Solving problem 4.2 in Tu's Introduction to manifolds. Not homework, voluntarily problem solving related to a course I am following.
A bilinear function $\omega_p$ on the tangent space $T_p(\mathbb{R}^3)$ is defined as \begin{align} \omega_p(a,b) &= \omega_p([a_1, a_2, a_3]^T, [b_1,b_2,b_3]^T) \\ &= p_3 det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{bmatrix}. \end{align} The point $p = (p_1, p_2, p_3)$ and tangent vectors $a,b$. Write $\omega$ in terms of basis 2-forms $dx_i \wedge dx_j$ at each point.
I am thinking that the given determinant corresponds to the area form $dx \wedge dy$. So my answer is that $$\omega = p_3 dx \wedge dy$$ at all points. The intuition is that $p_3$ is projected to a parallellogram in the $x-y$ plane. Is that correct?
Quite simply, your answer $\omega_p = p_3 dx\wedge dy$ is correct. When you apply this to a pair of vectors, $\omega_p(a,b)$ gives the signed area of the projection of the parallelogram spanned by $a$ and $b$ onto the $xy$-plane, scaled by a factor of $p_3$.