I think that, in $\mathbb{R}^3$, 1-vectors and 2-vectors have 3 components. Here, also, the cross product $\vec a \times \vec b$ and the exterior product $\vec a \wedge \vec b$ of vectors $\vec a$ and $\vec b$ have the same components (and so their norm is the area of the parallelogram delimited by $\vec a$ and $\vec b$). I also think the're related by the following:
$$\vec a \times \vec b = *(\vec a \wedge \vec b)$$
where $*$ is the Hodge star, which means they're Hodge duals.
All of this means that in $\mathbb{R}^3$ it's easy to mistake a 1-vector for a 2-vector (and conversely).
So, my question is: does the cross product produce a 1-vector (ie. an "oriented length"), a 2-vector (ie. an "oriented area"), or neither?
One way of viewing the differences and relationships is through the formalism of Geometric Algebra. There the 3D cross product and the wedge product can be related by an explicit multiplicative duality relationship
$$\begin{aligned}\mathbf{a} \wedge \mathbf{b}&= \mathbf{e}_1 \mathbf{e}_2\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{vmatrix}+ \mathbf{e}_1 \mathbf{e}_3\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \\ \end{vmatrix}+ \mathbf{e}_2 \mathbf{e}_3\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix} \\ &= \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_3\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{vmatrix}+ \mathbf{e}_1 \mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_2\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \\ \end{vmatrix}+ \mathbf{e}_1 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix} \\ &=\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3\left( { \mathbf{e}_3\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \\ \end{vmatrix}- \mathbf{e}_2\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \\ \end{vmatrix}+ \mathbf{e}_1\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix}} \right) \\ &= I (\mathbf{a} \times \mathbf{b}).\end{aligned}$$
where $ I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 $ is a unit pseudoscalar for the 3D space.
The wedge (exterior) product, a bivector, or grade-2 multivector, has a direct interpretation as an oriented area. The cross product, as the dual of that bivector, has a magnitude and direction that can also be interpretted as an oriented area, but it is a different beast since it is a grade-1 multivector. The pseudoscalar factor $I$ that relates the wedge and cross products is geometrically different than a unit scalar (i.e. $I^2 = -1$), which has implications for the applications of the products.
This distinction is one that is important in some contexts. For example, in physics the cross product isn't called a vector, but is called a pseudovector, since it has different transformation properties than a normal vector.