Parametrize the surface $(x-z)^2+(y-z)^2+(y-x)^2=1$

Question

So I know this equation is the equation for a cylinder. And I know that the equation $x^2+y^2=1$ can be parametrized to $x=\cos(t)$,$y=\sin(t)$, with $z=t$ making a helix.

I want to know how would you go about parametrizing this equation $$(x-z)^2+(y-z)^2+(y-x)^2=1 .$$ I tired to expand this equation and solve for x,y and z but it gave me some square roots.

Answer 1

Let $X:=\begin{pmatrix}x\\y\\z\end{pmatrix} \ \text{and} \ U:=\begin{pmatrix}1\\1\\1\end{pmatrix}, \ \text{with} \ \|U\|=\sqrt{3}. \tag{*}$

We can write :

$$(z-y)^2+(x-z)^2+(y-x)^2=\|X \times U\|^2=1 \ \iff \ \|X \times U\|=1.\tag{1}$$

Let us denote by $U^{\perp}$ the plane orthogonal to $U$.

Let us decompose

$$X=aU+X'\tag{2}$$

where $X' \in U^{\perp}$ is the orthogonal projection of $X$ onto $U^{\perp}.$

Plugging (2) into (1), we obtain :

$$\|(aU+X') \times U\|=\|X' \times U\|=1.\tag{3}$$

Otherwise said, $\|X'\| .\|U\| \sin(\frac{\pi}{2})=1.\tag{4}$

Thus using (*), (4) becomes :

$$\|X'\|=1/\sqrt{3}.$$

Therefore, the solution is the set of vectors $X$ whose orthogonal projection $X'$ onto $U^{\perp}$ has norm $1/\sqrt{3}$, i.e. belongs to the circle centered in $0$ with radius $1/\sqrt{3}$ in plane $U^{\perp}$.

Thus the locus of $X$ is the cylinder with axis defined by $U$ and radius $1/\sqrt{3}$.

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Let us come now at the parametrization of this surface with 2 independant parameters $a$ and $\theta$; it is obtained by writing (2) under the form :

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=X=aU+\frac{1}{\sqrt{3}}(\cos(\theta)V+\sin(\theta)W)\tag{4}$$

where $$V:=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\\0\end{pmatrix} \ \text{and} \ W:=\frac{1}{\sqrt{6}}\begin{pmatrix}1\\1\\-2\end{pmatrix}$$ is an orthonormal basis of plane $U^{\perp}$.

Finally, (4) gives the final parameterization of the cylinder by plugging in it the previous expressions of $V$ and $W$:

Answer 2

Hint Notice that the defining equation $$(y - z)^2 + (z - x)^2 + (x - y)^2 = 1$$ is invariant under all permutations of $(x, y, z)$. Any vector invariant under these permutations---in particular any vector parallel to the axis of the cylinder---is a multiple of the unit vector ${\bf e_3} = \frac{1}{\sqrt 3} (1, 1, 1)$, which suggests writing the defining equation in some extension of ${\bf e}_3$ to an orthonormal basis $({\bf e}_1, {\bf e}_2, {\bf e}_3)$. By construction, in the coordinates $(u, v, w)$ defined by this basis, the equation is $$u^2 + v^2 = c$$ for some $c$.