Proof of the definition of vector cross product

A proof of the definition of the vector dot product was written in front of it, because the proof is relatively simple, so it does not cause deep thinking. Later, when I was going to write a proof of cross-product, I found something really difficult to understand.

Set two vectors $\mathbf{a} = (x_1, y_1, z_1), \mathbf{b} = (X_2, y_2, Z_2) $, the two vectors have an angle of $\theta$, and many textbooks include Wikipedia (crossProduct) The definitions given are:

$$\mathbf{c} = \mathbf{a} \times \mathbf{b} = \mathbf{n} |\mathbf{a}| |\mathbf{b}| \sin{\theta}$$

Where $\mathbf{n}$ is a unit vector perpendicular to the vector $\mathbf{a},\mathbf{b}$, the direction is determined by the right hand rule. There seems to be nothing wrong with this definition, but I'm thinking about the question: How did he find out that the result of the cross product was perpendicular to the two vectors, given the definition of the mathematician? Why is the modulus of a vector exactly equal to $|\mathbf{a}| |\mathbf{b}| \sin{\theta}$? Here's my understanding of these issues.

I think mathematicians are just beginning to define the vector's fork multiplication ($\times$), the only basic definition given is: $\mathbf{a} \times \mathbf{b}$ result $\mathbf{c}$ is perpendicular to the vector $\mathbf{a},\mathbf{ A vector of b}$, whose direction is determined by the right-hand rule, and if the vector $\mathbf{a},\mathbf{b}$ parallel, the cross product results in a zero vector. With this definition, the coordinate expression of the cross product operation can be obtained according to the ratio of multiplication to addition:

 \begin{align} \mathbf{a} \times \mathbf{b} = & (X_1\mathbf{i} + y_1\mathbf{j} + z_1\mathbf{k}) \ Times (X_2\mathbf{i} + y_2\mathbf{j} + z_2\mathbf{k}) \ = &\ x_1\mathbf{i} \times  (X_2\mathbf{i} + Y_2\mat HBF{J} + z_2\mathbf{k}) + y_1\mathbf{j} \times  (X_2\mathbf{i} + y_2\mathbf{j} + z_2\mathbf{k}) + Z_1\mathbf{k} \times   (X_2\mathbf{i} + y_2\mathbf{j} + z_2\mathbf{k}) \ = &\ x_1 x_2 (\mathbf{i} \times \mathbf{i}) + x_1 y _2 (\mathbf{i} \times \mathbf{j}) + x_1 z_2 (\mathbf{i} \times \mathbf{k}) + \ &\ y_1 x_2 (\mathbf{j} \times \mathbf{i}) + Y_1 y_2 (\mathbf{j} \times \mathbf{j}) + Y_1 z_2 (\mathbf{j} \times \mathbf{k}) + \ \ & Amp;\ z_1 x_2 (\mathbf{k} \times \mathbf{i}) + Z_1 y_2 (\mathbf{k} \times \mathbf{j}) + Z_1 z_2 (\mathbf{k} \tim Es \mathbf{k}) \end{align}

Where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ represent the unit vectors of the x, y, and z axes respectively. So according to the definition of the vector cross product: $\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}$,$\ma Thbf{i} \times \mathbf{j} = \mathbf{k}$,$\mathbf{j} \times \mathbf{k} = \mathbf{i}$,$\mathbf{k} \times \mathbf{i} = \MATHB F{J}$,$\MATHBF{J} \times \mathbf{i} =-\mathbf{k}$,$\mathbf{k} \times \mathbf{j} =-\mathbf{i}$,$\mathbf{i} \times \ Mathbf{k} =-\mathbf{j}$, so you get:

\begin{align} \mathbf{a} \times \mathbf{b} &= (y_1 z_2-z_1 y_2) \mathbf{i} + (Z_1 x_2-x_1 z_2) \mathbf{j} + (X_1 y _2-y_1 x_2) \mathbf{k} \ &= (y_1 z_2-z_1 y_2, \ z_1 x_2-x_1 z_2, \ x_1 y_2-y_1 x_2) \end{align}

below to prove $|\mathbf{c}| = |\mathbf{a}| |\mathbf{b}| \sin{\theta}$:

\begin{align} |\mathbf{c}|^2 = &\ (y_1 z_2-z_1 y_2) ^2 + (z_1 x_2-x_1 z_2) ^2 + (x_1 y_2-y_1 x_2) ^2 \ = &\ Y_ 1^2 z_2^2 + z_1^2 y_2^2-2y_1 y_2 z_1 z_2 + z_1^2 x_2^2 + x_1^2 z_2^2-2x_1 x_2 z_1 z_2 + \ &\ x_1^2 y_2^2 + y_1^2 X_2^2-2x_1 x_2 y_1 y_2 \end{align}

It is also based on the definition of the vector dot product:

\begin{align} (|\mathbf{a}| |\mathbf{b}| \sin{\theta}) ^2 &= (|\mathbf{a}| |\mathbf{b}|) ^2 \sin^{2}{\theta} \ &= (|\mathbf{a}| |\mathbf{b}|) ^2 (1-\cos^{2}{\theta}) \ &= (|\mathbf{a}| |\mathbf{b}|) ^2 \left (1-\frac{(\mathbf{a} \cdot \mathbf{b}) ^2}{(|\mathbf{a}| |\mathbf{b}|) ^2} \right) \ &= (|\mathbf{a}| |\mathbf{b}|) ^2-(\mathbf{a} \cdot \mathbf{b}) ^2 \end{align}

Because:

\begin{align} (|\mathbf{a}| |\mathbf{b}|)  ^2 = &\ \left (\sqrt{x_1^2 + y_1^2 + z_1^2} \sqrt{x_2^2 + y_2^2 + z_2^2} \right) ^2 \ \ = &\ (x_1^2 + y_1^2 + z_1^2) (x_2^2 + y_2^2 + z_2^2) \ \ = &\ x_1^2 x_2^2 + y_1^2 y_2^2 + z_1^2 z_2^2 + x_1^2 y_2^2 + x_1^2 z_2^2 + y_1^2 x_2^2 + y _1^2 z_2^2 + z_1^2 x_2^2 + z_1^2 y_2^2 \end{align}

And

\begin{align} (\mathbf{a} \cdot \mathbf{b}) ^2 = &\ (x_1 x_2 + y_1 y_2 + z_1 z_2) ^2 \ \ &\ x_1^2 x_2^2 + y_1^2 y_2 ^2 + z_1^2 z_2^2 + 2x_1 x_2 y_1 y_2 + 2x_1 x_2 z_1 z_2 + 2y_1 y_2 z_1 z_2 \end{align}

Easy to see: $ (|\mathbf{a}| |\mathbf{b}|) ^2-(\mathbf{a} \cdot \mathbf{b}) ^2 = |\mathbf{c}|^2$, i.e.:

$$|\mathbf{c}| = |\mathbf{a}| |\mathbf{b}| \sin{\theta}$$

Proof of the definition of vector cross product