This has been bugging me for years so I finally decided to "derive" (for lack of a better term) the definition of the cross product in $\mathbb R{^3}$. Here was my method for finding a vector: $\mathbf w = \mathbf u \times \mathbf v$ such that $\mathbf w \cdot \mathbf u = \mathbf w \cdot \mathbf v = 0$, where $\mathbf u = [$$ \begin{matrix} a & b & c \\ \end{matrix}$$ ]$ and $\mathbf v = [$$ \begin{matrix} d & e & f \\ \end{matrix}$$ ]$. This of course shows orthogonality between $\mathbf w$ and $\mathbf u$, as well as $\mathbf v$. I set up the 2x3 matrix to solve for $\mathbf w = [$$ \begin{matrix} w_1 & w_2 & w_3 \\ \end{matrix}$$ ]$ as follows: $ $$ \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$$ \cdot \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}$$ = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ $ Of course this is 3 unknowns and 2 equations, so I knew there would have to be an arbitrary parameter. I was fine with this for the time being and after some dirty work, ended up with the following:

$$\begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} = t \begin{bmatrix} \frac{\begin{vmatrix}b & c \\ e & f\end{vmatrix}}{\begin{vmatrix}a & b \\ d & e\end{vmatrix}} \\ -\frac{\begin{vmatrix}a & c \\ d & f\end{vmatrix}}{\begin{vmatrix}a & b \\ d & e\end{vmatrix}} \\ 1 \end{bmatrix}$$

This looked very much like the "traditional" definition of the cross product, so I chose $t = \begin{vmatrix}a & b \\ d & e\end{vmatrix}$ and I finally ended up with $\mathbf w = $$\begin{pmatrix} \begin{vmatrix}b & c \\ e & f\end{vmatrix}\\-{\begin{vmatrix}a & c \\ d & f\end{vmatrix}} \\ \begin{vmatrix}a & b \\ d & e\end{vmatrix} \end{pmatrix}$$ $ which is the definition of the cross product that I've seen in pretty much all of my calculus and physics texts (also shown in determinant form with unit vectors). But where does that value for $t$ come from? Why does that particular value of $t$ work, besides my hunch to make it look like a definition that is universally accepted? Is the rationale behind $t$ being negative for $\mathbf w = \mathbf v \times \mathbf u$ just to satisfy the right-hand-rule?

Sorry if anything is messed up, this is my first ever time using MathJax.

By the way, I've checked similar questions which ask for the rationale for the cross product *existing* which I have learned from studying electromagnetics myself. But I wanted to see the rational behind the length of the vector, hence my value for $t$. Thanks for any help you can offer!