Prove the algebraic definition of cross product

Question

My textbook says that

Given $$\vec{u}=\langle a_1,a_2,a_3\rangle,\space \vec{v}=\langle b_1,b_2,b_3\rangle$$ We need to find vector $\vec{w}=\langle x,y,z\rangle$ that is perpendicular to both vectors $\vec{u}$ and $\vec{v}$. Then $\vec{u}\cdot\vec{w}=0$ and $\vec{v}\cdot\vec{w}=0$. Then we have the following simultaneous equations. $$\begin{cases}a_1x+a_2y+a_3z=0\\b_1x+b_2y+b_3z=0\end{cases}$$ Then by solving we can reach the final solution $$\vec{w}=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle$$

Now, I wonder how did he reach the final result with just two simultaneous equations involving 3 variables

Answer 1

The system of equations has infinitely many solutions as one should expect, since the orthogonal space $<\vec{u},\vec{v}>^{\perp}$ is one dimensional, the cross product is the unique vector in this orthogonal complement that has as norm $||\vec{u}||||\vec{v}||\sin(\angle(\vec{u},\vec{v}))$ which is the area of the parallelogram spanned by these two vectors and that fullfills the right hand rule. This makes sense for example when desribing a torque:https://en.wikipedia.org/wiki/Torque

Answer 2

The solution you write is a particular solution of the system of equation written in your post. Any vector colinear to $\vec{w}$ will also be a solution of the equation system (if $\vec{v}$ is a solution of the system then $\lambda \vec{v}$ is a solution as well).

Answer 3

You can move the $z$ terms to the right side and solve for $x$ and $y$ in terms of $z$: $$\begin{cases}a_1x+a_2y+a_3z=0\\b_1x+b_2y+b_3z=0\end{cases} \implies \begin{cases}a_1x+a_2y=-a_3z\\b_1x+b_2y=-b_3z\end{cases} \\\implies \begin{pmatrix} a_1&a_2\\b_1&b_2 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} =-\begin{pmatrix} a_3\\b_3 \end{pmatrix}z.$$ If $a_1 b_2 -a_2 b_1 \neq 0$, then $$\begin{align} \begin{pmatrix} x\\y \end{pmatrix} &=-\frac{1}{a_1 b_2 -a_2 b_1} \begin{pmatrix} b_2&-a_2\\-b_1&a_1 \end{pmatrix} \begin{pmatrix} a_3\\b_3 \end{pmatrix}z\\ &=\frac{1}{a_1 b_2 -a_2 b_1} \begin{pmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3 \end{pmatrix} z.\\ \end{align}.$$ The system of equations has then an infinite number of solutions, namely one for every choice of $z$. Choosing $z=a_1 b_2 -a_2 b_1$ gives $x=a_2b_3-a_3b_2$ and $y=a_3b_1-a_1b_3$.