Let $x,y,z$ be the standard coordinates on $\mathbb{R}^3$. A plane in $\mathbb{R}^3$ is vertical if $ax + by = 0$ for some $a,b \ne 0$. Prove that restricted to a vertical plane $dx \wedge dy = 0$.
What is the intuition for this problem? If $S$ is the plane in question and denote $\omega : = dx \wedge dy$, then $$\omega|_S=i^*\omega$$ where $i : S \to \mathbb{R}^3$ is the inclusion and $i^*$ the pullback. So $$i^*(dx \wedge dy)=i^*(dx) \wedge i^*(dy) = d(x \circ i) \wedge d(y \circ i).$$
Can we show that $$d(x \circ i) \wedge d(y \circ i) = 0?$$
Algebraic reasons have been given in the comments (y and x being linearly dependent etc.)
A vertical plane casts no 2-dimensional shadow on the xy-coordinate plane. Its projection instead is a line (on which y is a linear function of x). The $dx\wedge dy$ which would be a top, "volume" form on a 2-manifold is null on a 1-manifold. Indeed surface area of a line is 0, volume of a surface is 0 and so on.
The volume form of the ambient space $dx\wedge dy \wedge dz$, given a normal vector field to a surface, induces the following surface area form:
$ds=n_x\ dy\wedge dz+n_y\ dz\wedge dx+n_z\ dx\wedge dy = \dfrac{dy\wedge dz}{n_x} = \dfrac{dz\wedge dx}{n_y} = \dfrac{dx\wedge dy}{n_z}$
As a plane approaches the vertical $n_z\to 0$, and the xy-element of the surface area, $dx\wedge dy$ has to approach a zero 2-form.