Is joint variation a plane? This is probably a foolish question, but I am going over some old math problems and started thinking about joint variation. My question is:
Can joint variation $y = kxz$ be graphed as plane in $3D$ space? Such that if $k$ remains constant, and $x$ and $z$ vary, will $(x, y, z)$ make a plane? It seems intuitively like it will... but I was never made to graph direct relationships that way. 
Aside: Is there any application to this you have encountered? I googled and didn't get any straight answers but did see some snippets about virtual reality...
 A: 
For some constant $k$, is the set $S$ coplanar, when it is defined as $S:=\{(x,y,z)\in \Bbb R^3:y=k xz\}$ ?

Consider that for all $x=0$, then for any $z$ (and any $k$) we have $y=0$. $$\{(0,0,z):z\in\Bbb R\}\subset S$$
Likewise that for all $z=0$, then for any $x$ (and any $k$) we have $y=0$.$$\{(x,0,0):x\in\Bbb R\}\subset S$$
Given these two facts, you can now identify what plane $S$ must be, if it is to be coplanar. 
Then too, you can determine what this implies about $k$.
A: I randomized some $(x,z)$ pairs, calculated $y = xz$ for each pair and then made a plot, but it is too big for a comment.

We should not be able to rotate a plane to have it look like this so it must be twisted.
In fact, $y=kxz$ or $kxz-y=0$ is example of one of the lowest degrees of an algebraic variety: a set of equations of multivariate polynomials. In our case we have just 1 equation and the degree is only 2, since the highest sum of exponent for any term is 2 ($kx^1z^1$ and $1+1=2$).
These algebraic varieties are often non-linear differentiable manifolds of some dimensionality and have lots of applications in science and engineering.
Both the fields of algebraic geometry and differential geometry deals with these things.
