Equation Derivation Proof of Vector Triple Product

目标：最近在看论文,Some basic formula reasoning is required,Equations for triple products are often encountered.为了更深入的理解.So derive this formula.

定义：
向量三重积
$$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\cdot \overrightarrow{c})\cdot \overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\cdot \overrightarrow{c}$$

其中$\overrightarrow{a}=(a_0,a_1,...,a_n)$; $\overrightarrow{b}=(b_0,b_1,...,b_n)$;$\overrightarrow{c}=(c_0,c_1,...,c_n)$

General in space vector$n=3$

证明,It can be proved in two ways

第一种是最简单的方式,Expand the left and right items directly.
In general, the cross product can be converted to the product of a matrix and a vector.The transformation is shown below
$$Skew(a)=\left[\begin{array}{c}0-a_2{a}_1\\ a_{2}0-a_0\\ -a_1{a}_{0}0\end{array}\right]$$
So the formula on the left is ：
$$\begin{gathered} \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c}) \\ =skew(\overrightarrow{a})(skew(\overrightarrow{b})\overrightarrow{c}) \\ =\left[\begin{array}{c}0-a_2{a}_1\\ a_{2}0-a_0\\ -a_1{a}_{0}0\end{array}\right]\left[\begin{array}{c}0-b_2{b}_1\\ b_{2}0-b_0\\ -b_1{b}_{0}0\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\end{array}\right]=\left[\begin{array}{c}0-a_2{a}_1\\ a_{2}0-a_0\\ -a_1{a}_{0}0\end{array}\right]\left[\begin{array}{c}{b}_1{c}_2-b_2{c}_1\\ b_2{c}_0-b_0{c}_2\\ b_0{c}_1-b_1{c}_0\end{array}\right] \\ =\left[\begin{array}{c}-a_2(b_2{c}_0-b_0{c}_2)+a_1(b_0{c}_1-b_1{c}_0)\\ a_2(b_1{c}_2-b_2{c}_1)-a_0(b_0{c}_1-b_1{c}_0)\\ -a_1(b_1{c}_2-b_2{c}_1)+a_0(b_2{c}_0-b_0{c}_2)\end{array}\right]=\left[\begin{array}{c}(a_1{c}_1+a_2{c}_2)b_0-(a_1{b}_1+a_2{b}_2)c_0\\ (a_0{c}_0+a_2{c}_2)b_1-(a_0{b}_0+a_2{b}_2)c_1\\ (a_0{c}_0+a_1{c}_1)b_2-(a_0{b}_0+a_1{b}_1)c_2\end{array}\right] \\ =\left[\begin{array}{c}(a_0{c}_0+a_1{c}_1+a_2{c}_2)b_0-(a_0{b}_0+a_1{b}_1+a_2{b}_2)c_0\\ (a_0{c}_0+a_1{c}_1+a_2{c}_2)b_1-(a_0{b}_0+a_1{b}_1+a_2{b}_2)c_1\\ (a_0{c}_0+a_1{c}_1+a_2{c}_2)b_2-(a_0{b}_0+a_1{b}_1+a_2{b}_2)c_2\end{array}\right]=(\overrightarrow{a}\cdot \overrightarrow{c})\cdot \overrightarrow{b}-(\overrightarrow{a}\cdot \overrightarrow{b})\cdot \overrightarrow{c} \end{gathered}$$

证明完毕.

第二种方法: in a geometrically meaningful way.It has the heuristic kind.It is recommended to use this method to prove the equation for the triple product above.
To make it easier for everyone to understand,画出图像,得到下图

Because of the cross product,is the vertical vector of the two vectors.比如$\overrightarrow{b}\times \overrightarrow{c}$,it is perpendicular to$\overrightarrow{b},\overrightarrow{c}$The plane in which the two vectors of .
Arbitrary vectors at the same time$\overrightarrow{a}$和$\overrightarrow{b}\times \overrightarrow{c}$叉乘.The resulting vector must be parallel to $\overrightarrow{b},\overrightarrow{c}$The plane in which the two vectors of .(The red line segment represents).因此可以写成如下：
$$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=m\overrightarrow{b}+n\overrightarrow{c}$$
Because the red vector$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})$和$\overrightarrow{a}$垂直.因此：
$$\begin{gathered} \overrightarrow{a}\cdot (\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c}))=0 \\ =>\overrightarrow{a}\cdot (m\overrightarrow{b}+n\overrightarrow{c})=0 \\ =>m(\overrightarrow{a}\cdot \overrightarrow{b})+n(\overrightarrow{a}\cdot \overrightarrow{c})=0 \end{gathered}$$
To solve the above formula,We use construction,Construct two numbers such that the above formula holds.构造如下：
存在$p\in R$;且$m=p(\overrightarrow{a}\cdot \overrightarrow{c})$;$n=-p(\overrightarrow{a}\cdot \overrightarrow{b})$,Make the above formula hold constant.为了方便理解,The constructed terms are brought into the formula$m(\overrightarrow{a}\cdot \overrightarrow{b})+n(\overrightarrow{a}\cdot \overrightarrow{c})=0$
$$\begin{gathered} m(\overrightarrow{a}\cdot \overrightarrow{b})+n(\overrightarrow{a}\cdot \overrightarrow{c})=0 \\ =p(\overrightarrow{a}\cdot \overrightarrow{c})(\overrightarrow{a}\cdot \overrightarrow{b})-p(\overrightarrow{a}\cdot \overrightarrow{b})(\overrightarrow{a}\cdot \overrightarrow{c})=0 \end{gathered}$$
The substituted equation can see that the above is a middle identity.

The above equation and vector$\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$取值无关.
因此将$m=p(\overrightarrow{a}\cdot \overrightarrow{c})$;$n=-p(\overrightarrow{a}\cdot \overrightarrow{b})$,代入到$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=m\overrightarrow{b}+n\overrightarrow{c}$,得到如下公式：
$$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=m\overrightarrow{b}+n\overrightarrow{c}=p(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-p(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{c}$$

Because of the above formula and vector$\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$无关,Simple vectors can be used$\overrightarrow{a}=[1,1,1]$;$\overrightarrow{b}=[0,1,0]$;$\overrightarrow{c}=[0,0,1]$
Bring in the corresponding simple formula,就可以得到$p=1$
证明完毕.

参考资料如下：
https://www.youtube.com/watch?v=4U5fkwYDvZg