Wedge product = set intersection? In a research article [1] I found the following formulation:

The wedge product may be considered as set intersection. For
  example, surfaces of constant $f(x,y,z)$ and surface of constant $g(x,y,z)$
  intersects along the lines given by $df \wedge dg$. The notion
  of interpreting the wedge product as set intersection is appealing
  from a topological standpoint.

The article almost does not make use of exterior differential systems or other sophisticated math. It also gives no precise reference for the above statement.
Questions:


*

*Does this viewpoint (wedge product = set intersection) make sense at all?

*What is meant with "the lines given by $df \wedge dg$"? Edit: E.g. When I set $f(x,y,z) = \frac{1}{2}(x^2 + y^2 + z^2) $ and $g(x,y,z) = ax + by + cz$ with real constants $a,b,c$. Then we have $df\wedge dg = (x dx + y dy + z dz)\wedge(a dx + b dy + c dz)$. But what line does correspond to this 2-form. (I would suspect that we obtain circles.)


[1] http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=990890

At a non-critical point $x$ of $f$, the kernel of the $1$-form $\text{d}f$ is the tangent space to the surface $X := f^{-1}(f(x))$ at $x$; similarly, for a non-critical point $x$ of $g$, $\text{d}g_x$ has the tangent space of $Y := g^{-1}(g(x))$ at $x$ as its kernel. Putting both together, at non-critical points $x$ for both $f$ and $g$ the radical of the skew-symmetric $2$-form $\text{d}f\wedge\text{d}g$ contains the intersection $\text{T}_x X\cap\text{T}_x Y$, which at a point of transversal intersection of $X$ and $Y$ is just $\text{T}_x (X\cap Y)$. In this sense, the $2$-form $\text{d}f\wedge\text{d}g$ can be viewed as 'cutting out' the intersection of level sets of $f$ and $g$.
Also, you might want to look at Poincaré Duality between the $\cup$/$\wedge$-product in cohomology and the intersection product in homology.
Edit concerning the added example
Wlog we take $g(x,y,z) := z$, so that $\text{d}g = \text{d}z$, pick any ${\mathbf x}_0=(x_0,y_0,z_0)\in {\mathbb R}^3$ and put $r_0 := f({\mathbf x}_0)=\frac{1}{2}(x_0^2+y_0^2+z_0^2)$. Building on the intuition above, we  expect $(\text{d}f\wedge\text{d}g)_{{\mathbf x}_0}$ to contain the tangent space of the intersection $\{z = z_0\}\cap\{\frac{1}{2}(x^2+y^2+z^2)=r_0\}$ at ${\mathbf x}_0$ in its radical, with equality at points ${\mathbf x}_0$ of transversal intersection, which are precisely those ${\mathbf x}_0$ where $(x_0,y_0)\neq (0,0)$. 
Now, $(\text{d}f\wedge \text{d} g)_{{\mathbf x}_0} = (x_0 \text{d}x + y_0\text{d}y + z_0\text{d}z)\wedge \text{d}z = x_0\text{d}x\wedge\text{d}z + y_0\text{d}y\wedge\text{d}z$. Let's call this form $\omega$; on a tangent vector ${\mathbf v} := v_x\frac{\partial}{\partial x} + v_y\frac{\partial}{\partial y} + v_z\frac{\partial}{\partial z}$ its restriction is given by $\omega({\mathbf v},-) = (v_x x_0 + v_y y_0)\text{d} z - v_z x_0 \text{d} x - v_z y_0\text{d}y$. For $(x_0,y_0)=(0,0)$, this vanishes completely, while for $(x_0,y_0)\neq (0,0)$ the radical is given by those $\mathbf v$ where $v_z=0$ and $(v_x,v_y)\perp (x_0,y_0)$. This is precisely the expected tangent line.
