The Wikipedia article https://en.wikipedia.org/wiki/Bivector#Simple_bivectors states that
"A bivector that can be written as the exterior product of two vectors is simple. In two and three dimensions all bivectors are simple."
This implies that if we have a bivector in 3D given by $$B :=\alpha(e_1\wedge e_2) + \beta(e_2\wedge e_3) + \gamma(e_3\wedge e_1)$$
then there are the two vectors $v, w$ such that $B = v \wedge w$.
I am wondering how we can construct such $v, w$ from $\alpha, \beta, \gamma$? I understand that there is not a unique choice of $v$ and $w$ however is there a 'nicest' choice? One that is the most symmetric?
I attempted the following manipulation on the definition of $B$:
$$\begin{eqnarray}B &=&\alpha(e_1\wedge e_2) + \beta(e_2\wedge e_3) + \gamma(e_3\wedge e_1) \\ &=&\alpha(e_1\wedge e_2) – \beta(e_3\wedge e_2) – \gamma(e_1\wedge e_3) \\ &=& (\alpha e_1 – \beta e_3)\wedge e_2 – \left(\frac{\gamma}{\alpha}\right)(\alpha e_1 \wedge e_3) + \frac{\beta\gamma}{\alpha}(e_3\wedge e_3)\\ &=& (\alpha e_1 – \beta e_3)\wedge e_2 – (\alpha e_1) \wedge\left(\frac{\gamma}{\alpha}e_3\right) + (\beta e_3)\wedge\left(\frac{\gamma}{\alpha} e_3\right) \\ &=& (\alpha e_1 – \beta e_3)\wedge e_2 – (\alpha e_1 – \beta e_3) \wedge\left(\frac{\gamma}{\alpha}e_3\right) \\ &=& (\alpha e_1 – \beta e_3)\wedge \left(e_2 – \frac{\gamma}{\alpha}e_3\right)
\end{eqnarray}$$
This shows that the decomposition of a 3D bivector into the wedge product of two vectors is possible (and thus that every 3D bivector is simple) however it is not a very satisfying end result in that it is not symmetric and doesn't offer any insight into the nature of the decomposition.
For example, a nicer decomposition would be one of the form
$$(ae_1 + be_2 + ce_3)\wedge (a'e_1 + b'e_2 + c'e_3)$$
in which there is some symmetry in the coefficients $a, b, c, a', b', c'$.
Best Answer
There's a difficulty in your solution in that it fails when $\alpha=0$, and you need to treat this case separately.
At least over $\Bbb R$, your problem is essentially the same as the following: given a vector $u$ in $\Bbb R^3$ write it as a vector product $v\times w$. One can get such a representation by taking $v$ to be any nonzero vector orthogonal to $u$. Then one has a one-dimensional family of allowable $w$. If $u$ and $v$ are chosen with unit length (there's no real loss in generality by doing this) then $w=u\times v+\lambda v$.
[In your solution you are essentially arguing that $(b,-a,0)$ is a vector orthogonal to $(a,b,c)$.]
The problem boils down to finding a unit vector orthogonal to a given unit vector. Whether a proposed solution satisfies your aesthetic criterion of "symmetry" is up to you to decide. I would say that given $(a,b,c)$ at least one of $(a^2+b^2)^{-1/2}(b,-a,0)$, $(a^2+c^2)^{-1/2}(c,0,-a)$ and $(b^2+c^2)^{-1/2}(0,c,-b)$ is well-defined and works.
But whether or not there is a solution that meets your aesthetic criteria, there is no continuous solution. More precisely there is no continuous map $F:S^2\to S^2$ (where $S^2$ is the unit sphere) with $F(u)$ always orthogonal to $u$. Such an $F$ would define a non-zero section of the tangent bundle on $S^2$ and so fall foul of the "hairy ball" theorem of algebraic topology.