Parametrize the surface $(x-z)^2+(y-z)^2+(y-x)^2=1$ 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. 
 A: 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 $1/\sqrt{3}$.

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 $W$:
A: 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 $c$.
