>\frac13$.

Hence $f$ is minimized at $X_1$ and $X_2$ and the minimal distance from $\Sigma$ to the orgin is $\sqrt{f(X_{1/2})}=\frac1{\sqrt{3}}$.

Answer 2

Apply Lagrange Multipliers Method to minimize $f(x,y,z)= x^2+y^2+z^2$ subject to $g(x,y,z)= 2x^2+2xy+4yz+z^2-1$, results in $x=y=z=\frac13$ or $x=y=z=-\frac13$ with the minimum value of $f(x,y,z)=\frac13$. Therefor the minimum distance is $\sqrt{\frac13}$.