How to perform wedge product

I assume your notation $[v^1,v^2,v^3]$ means a vector $\mathbf v \in V \equiv \mathbb R^3$ where $$\mathbf v = v^1\ \mathbf e_1 + v^2\ \mathbf e_2 + v^3\ \mathbf e_3. $$ The wedge operator is a bilinear, antisymmetric operator takes in two vectors $\mathbf v, \mathbf w \in V$ and outputs a bivector $\mathbf v \wedge \mathbf w \in \Lambda^2(V)$ such that $$\begin{split} \mathbf v \wedge \mathbf w &= (v^1\ \mathbf e_1 + v^2\ \mathbf e_2 + v^3\ \mathbf e_3) \wedge (w^1\ \mathbf e_1 + w^2\ \mathbf e_2 + w^3\ \mathbf e_3) \\ &= v^1 w^1\ \mathbf e_1 \wedge \mathbf e_1 + (v^1 w^2 - v^2 w^1)\ \mathbf e_1 \wedge \mathbf e_2 + (v^1 w^3 - v^3 w^1)\ \mathbf e_1 \wedge \mathbf e_3 \\ &\quad + v^2w^2\ \mathbf e_2 \wedge \mathbf e_2 + (v^2 w^3 - v^3 w^2)\ \mathbf e_2 \wedge \mathbf e_3 + v^3w^3\ \mathbf e_3 \wedge \mathbf e_3 \\ &= (v^1 w^2 - v^2 w^1)\ \mathbf e_1 \wedge \mathbf e_2 + (v^1 w^3 - v^3 w^1)\ \mathbf e_1 \wedge \mathbf e_3 + (v^2 w^3 - v^3 w^2)\ \mathbf e_2 \wedge \mathbf e_3. \end{split}\tag1$$ If you're wondering why I have made these simplifications:

• The wedge is bilinear, i.e. $(a\mathbf v + b \mathbf w) \wedge \mathbf x = a(\mathbf v \wedge \mathbf x) + b(\mathbf w \wedge \mathbf x)$ and the same in the second slot, so this allows us to distribute it through the linear combinations in parentheses after the first equal sign, and take out the coefficients;
• The wedge is antisymmetric, i.e. $\mathbf v \wedge \mathbf w = - \mathbf w \wedge \mathbf v$, which implies $\mathbf v \wedge \mathbf v = \mathbf 0$, so this allows us to "reorder" $a(\mathbf e_2 \wedge \mathbf e_1) = - a (\mathbf e_1 \wedge \mathbf e_2)$, etc., in the second line, and simplify $b (\mathbf e_1 \wedge \mathbf e_1) = b\ \mathbf 0 = \mathbf 0$, etc., in the third.

Notice that if you set $$\begin{split} \mathbf e_1 \wedge \mathbf e_2 &\mapsto \mathbf e_3, \\ \mathbf e_1 \wedge \mathbf e_3 &\mapsto - \mathbf e_2, \\ \mathbf e_2 \wedge \mathbf e_3 &\mapsto \mathbf e_1, \end{split}\tag2$$ you get the cross product $\mathbf v \times \mathbf w$! Since $\{\mathbf e_i \wedge \mathbf e_j,\ 1\leq i < j\leq n\}$ turns out to be a basis for $\Lambda^2(V)$ when $\dim V = n$, $(2)$ defines a bilinear map $\star : \Lambda^2(V) \to \Lambda^1(V) \equiv V$ which is called Hodge dual and can actually be generalized to higher dimensions. However it is basically only in dimension $3$ that it takes this simple, familiar form (this is because $\dim \Lambda^2(V) = \dim V$ when $n=3$).

In the specific case where $$\mathbf v = \mathbf e_1 + 3 \mathbf e_2 -2 \mathbf e_3, \qquad \mathbf w = 5 \mathbf e_1 + 2 \mathbf e_2 + 8 \mathbf e_3, $$ you get, from $(1)$, $$\begin{split} \mathbf v \wedge \mathbf w &= -13\ \mathbf e_1 \wedge \mathbf e_2 + 18\ \mathbf e_1 \wedge \mathbf e_3 +28\ \mathbf e_2 \wedge \mathbf e_3 . \end{split}$$ As a confirmation, notice $\star(\mathbf v \wedge \mathbf w) = \mathbf v \times \mathbf w$.

If you are interested in the "concrete" side of exterior algebra, I suggest you give a look at Sergei Winitzki's Linear Algebra via Exterior Products, which is available online for free. He introduces these very abstract concepts, like tensor products, wedge products, hodge duality, etc., but always manages to include simple computational examples that show how they work. You can find the book here.

Wedge product refers to the exterior algebra, not clifford algebra (though it can be emulated in the latter).

So, suppose you have vectors $x=(1,3,-2)$ and $y=(5,2,8)$. Let $\{e_1,e_2,e_3\}$ be the basis of our vector space $\mathbb R^3$, so $x=1\cdot e_1 + 3 \cdot e_2 - 2 \cdot e_3$ and $y=5\cdot e_1 + 2 \cdot e_2 + 8 \cdot e_3$.

The wedge product of two vectors, strictly speaking, is not itself a vector of the same space $V$, but of the exterior square $\Lambda ^2V$. If $\dim V=n$, then $\dim \Lambda^2 V = \frac{n(n-1)}{2}$. In three dimensions, however, it happens that $\dim \Lambda^2 V=\frac{3\cdot 2}{2}=3$.

The main rules for wedge products are $a \wedge a=0$ and $a \wedge b = - b \wedge a$. One common basis for $\Lambda^2 \mathbb R^3$ is $\{e_2 \wedge e_3, e_3 \wedge e_1, e_1 \wedge e_2\}$.

So, let's plug everything in (and remember that $\wedge$ is bilinear):

$$x\wedge y=(1\cdot e_1 + 3 \cdot e_2 - 2 \cdot e_3) \wedge (5\cdot e_1 + 2 \cdot e_2 + 8 \cdot e_3) = \\ 1\cdot 5 \cdot e_1 \wedge e_1 + 1\cdot 2 \cdot e_1 \wedge e_2 + 1\cdot 8 \cdot e_1 \wedge e_3 + \\ 3\cdot 5 \cdot e_2\wedge e_1 +3\cdot 2 \cdot e_2\wedge e_2 +3\cdot 8 \cdot e_2\wedge e_3 - \\ 2\cdot 5 \cdot e_3\wedge e_1 -2\cdot 2 \cdot e_3\wedge e_2 -2\cdot 8 \cdot e_3\wedge e_3 = \\ 5 \cdot \mathbb O + 2 \cdot e_1 \wedge e_2 - 8 \cdot e_3 \wedge e_1 - \\ 15 \cdot e_1\wedge e_2 +6 \cdot \mathbb O +24 \cdot e_2\wedge e_3 - \\ 10 \cdot e_3\wedge e_1 +4 \cdot e_2\wedge e_3 -16 \cdot \mathbb O = \\ 28 \cdot e_2\wedge e_3-18 \cdot e_3\wedge e_1 - 13 \cdot e_1\wedge e_2 $$

Here, $\mathbb O$ denotes the zero vector.