Exterior Product vs Cross Product

They are not equal: $\vec a\wedge\vec b$ is a 2-vector, while $\vec a\times \vec b$ is just a vector. They are related by the Hodge dual operator: $$ \star: \vec e_i\wedge\vec e_j\mapsto \text{sgn}(\sigma)\vec e_k $$ where $\sigma$ is the permutation $(1,2,3)\mapsto(i,j,k)$.

In Clifford algebra $\mathcal{Cl}_3$, they are related by: $$ \vec a\wedge\vec b=(\vec a\times\vec b)\vec e_{123},\quad \vec e_{123}=\vec e_1\vec e_2\vec e_3. $$ Here, the Clifford product is defined by: $$ \vec e_i\vec e_j=\begin{cases} -\vec e_j\vec e_i & i\neq j\\ 1 & i=j \end{cases} $$ Knowledge of such can be easily picked up from Lounesto's Clifford Algebras and Spinors.

These two products are definitely not the same, but they are dual to each other in $3$-D.

There are a couple giveaways. For one thing, for every $a,b\in \Bbb R^3$, $a\times b\in \Bbb R^3$, whereas $e_1\wedge e_2\notin \Bbb R^3$. Instead, this wedge product is in the exterior algebra, but outside $\Bbb R^3$.

The second giveaway is that the wedge is associative, whereas the cross product is not associative.

Thirdly you can just stumble across some different behaviors like this: $e_1\times(e_2\times e_1)=e_1\times(-e_3)=e_2$.

But $e_1\wedge(e_2\wedge e_1)=-e_1\wedge e_1\wedge e_2=0$.