【解析几何笔记】10.向量的外积

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10. 向量的外积

10.1 向量外积的定义

$\alpha \times \beta$是一个向量，$\alpha \times \beta$垂直于$\alpha$和$\beta$所在的平面，至于朝上还是朝下，取决于$\alpha ,\beta ,\alpha \times \beta$构成右手系。（用右手，四指朝向和$\alpha$一致，然后朝向$\beta$四指握紧，如果拇指向上和$\beta$就是$\alpha \times \beta$朝上，否则$\alpha \times \beta$朝下。）
$|\alpha \times \beta |=|\alpha ||\beta |\sin <\alpha ,\beta >$（恰好是$\alpha$与$\beta$所张成的平行四边形面积）

10.2 向量外积的性质

• $\alpha \times \beta =\vec{0}\iff \alpha //\beta$
• $\alpha \times \beta =-\beta \times \alpha$（反交换性）
• 假定$\alpha$是单位向量，$\beta \perp \alpha$，此时$|\beta \times \alpha |=|\beta |$（直角正弦为1）
  故$\beta \times \alpha$可看作$\beta$绕$\alpha$旋转$90^{\circ }$而得
  若$\alpha$不是单位向量，此时$|\beta \times \alpha |=|\alpha ||\beta |$（直角条件没变，正弦值还是1）
  记${\alpha }_0=\frac{\alpha }{|\alpha |}$
  所以$\alpha \times \beta =|\alpha |{\alpha }_0\times \beta$
  若$\beta$不与$\alpha$垂直，此时$\alpha \times \beta$可看作${\bar{P}}_{\alpha }\beta$绕$\alpha$旋转$90^{\circ }$再乘上$|\alpha |$
  
  【定理】如果$\alpha \ne \vec{0},\alpha \times \beta$可以看成${\bar{P}}_{\alpha }\beta$绕$\alpha$旋转$90^{\circ }$再乘上$|\alpha |$
  
  
  
  
  
  
  
• （双线性性）
  $\alpha \times (\lambda \beta )=\lambda (\alpha \times \beta )$
  $\alpha \times (\beta +\gamma )=\alpha \times \beta +\alpha \times \gamma$
  【证】（1）$\alpha \times (\lambda \beta )=|\alpha |{\alpha }_0\times (\lambda \beta )=|\alpha |{\alpha }_0\times {\bar{P}}_{\alpha }(\lambda \beta )=|\alpha |{\alpha }_0\times \lambda {\bar{P}}_{\alpha }\beta =\lambda |\alpha |{\alpha }_0\times {\bar{P}}_{\alpha }\beta =\lambda (\alpha \times \beta )$
  （2）$\alpha \times (\beta +\gamma )=|\alpha |{\alpha }_0\times {\bar{P}}_{\alpha }(\beta +\gamma )=|\alpha |{\alpha }_0\times ({\bar{P}}_{\alpha }\beta +{\bar{P}}_{\alpha }\gamma )=|\alpha |({\alpha }_0\times {\bar{P}}_{\alpha }\beta +{\alpha }_0\times {\bar{P}}_{\alpha }\gamma )=|\alpha |{\alpha }_0\times {\bar{P}}_{\alpha }\beta +|\alpha |{\alpha }_0\times {\bar{P}}_{\alpha }\gamma =\alpha \times \beta +\alpha \times \gamma$
  
  
  
  

10.3 用坐标计算外积

假设有空间仿射坐标系$[O:e_1,e_2,e_3],\alpha =a_1{e}_1+a_2{e}_2+a_3{e}_3,\beta =b_1{e}_1+b_2{e}_2+b_3{e}_3$
则$\alpha \times \beta =(a_1{e}_1+a_2{e}_2+a_3{e}_3)\times (b_1{e}_1+b_2{e}_2+b_3{e}_3)=a_1{b}_1{e}_1\times e_1+a_1{b}_2{e}_1\times e_2+a_1{b}_3{e}_1\times e_3+a_2{b}_1{e}_2\times e_1+a_2{b}_2{e}_2\times e_2+a_2{b}_3{e}_2\times e_3+a_3{b}_1{e}_3\times e_1+a_3{b}_2{e}_3\times e_2+a_3{b}_3{e}_3\times e_3=a_1{b}_2{e}_1\times e_2+a_1{b}_3{e}_1\times e_3+a_2{b}_1{e}_2\times e_1+a_2{b}_3{e}_2\times e_3+a_3{b}_1{e}_3\times e_1+a_3{b}_2{e}_3\times e_2=a_1{b}_2{e}_1\times e_2+a_1{b}_3{e}_1\times e_3-a_2{b}_1{e}_1\times e_2+a_2{b}_3{e}_2\times e_3-a_3{b}_1{e}_1\times e_3-a_3{b}_2{e}_2\times e_3=(a_1{b}_2-a_2{b}_1)e_1\times e_2+(a_1{b}_3-a_3{b}_1)e_1\times e_3+(a_2{b}_3-a_3{b}_2)e_2\times e_3=\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|e_1\times e_2+\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|e_1\times e_3+\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|e_2\times e_3$
若$[O:e_1,e_2,e_3]$是一个右手直角坐标系。

$e_1\times e_2=e_3$
$e_3\times e_1=e_2$
$e_2\times e_3=e_1$
则$\alpha \times \beta =\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|e_3-\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|e_2+\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|e_1$