叉乘

几何图示：

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设有

$$\mathbf{a}=\left(a_x,a_y,a_z\right),\mathbf{b}=\left(b_x,b_y,b_z\right)$$

i，j，k分别是X，Y，Z轴方向的单位向量，则：

$$\mathbf{a}\times \mathbf{b}=\left(a_y{b}_z-a_z{b}_y\right)\mathbf{i}+\left(a_z{b}_x-a_x{b}_z\right)\mathbf{j}+\left(a_x{b}_y-a_y{b}_x\right)\mathbf{k}$$

为了帮助记忆，利用三阶行列式，写成

$$\left|\begin{array}{lll}i& j& k\\ a_x& a_y& a_z\\ b_x& b_y& b_z\end{array}\right|$$

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设a=(l,m,n),b=(o,p,q)，则

$$\mathbf{a}\times \mathbf{b}=(l,m,n)\times (o,p,q)=(mq-np,no-lq,lp-mo)$$

利用三阶行列式，写成

$$\left|\begin{array}{ccc}i& j& k\\ l& m& n\\ o& p& q\end{array}\right|$$

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推导

设有三个单位向量i，j，k，则：
$$\begin{array}{rl}& \overrightarrow{\mathrm{i}}=(1,0,0)\\ & \overrightarrow{\mathrm{j}}=(0,1,0)\\ & \overrightarrow{\mathrm{k}}=(0,0,1)\end{array}$$

且
$$\begin{gathered} \overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}\times \overrightarrow{\mathrm{k}} \\ \overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{k}}\times \overrightarrow{\mathrm{i}} \\ \overrightarrow{\mathrm{k}}=\overrightarrow{\mathrm{i}}\times \overrightarrow{\mathrm{j}} \\ \overrightarrow{\mathrm{k}}\times \overrightarrow{\mathrm{j}}=-\overrightarrow{\mathrm{i}} \\ \overrightarrow{\mathrm{i}}\times \overrightarrow{\mathrm{k}}=-\overrightarrow{\mathrm{j}} \\ \overrightarrow{\mathrm{j}}\times \overrightarrow{\mathrm{i}}=-\overrightarrow{\mathrm{k}} \\ \overrightarrow{\mathrm{i}}\times \overrightarrow{\mathrm{i}}=\overrightarrow{\mathrm{j}}\times \overrightarrow{\mathrm{j}}=\overrightarrow{\mathrm{k}}\times \overrightarrow{\mathrm{k}}=\overrightarrow{0}=0 \end{gathered}$$

i，j，k是三个相互垂直的向量。它们刚好可以构成一个坐标系。
对于处于i，j，k构成的坐标系中的向量u，v我们可以如下表示：
$$\begin{array}{rl}& \overrightarrow{\mathrm{u}}=\left({\mathrm{x}}_{\mathrm{u}},{\mathrm{y}}_{\mathrm{u}},{\mathrm{z}}_{\mathrm{u}}\right)\\ & \overrightarrow{\mathrm{v}}=\left({\mathrm{x}}_{\mathrm{v}},{\mathrm{y}}_{\mathrm{v}},{\mathrm{z}}_{\mathrm{v}}\right)\end{array}$$

$$\begin{array}{rl}& \overrightarrow{\mathrm{u}}={\mathrm{X}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{i}}+{\mathrm{y}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{j}}+{\mathrm{z}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{k}}\\ & \overrightarrow{\mathrm{v}}={\mathrm{X}}_{\mathrm{v}}\cdot \overrightarrow{\mathrm{i}}+{\mathrm{y}}_{\mathrm{v}}\cdot \overrightarrow{\mathrm{j}}+{\mathrm{z}}_{\mathrm{v}}\cdot \overrightarrow{\mathrm{k}}\end{array}$$

$$\overrightarrow{\mathrm{u}}\times \overrightarrow{\mathrm{v}}=\left({\mathrm{x}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{i}}+{\mathrm{y}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{j}}+{\mathrm{z}}_{\mathrm{u}}\cdot \overrightarrow{\mathrm{k}}\right)\times \left({\mathrm{x}}_{\mathrm{v}}\cdot \overrightarrow{\mathrm{i}}+{\mathrm{y}}_{\mathrm{v}}\overrightarrow{\mathrm{j}}+{\mathrm{z}}_{\mathrm{v}}\cdot \overrightarrow{\mathrm{k}}\right)$$

展开，得：

$$\begin{array}{rl}\overrightarrow{\mathrm{u}}\times \overrightarrow{\mathrm{v}}=& {\mathrm{x}}_{\mathrm{u}}\cdot {\mathrm{x}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{i}}\cdot \overrightarrow{\mathrm{i}})+{\mathrm{x}}_{\mathrm{u}}\cdot {\mathrm{y}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{i}}\cdot \overrightarrow{\mathrm{j}})+{\mathrm{x}}_{\mathrm{u}}\cdot {\mathrm{z}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{i}}\cdot \overrightarrow{\mathrm{k}})\\ & +{\mathrm{y}}_{\mathrm{u}}\cdot {\mathrm{x}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{j}}\cdot \overrightarrow{\mathrm{i}})+{\mathrm{y}}_{\mathrm{u}}\cdot {\mathrm{y}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{j}}\cdot \overrightarrow{\mathrm{j}})+{\mathrm{y}}_{\mathrm{u}}\cdot {\mathrm{z}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{j}}\cdot \overrightarrow{\mathrm{k}})\\ & +{\mathrm{z}}_{\mathrm{u}}\cdot {\mathrm{x}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{k}}\cdot \overrightarrow{\mathrm{i}})+{\mathrm{z}}_{\mathrm{u}}\cdot {\mathrm{y}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{k}}\cdot \overrightarrow{\mathrm{j}})+{\mathrm{z}}_{\mathrm{u}}\cdot {\mathrm{z}}_{\mathrm{v}}\cdot (\overrightarrow{\mathrm{k}}\cdot \overrightarrow{\mathrm{k}})\end{array}$$

化简，得：

$$\begin{array}{c}\vec{u}\times \vec{v}=\overrightarrow{0}+x_u\cdot y_v\cdot (\vec{k})+x_u\cdot z_v\cdot (-\vec{j})\\ +y_u\cdot x_v\cdot (-\vec{k})+\overrightarrow{0}+y_u\cdot z_v\cdot (\vec{i})\\ +z_u\cdot x_V\cdot (\vec{j})+z_u\cdot y_v\cdot (-\vec{i})+\overrightarrow{0}\end{array}$$

合并，得：

$$\overrightarrow{\mathrm{u}}\times \overrightarrow{\mathrm{v}}=\left({\mathrm{y}}_{\mathrm{u}}\cdot {\mathrm{z}}_{\mathrm{v}}-{\mathrm{z}}_{\mathrm{u}}\cdot {\mathrm{y}}_{\mathrm{v}}\right)\cdot (\overrightarrow{\mathrm{i}})+\left({\mathrm{z}}_{\mathrm{u}}\cdot {\mathrm{x}}_{\mathrm{v}}-{\mathrm{x}}_{\mathrm{u}}\cdot {\mathrm{z}}_{\mathrm{v}}\right)\cdot (\overrightarrow{\mathrm{j}})+\left({\mathrm{x}}_{\mathrm{u}}\cdot {\mathrm{y}}_{\mathrm{v}}-{\mathrm{y}}_{\mathrm{u}}\cdot {\mathrm{x}}_{\mathrm{v}}\right)\cdot (\overrightarrow{\mathrm{k}})$$

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叉乘的几何意义

设
$$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$$

则c的方向垂直于a与b所决定的平面，c的指向按右手定则从a转向b来确定。
且c的长度在数值上等于以a，b，夹角为θ组成的平行四边形的面积。
$$|c|=|a||b|\cdot \sin \theta$$

右手定则：
若坐标系是满足右手定则的，当右手的四指从a以不超过180度的转角转向b时，竖起的大拇指指向是c的方向。

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代数规则

• 反交换律
  $$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$$

• 加法的分配律

$$\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$$

• 与标量乘法兼容
  $$r\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}}\times r\overrightarrow{\mathrm{b}}=r\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})$$

• 不满足结合律，但满足雅可比恒等式
  $$\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+\overrightarrow{\mathrm{b}}\times (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})+\overrightarrow{\mathrm{c}}\times (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})=0$$

• 分配律，线性性和雅可比恒等式别表明：具有向量加法和叉积的R3构成了一个李代数。

• 两个非零向量a和b平行，当且仅当a×b=0。

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拉格朗日公式

设向量坐标 $\vec{a}=(x1,y1,z1),\vec{b}=(x2,y2,z2),\vec{c}=(x3,y3,z3)$, 则
$$\begin{array}{rl}& (\vec{a}\times \vec{b})\times \vec{c}=\left(y_1{z}_2-z_1{y}_2,z_1{x}_2-x_1{z}_2,x_1{y}_2-x_2{y}_1\right)\times \left(x_3,y_3,z_3\right)\\ & =(z_1{z}_3{x}_2-x_1{z}_2{z}_3-x_1{y}_2{y}_3+x_2{y}_1{y}_3,x_1{x}_3{y}_2-x_2{x}_3{y}_1\\ & -y_1{z}_2{z}_3+z_1{z}_3{y}_2,y_1{y}_3{z}_2-y_2{y}_3{z}_1-x_2{x}_3{z}_1+x_1{x}_3{z}_2)\\ & =\left(x_2\left(x_1{x}_3+y_1{y}_3+z_1{z}_3\right),y_2\left(x_1{x}_3+y_1{y}_3+z_1{z}_3\right),z_2\left(x_1{x}_3+y_1{y}_3+z_1{z}_3\right)\right)\\ & -\left(x_1\left(x_2{x}_3+y_2{y}_3+z_2{z}_3\right),y_1\left(x_2{x}_3+y_2{y}_3+z_2{z}_3\right),z_1\left(x_2{x}_3+y_2{y}_3+z_2{z}_3\right)\right)\\ & =\left(x_2(\vec{a}\cdot \vec{c}),y_2(\vec{a}\cdot \vec{c}),z_2(\vec{a}\cdot \vec{c})\right)-\left(x_1(\vec{b}\cdot \vec{c}),y_1(\vec{b}\cdot \vec{c}),z_1(\vec{b}\cdot \vec{c})\right)\\ & =\vec{b}(\vec{a}\cdot \vec{c})-\vec{a}(\vec{b}\cdot \vec{c})\end{array}$$
这就是二重向量叉乘化简公式 $(\vec{a}\times \vec{b})\times \vec{c}=\vec{b}(\vec{a}\cdot \vec{c})-\vec{a}(\vec{b}\cdot \vec{c})$

转化一下，得
$$\vec{a}\times (\vec{b}\times \vec{c})=\vec{b}(\vec{a}\cdot \vec{c})-\vec{c}(\vec{a}\cdot \vec{b})$$

向量积（矢积）与数量积（标积）的区别

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