The Geometric Meaning of Vector Cross Product and the Calculation of its Modulus

Purpose: In the traditional vector cross product calculation, cross product is often encountered. Defined as a vector. The direction of this vector satisfies the right-hand rule. Its modulus size is generally ignored. So speculate.

Vector cross product definition:
Outer product (English: Cross product ), also known as vector product (English: Vector product t), is a binary operation on two vectors in three-dimensional space, with the symbol: $\times$ means. can be defined as:
$$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\left(1\right)$$
Suppose two vectors$\overrightarrow{a}\times \overrightarrow{b}$Outer product, its direction is $\overrightarrow{c}$. Its direction is determined by the right-hand rule. The modulus length is equal to the area of ​​the parallelogram of these two vector sides.
Its definition can also be written as:
$$\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|sin\left(\theta \right)\overrightarrow{n}\left(2\right)$$
θ \$\theta$ is the angle between two vectors$0\le i\le 180$；$|\overrightarrow{a}||\overrightarrow{b}|Respectively$ , two vectors$\overrightarrow{a}\overrightarrow{b}$The mold length. $\overrightarrow{n}$is perpendicular to $\overrightarrow{a}\overrightarrow{b}$The normal vector of the plane where it satisfies the right-hand rule. As shown in the figure below:

the above definition is easy to understand. But generally in algebra to calculate the cross product of two vectors, determinant calculation will be used. As a set of unit products $(\overrightarrow{i},\overrightarrow{j},\overrightarrow{k})$;其中$\overrightarrow{a}=a_0\overrightarrow{i}+a_1\overrightarrow{j}+a_2\overrightarrow{k}$;$\overrightarrow{b}=b_0\overrightarrow{i}+b_1\overrightarrow{j}+b_2\overrightarrow{k}$
When calculating the cross product of two vectors, the general algebraic method is:
$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=(a_0\overrightarrow{i}+a_1\overrightarrow{j}+a_2\overrightarrow{k})\times (\overrightarrow{b}=b_0\overrightarrow{i}+b_1\overrightarrow{j}+b_2\overrightarrow{k}) \\ =a_0{b}_0(\overrightarrow{i}\times \overrightarrow{i})+a_0{b}_1(\overrightarrow{i}\times \overrightarrow{j})+a_0{b}_2(\overrightarrow{i}\times \overrightarrow{k})+ \\ {a}_1{b}_0(\overrightarrow{j}\times \overrightarrow{i})+a_1{b}_1(\overrightarrow{j}\times \overrightarrow{j})+a_1{b}_2(\overrightarrow{j}\times \overrightarrow{k})+ \\ {a}_2{b}_0(\overrightarrow{k}\times \overrightarrow{i})+a_2{b}_1(\overrightarrow{k}\times \overrightarrow{j})+a_2{b}_2(\overrightarrow{k}\times \overrightarrow{k})(3) \end{gathered}$$

Because the basis vector $(\overrightarrow{i},\overrightarrow{j},\overrightarrow{k})$ are perpendicular to each other and are unit vectors. $\overrightarrow{0}$means all are $0$的向量。所以得到：
$$\begin{gathered} \overrightarrow{i}\times \overrightarrow{i}=\overrightarrow{0}(4) \\ \overrightarrow{j}\times \overrightarrow{j}=\overrightarrow{0}(5) \\ \overrightarrow{k}\times \overrightarrow{k}=\overrightarrow{0}(6) \\ \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k}(7) \\ \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i}(8) \\ \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j}\left(9\right) \end{gathered}$$
will$\left(4\right)\left(5\right)\left(6\right)\left(7\right)\left(8\right)\left(9\right)$ into formula$(3)$得到如下：
$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=-a_0{b}_0\overrightarrow{0}+a_0{b}_1\overrightarrow{k}-a_0{b}_2\overrightarrow{j} \\ -a_1{b}_0\overrightarrow{k}-a_1{b}_1\overrightarrow{0}+a_1{b}_2\overrightarrow{i} \\ +a_2{b}_0\overrightarrow{j}-a_2{b}_1\overrightarrow{i}-a_2{b}_2\overrightarrow{0} \\ =(a_1{b}_2-a_2{b}_1)\overrightarrow{i}+(a_2{b}_0-a_0{b}_2)\overrightarrow{j}+(a_0{b}_1-a_1{b}_0)\overrightarrow{k}(10) \end{gathered}$$

Formula $\left(10\right)$ is expressed by determinant calculation in daily life. Use$(\overrightarrow{i},\overrightarrow{j},\overrightarrow{k})$ matrixremainder formulacalculation method. It is equivalent to algebraic computation.
$$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ a_0& a_1& a_2\\ b_0& b_1& b_2\end{array}\right]=(a_1{b}_2-a_2{b}_1)\overrightarrow{i}+(a_2{b}_0-a_0{b}_2)\overrightarrow{j}+(a_0{b}_1-a_1{b}_0)\overrightarrow{k}$$

Because it is a base vector, in Euclidean geometry, its expression is:
$$\overrightarrow{i}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right];\overrightarrow{j}=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right];\overrightarrow{k}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\left(11\right)$$
so$\left(11\right)$ into$(10)$得到：
$$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{array}{c}{a}_1{b}_2-a_2{b}_1\\ a_2{b}_0-a_0{b}_2\\ a_0{b}_1-a_1{b}_0\end{array}\right](12)$$

The above is an expression based on the basis vector, which corresponds to the above formula, so it can be obtained:
$$\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|sin\left(\theta \right)\overrightarrow{n}=\left[\begin{array}{c}{a}_1{b}_2-a_2{b}_1\\ a_2{b}_0-a_0{b}_2\\ a_0{b}_1-a_1{b}_0\end{array}\right](13)$$

In some applications, often vector representations are transformed into matrix operations. So (13) formula can represent the multiplication of matrix and vector.
$$\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|sin\left(\theta \right)\overrightarrow{n}=\left[\begin{array}{c}{a}_1{b}_2-a_2{b}_1\\ a_2{b}_0-a_0{b}_2\\ a_0{b}_1-a_1{b}_0\end{array}\right]=\left[\begin{array}{ccc}0& -a_2& a_1\\ a_2& 0& -a_0\\ -a_1& a_0& 0\end{array}\right]\left[\begin{array}{c}{b}_0\\ b_1\\ b_2\end{array}\right]=\overrightarrow{a}\times \overrightarrow{b}(14)$$

The cross product of two vectors is only defined in three dimensions. Not defined in two dimensions.

The following introduces the relationship between the determinant of a vector and the area of ​​a parallelogram formed by a vector.
Suppose $\overrightarrow{a},\overrightarrow{b}$is a two-dimensional vector. That's easy to explain. So the plot looks like this:

Calculate the triangle area as:
$$|area|=\frac{1}{2}|\overrightarrow{a}||\overrightarrow{b}|sin\left(\theta \right)\left(15\right)$$

Transform the expression, because $sin\left(\theta \right)$ is not easy to calculate, you need to calculate$cos\left(\theta \right)$ .

其中$|{\overrightarrow{a}}^{\prime }|=|\overrightarrow{a}|$;$|\overrightarrow{b}|sin\left(\theta \right)=|{\overrightarrow{b}}^{\prime }|cos(i^{\prime })$;

$$|area|=\frac{1}{2}|\overrightarrow{a}||\overrightarrow{b}|sin\left(\theta \right)=\frac{1}{2}|\overrightarrow{b}||\overrightarrow{a}|cos\left(i^{\prime }\right)\left(16\right)$$
in which$i^{\prime }+i=90$.且$|{\overrightarrow{a}}^{\prime }|=|\overrightarrow{a}|$ , it is easy to get the simplification of the formula. Simplifying the above equation is:
$$|area|=\frac{1}{2}|\overrightarrow{b}||{\overrightarrow{a}}^{\prime }|cos(i^{\prime })=\frac{1}{2}\overrightarrow{b}\cdot {\overrightarrow{a}}^{\prime }=\frac{1}{2}{\overrightarrow{a}}^{\prime }\cdot \overrightarrow{b}(17)$$

because ${\overrightarrow{a}}^{\prime }$ is via$\overrightarrow{a}$Rotate 90 degrees to get, as shown below.

So suppose $\overrightarrow{a}=\left[\begin{array}{c}{a}_0\\ a_1\end{array}\right]$ 得到${\overrightarrow{a}}^{\prime }=\left[\begin{array}{c}-a_1\\ a_0\end{array}\right]$

因此得到公式：
$$2|area|={\overrightarrow{a}}^{\prime }\cdot \overrightarrow{b}=\left[\begin{array}{c}-a_1\\ a_0\end{array}\right]\cdot \left[\begin{array}{c}{b}_0\\ b_1\end{array}\right]=a_0{b}_1-a_1{b}_0(18)$$

It can be seen that the determinant is an expression of area.
$$2|area|=\left|\begin{array}{cc}{a}_0& a_1\\ b_0& b_1\end{array}\right|$$