Two-form that evaluates to zero when input nonzero vectors are not linearly dependent

Question

I am currently reading the book "A Geometric Approach to Differential Forms" from Bachman (edition 2). There, exercise 3.21 states the following:

Show that if $\omega$ is a 2-form on $T_p \mathbb{R}^3$, then there is a line $l$ in $T_p\mathbb{R}^3$ such that if the plane spanned by $V_1$ and $V_2$ contains $l$, then $\omega(V_1,V_2)=0$.

I am confused on several levels here.

What I think I understand

My intuition is that there are nonzero $V_1$ and $V_2$ vectors such that $V_1\neq c V_2$ ($c$ being an arbitrary constant; vectors are not linearly dependent), but $\omega(V_1,V_2)=0$.

In other words, there exists some plane $T_p \mathbb{R}^3$ where $(V_1,V_2)$ are projected and they lie in top of each other, as in the drawing below where the two form is instead $\omega \wedge \nu$, $\omega$ and $\nu$ being 1-forms:

In this configuration, $V_1$ and $V_2$ are obviously on a line $l$. So, in this case, when the plane spanned by $(V_1,V_2)$ is equal to the line $l$, then we get $\omega(V_1,V_2)=0$ by necessity.

What confuses me

The word "contains" is confusing me because there may exist a plane on $T_p \mathbb{R}^3$, where the projection of $(V_1,V_2)$ spans a plane that contains the line $l$, but such that $\omega(V_1,V_2)\neq 0$ as in the following drawing:

The area of the projected parallelogram is evidently nonzero ($\omega(V_1,V_2)\neq 0$) in this case, and the second part of the proof I am looking for does not apply:

... if the plane spanned by $V_1$ and $V_2$ contains $l$, then $\omega(V_1,V_2)=0$.

What am I missing here? Why is it that when there is a line $l$ in $T_p \mathbb{R}^3$ contained in the plane spanned by $(V_1,V_2)$, $\omega(V_1,V_2)=0$?

Answer 1

I quote Bachman, because the context is missing from your question and this makes it hard to understand your approach. He says:

Every 2-form on $T_p\mathbb{R}^3$ projects pairs of vectors onto some plane and returns the area of the resulting parallelogram, scaled by some constant

I will give you a solution based on this fact, because I can't fully understand your reasoning. Hopefully, you can work out how this relates to your approach.

Let $\omega$ be the given 2-form, and let $\Pi_\omega$ be the plane given by the observation above. There is arguably one canonical choice of a line with the data we have, the normal to $\Pi_\omega$ at $p$. We call it $\ell_p$. And now, if $\pi$ is any other plane containing $\ell_p$, the intersection of $\Pi_\omega$ and $\pi$ is a line $L_p$ in $\Pi_\omega$. We “see” (or you can show it) that when a vector in $\pi$ is projected onto $\Pi_\omega$, the result lies in the line $L_p$. So any pair of vectors in $\pi$ project onto a degenerate parallelogram, which has area zero.