There is a unique bi-linear map $\R_1^3\times\R_1^3\rar\R_1^3$, $(x,y)\mapsto x*y$ with the following properties:
  1. For all $x,y\in\R_1^3$: $y*x=-x*y$.
  2. $e_0*e_1=e_2$, $e_1*e_2=-e_0$, $e_2*e_0=e_1$.
1. Verify that for all $x,y\in\R_1^3$: $\la x,x*y\ra=\la y,x*y\ra=0$. Thus given any pair of vectors $(x,y)$ the vector $x*y$ is orthogonal to both $x$ and $y$. 2. For all $x\in\R_1^3$ the linear map $y\mapsto x*y$ is skew-symmetric on $\R_1^3$. It follows that $\o:(x,y,z)\mapsto\la x*y,z\ra$ is a volume-form on $\R_1^3$; show that $\o(e_0,e_1,e_2)=+1$, i.e. by lemma: $\o(x_0,x_1,x_2)=\det(e_j^*(x_k))$. 3. Verify Jacobi's identity: for all $x,y,z$: $x*(y*z)+y*(z*x)+z*(x*y)=0$.
1. These conditions imply that for $x=\sum x_je_j$ and $y=\sum y_ke_k$: $$ x*y =-(x_1y_2-x_2y_1)e_0+(x_2y_0-x_0y_2)e_1+(x_0y_1-x_1y_0)e_2 $$ i.e. $(x,y)\mapsto x*y$ is unique. Moreover $$ \la x,x*y\ra =(x_0y_1-x_1y_0)x_2+(x_1y_2-x_2y_1)x_0+(x_2y_0-x_0y_2)x_1 =0 $$ and $\la y,x*y\ra=-\la y,y*x\ra=0$.
2. Put $Ay\colon=x*y$, then by 1. $\la y,Ay\ra=0$ ant therefore $$ 0 =\la y+z,A(y+z)\ra =\la y,Ay\ra+\la y,Az\ra+\la z,Ay\ra+\la z,Az\ra =\la y,Az\ra+\la z,Ay\ra $$ i.e. $A$ is skew-symmetric. Finally $\o(e_0,e_1,e_2)=\la e_0*e_1,e_2\ra=\la e_2,e_2\ra=+1$.
3. The $3$-linear map $\eta(x,y,z)\colon=x*(y*z)+y*(z*x)+z*(x*y)$ is alternating, e.g.: $$ \eta(y,x,z) =y*(x*z)+x*(z*y)+z*(y*x) =-y*(z*x)-x*(y*z)-z*(x*y) =-\eta(x,y,z) $$ and analoguesly $\eta(x,z,y)=\eta(z,y,x)=-\eta(x,y,z)$. Finally $$ \eta(e_0,e_1,e_2) =e_0*(-e_0)+e_1*e_1+e_2*e_2 =0 $$ i.e. $\eta=0$.