# 向量叉乘的线性性质 几何解释

$| \vec a \times \vec b | = |\vec a| \cdot |\vec b| \cdot \sin \theta$

1. $\vec a \times \vec a = \vec0$, 因为夹角是0, 所以平行四边形面积也是0, 即叉积长度为0
2. $\vec a \times \vec b = - (\vec b \times \vec a)$, 等式两边的叉积等大反向, 模长因为平行四边形不变而相同, 方向因为右手法则旋转方向相反而相反
3. $(\lambda \vec a)\times b = \lambda (\vec a \times \vec b)$, 这点比较好想, 因为: ①正数$\lambda$数量乘不会影响$\vec a$的方向, 所以左右的叉积方向一样; 负数$\lambda$使得$\vec a$反向了, 但也使得左右叉积方向相反. ②对$\vec a$进行缩放, 平行四边形面积也同等缩放.
4. $(\vec a+\vec b) \times \vec c = \vec a \times \vec c + \vec b \times \vec c$, 这种分配率是以前最难想象的了.

### 结论4的证明

$\vec a \times \vec b = \vec a \times \vec b \, '$

1. $\vec a \times \vec b$就是$\vec b \,'$逆时针旋转90度, 并且伸缩$|\vec a|$(蓝色的向量)
2. $\vec a \times \vec c$就是$\vec c\,'$逆时针旋转90度, 并且伸缩$|\vec a|$(绿色的向量)
3. $\vec a \times (\vec b +\vec c)​$就是$\vec b\,'+\vec c\,'​$逆时针旋转90度, 并且伸缩$|\vec a|​$(红色的向量)

$\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec b.$

### 几何证明以后...

{\begin{aligned}\mathbf {u} \times \mathbf {v} ={}&(u_{1}\mathbf {i} +u_{2}\mathbf {j} +u_{3}\mathbf {k} )\times (v_{1}\mathbf {i} +v_{2}\mathbf {j} +v_{3}\mathbf {k} )\\={}&u_{1}v_{1}(\mathbf {i} \times \mathbf {i} )+u_{1}v_{2}(\mathbf {i} \times \mathbf {j} )+u_{1}v_{3}(\mathbf {i} \times \mathbf {k} )+{}\\&u_{2}v_{1}(\mathbf {j} \times \mathbf {i} )+u_{2}v_{2}(\mathbf {j} \times \mathbf {j} )+u_{2}v_{3}(\mathbf {j} \times \mathbf {k} )+{}\\&u_{3}v_{1}(\mathbf {k} \times \mathbf {i} )+u_{3}v_{2}(\mathbf {k} \times \mathbf {j} )+u_{3}v_{3}(\mathbf {k} \times \mathbf {k} )\\\end{aligned}}

$\mathbf {i} \times \mathbf {i} =\mathbf {j} \times \mathbf {j} =\mathbf {k} \times \mathbf {k} =\mathbf {0}$

\left\{ {\displaystyle {\begin{aligned}\mathbf {i} \times \mathbf {j} &=\mathbf {k} \\\mathbf {j} \times \mathbf {k} &=\mathbf {i} \\\mathbf {k} \times \mathbf {i} &=\mathbf {j} \end{aligned}}} \right. \quad\quad \left\{ {\displaystyle {\begin{aligned}\mathbf {j\times i} &=-\mathbf {k} \\\mathbf {k\times j} &=-\mathbf {i} \\\mathbf {i\times k} &=-\mathbf {j} \end{aligned}}} \right.

{\displaystyle {\begin{aligned} \mathbf {u\times v} &=(u_{2}v_{3}-u_{3}v_{2})\mathbf {i} +(u_{3}v_{1}-u_{1}v_{3})\mathbf {j} +(u_{1}v_{2}-u_{2}v_{1})\mathbf {k} \\ &={\begin{vmatrix}u_{2}&u_{3}\\v_{2}&v_{3}\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}u_{1}&u_{3}\\v_{1}&v_{3}\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}u_{1}&u_{2}\\v_{1}&v_{2}\end{vmatrix}}\mathbf {k} \\ &={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\\\end{vmatrix}} \end{aligned}}}

https://en.wikipedia.org/wiki/Cross_product

posted @ 2017-10-09 22:04  zzdyyy  阅读(89976)  评论(3编辑  收藏  举报