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

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 $$(V_1,V_2)$$, $$\omega(V_1,V_2)=0$$?

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
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.
• I think this is closest to what the author expected for an answer. Another way to think about it, using his notation: If $\omega = \langle V \rangle^{-1} \wedge \langle W\rangle^{-1}$, then $\ell$ is the line through $p$ spanned by $V \times W$. But it seems like this group of exercises is designed to give a geometric definition of the cross product. Commented Aug 5 at 18:57