I am an undergraduate physics student trying to learn this on my own, so it's been a bear digging through either very general resources or resources that don't answer my questions.
What I am trying to prove is:
$$a \wedge b = (e_{x} \wedge e_{y} \wedge e_{z}) \cdot (a \times b)$$
...by direct computation in components and term identification where $$e_{i}$$ is a basis vector and $$a,b$$ are unit vectors in some arbitrary direction (describing a detector direction chosen experimentally).
My understanding is that it should look something like this:
$$(e_{x} \wedge e_{y} \wedge e_{z}) \cdot ((a_{2}b_{3}-a_{3}b_{2})e_{x} + (a_{1}b_{3}-a_{3}b_{1})e_{x}) + (a_{1}b_{2}-a_{2}b_{1})e_{x})$$
But I don't know how to actually go through the steps to get there since I have never even taken abstract algebra.
Let $uv$ represent the Clifford product of multivectors $u$ and $v$. I will also assume that $e_x, e_y, e_z$ is an orthonormal basis with respect to the usual inner product on $\Bbb R^3$. This implies that $e_i\wedge e_j = e_ie_j$ for all $i \ne j$. Then \begin{align}&(e_{x} \wedge e_{y} \wedge e_{z}) \cdot \big[(a_{2}b_{3}-a_{3}b_{2})e_{x} + (a_{3}b_{1}-a_{1}b_{3})e_{y} + (a_{1}b_{2}-a_{2}b_{1})e_{z}\big] \\ &= (e_xe_ye_z)\cdot \big[(a_{2}b_{3}-a_{3}b_{2})e_{x} + (a_{3}b_{1}-a_{1}b_{3})e_{y} + (a_{1}b_{2}-a_{2}b_{1})e_{z}\big] \\ &= (a_2b_3-a_3b_2)\big[(e_xe_ye_z)\cdot e_x\big] + (a_3b_1-a_1b_3)\big[(e_xe_ye_z)\cdot e_y\big] + (a_1b_2-a_2b_1)\big[(e_xe_ye_z)\cdot e_z\big] \\ &= (a_2b_3-a_3b_2)\langle (e_xe_ye_z)e_x\rangle_2 + (a_3b_1-a_1b_3)\langle (e_xe_ye_z)e_y\rangle_2 + (a_1b_2-a_2b_1)\langle(e_xe_ye_z)e_z\rangle_2 \\ &= (a_2b_3-a_3b_2)\langle e_xe_xe_ye_z\rangle_2 + (a_3b_1-a_1b_3)\langle -e_xe_ye_ye_z\rangle_2 + (a_1b_2-a_2b_1)\langle e_xe_ye_ze_z\rangle_2 \\ &= (a_2b_3-a_3b_2)e_ye_z - (a_3b_1-a_1b_3) e_xe_z + (a_1b_2-a_2b_1)e_xe_y \\ &= (a_2b_3-a_3b_2)e_y\wedge e_z + (a_3b_1-a_1b_3) e_z\wedge e_x + (a_1b_2-a_2b_1)e_x\wedge e_y \\ &= a\wedge b\end{align}