Geometrical meaning of $x^2+y^2+z^2-xy-xz-yz$

Question

I am looking for a geometrical interpretation of the symmetrical expression

$$f=x^2+y^2+z^2-xy-xz-yz\tag{1}$$

with $x,y,z \in \mathbb{R}$. I could for example $f$ interprete as dot products of a vector with its permuted vector

$$f=\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}-\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}y\\z\\x\end{pmatrix}\tag{2}$$

however I think there are more symmetrical ways to represent $f$ geometrically. Maybe $x,y,z$ can be thought of sides of a triangle, etc. Geometrical interpretations in $\mathbb{R},\mathbb{R^2}, \mathbb{R^3}$ are of interest.

Answer 1

Render $(x-y)^2=x^2-2xy+y^2$ and similarly for the $x,z$ and $y,z$ pairs. When you add up the squares you get

$2(x^2+y^2+z^2-xy-xz-yz).$

So $x^2+y^2+z^2-xy-xz-yz$ is half the sum of squared distances between three points on a line.

Answer 2

Let $L \subset \Bbb R^3 $ be the line $L = \{ (t,t, t) \mid t \in \Bbb R \}$. The projection of the point $P = (x, y, z) \in \Bbb R^3$ onto $L$ is $(s, s, s)$ with $s = (x+y+z)/3$, and the squared distance from $P$ to $L$ is $$ (x-s)^2+(y-s)^2+(z-s)^2 = \frac 23 (x^2+y^2+z^2-xy -yz-zx) \, . $$

So $f$ is $3/2$ times the squared distance of $(x, y, z)$ to the line $L$.

Remark: As mentioned in the comments, $f=\frac{1}{2}\left((x-y)^2+(y-z)^2+(z-x)^2\right)$, so that $f$ is zero exactly if $x=y=z$, that is if $P=(x, y, z)$ lies on the line $L$. That suggests to express $f$ in terms of the distance of $P$ to $L$.