叉乘、向量积的计算以及推导
叉乘
几何图示：
设有
a
=
(
a
x
,
a
y
,
a
z
)
,
b
=
(
b
x
,
b
y
,
b
z
)
\mathbf{a}=\left(a_{x}, a_{y}, a_{z}\right), \mathbf{b}=\left(b_{x}, b_{y}, b_{z}\right)
a=(ax,ay,az),b=(bx,by,bz)
i，j，k分别是X，Y，Z轴方向的单位向量，则：
a
×
b
=
(
a
y
b
z
−
a
z
b
y
)
i
+
(
a
z
b
x
−
a
x
b
z
)
j
+
(
a
x
b
y
−
a
y
b
x
)
k
\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}
a×b=(aybz−azby)i+(azbx−axbz)j+(axby−aybx)k
为了帮助记忆，利用三阶行列式，写成
∣
i
j
k
a
x
a
y
a
z
b
x
b
y
b
z
∣
\left|\begin{array}{lll} i & j & k \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{array}\right|
∣∣∣∣∣∣iaxbxjaybykazbz∣∣∣∣∣∣
设a=(l,m,n),b=(o,p,q)，则
a
×
b
=
(
l
,
m
,
n
)
×
(
o
,
p
,
q
)
=
(
m
q
−
n
p
,
n
o
−
l
q
,
l
p
−
m
o
)
\mathbf{a} \times \mathbf{b}=(l, m, n) \times(o, p, q)=(m q-n p, n o-l q, l p-m o)
a×b=(l,m,n)×(o,p,q)=(mq−np,no−lq,lp−mo)
利用三阶行列式，写成
∣
i
j
k
l
m
n
o
p
q
∣
\left|\begin{array}{ccc} i & j & k \\ l & m & n \\ o & p & q \end{array}\right|
∣∣∣∣∣∣ilojmpknq∣∣∣∣∣∣
推导
设有三个单位向量i，j，k，则：
i
→
=
(
1
,
0
,
0
)
j
→
=
(
0
,
1
,
0
)
k
→
=
(
0
,
0
,
1
)
\begin{aligned} &\overrightarrow{\mathrm{i}}=(1,0,0) \\ &\overrightarrow{\mathrm{j}}=(0,1,0) \\ &\overrightarrow{\mathrm{k}}=(0,0,1) \end{aligned}
i
=(1,0,0)j
=(0,1,0)k
=(0,0,1)
且
i
→
=
j
→
×
k
→
j
→
=
k
→
×
i
→
k
→
=
i
→
×
j
→
k
→
×
j
→
=
−
i
→
i
→
×
k
→
=
−
j
→
j
→
×
i
→
=
−
k
→
i
→
×
i
→
=
j
→
×
j
→
=
k
→
×
k
→
=
0
→
=
0
\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{\mathrm{0}} = 0
i
=j
×k
j
=k
×i
k
=i
×j
k
×j
=−i
i
×k
=−j
j
×i
=−k
i
×i
=j
×j
=k
×k
=0
=0
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：
u
→
=
(
x
u
,
y
u
,
z
u
)
v
→
=
(
x
v
,
y
v
,
z
v
)
\begin{aligned} &\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{aligned}
u
=(xu,yu,zu)v
=(xv,yv,zv)
u
→
=
X
u
⋅
i
→
+
y
u
⋅
j
→
+
z
u
⋅
k
→
v
→
=
X
v
⋅
i
→
+
y
v
⋅
j
→
+
z
v
⋅
k
→
\begin{aligned} &\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{aligned}
u
=Xu⋅i
+yu⋅j
+zu⋅k
v
=Xv⋅i
+yv⋅j
+zv⋅k
u
→
×
v
→
=
(
x
u
⋅
i
→
+
y
u
⋅
j
→
+
z
u
⋅
k
→
)
×
(
x
v
⋅
i
→
+
y
v
j
→
+
z
v
⋅
k
→
)
\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)
u
×v
=(xu⋅i
+yu⋅j
+zu⋅k
)×(xv⋅i
+yvj
+zv⋅k
)
展开，得：
u
→
×
v
→
=
x
u
⋅
x
v
⋅
(
i
→
⋅
i
→
)
+
x
u
⋅
y
v
⋅
(
i
→
⋅
j
→
)
+
x
u
⋅
z
v
⋅
(
i
→
⋅
k
→
)
+
y
u
⋅
x
v
⋅
(
j
→
⋅
i
→
)
+
y
u
⋅
y
v
⋅
(
j
→
⋅
j
→
)
+
y
u
⋅
z
v
⋅
(
j
→
⋅
k
→
)
+
z
u
⋅
x
v
⋅
(
k
→
⋅
i
→
)
+
z
u
⋅
y
v
⋅
(
k
→
⋅
j
→
)
+
z
u
⋅
z
v
⋅
(
k
→
⋅
k
→
)
\begin{aligned} \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{aligned}
u
×v
=xu⋅xv⋅(i
⋅i
)+xu⋅yv⋅(i
⋅j
)+xu⋅zv⋅(i
⋅k
)+yu⋅xv⋅(j
⋅i
)+yu⋅yv⋅(j
⋅j
)+yu⋅zv⋅(j
⋅k
)+zu⋅xv⋅(k
⋅i
)+zu⋅yv⋅(k
⋅j
)+zu⋅zv⋅(k
⋅k
)
化简，得：
u
⃗
×
v
⃗
=
0
→
+
x
u
⋅
y
v
⋅
(
k
⃗
)
+
x
u
⋅
z
v
⋅
(
−
j
⃗
)
+
y
u
⋅
x
v
⋅
(
−
k
⃗
)
+
0
→
+
y
u
⋅
z
v
⋅
(
i
⃗
)
+
z
u
⋅
x
V
⋅
(
j
⃗
)
+
z
u
⋅
y
v
⋅
(
−
i
⃗
)
+
0
→
\begin{gathered} \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{gathered}
u
×v
=0
+xu⋅yv⋅(k
)+xu⋅zv⋅(−j
)+yu⋅xv⋅(−k
)+0
+yu⋅zv⋅(i
)+zu⋅xV⋅(j
)+zu⋅yv⋅(−i
)+0
合并，得：
u
→
×
v
→
=
(
y
u
⋅
z
v
−
z
u
⋅
y
v
)
⋅
(
i
→
)
+
(
z
u
⋅
x
v
−
x
u
⋅
z
v
)
⋅
(
j
→
)
+
(
x
u
⋅
y
v
−
y
u
⋅
x
v
)
⋅
(
k
→
)
\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}})
u
×v
=(yu⋅zv−zu⋅yv)⋅(i
)+(zu⋅xv−xu⋅zv)⋅(j
)+(xu⋅yv−yu⋅xv)⋅(k
)
叉乘的几何意义
设
c
→
=
a
→
×
b
→
\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}
c
=a
×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
∣
⋅
sin
θ
|c|=|a||b|·\sin \theta
∣c∣=∣a∣∣b∣⋅sinθ
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。
代数规则
反交换律
a
→
×
b
→
=
−
b
→
×
a
→
\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}
a
×b
=−b
×a
加法的分配律
a
⃗
×
(
b
⃗
+
c
⃗
)
=
a
⃗
×
b
⃗
+
a
⃗
×
c
⃗
\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}
a
×(b
+c
)=a
×b
+a
×c
与标量乘法兼容
r
a
→
×
b
→
=
a
→
×
r
b
→
=
r
×
(
a
→
×
b
→
)
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}})
ra
×b
=a
×rb
=r×(a
×b
)
不满足结合律，但满足雅可比恒等式
a
→
×
(
b
→
×
c
→
)
+
b
→
×
(
c
→
×
a
→
)
+
c
→
×
(
a
→
×
b
→
)
=
0
\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
a
×(b
×c
)+b
×(c
×a
)+c
×(a
×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。
拉格朗日公式
设向量坐标
a
⃗
=
(
x
1
,
y
1
,
z
1
)
,
b
⃗
=
(
x
2
,
y
2
,
z
2
)
,
c
⃗
=
(
x
3
,
y
3
,
z
3
)
\vec{a}=(x 1, y 1, z 1), \vec{b}=(x 2, y 2, z 2), \vec{c}=(x 3, y 3, z 3)
a
=(x1,y1,z1),b
=(x2,y2,z2),c
=(x3,y3,z3), 则
(
a
⃗
×
b
⃗
)
×
c
⃗
=
(
y
1
z
2
−
z
1
y
2
,
z
1
x
2
−
x
1
z
2
,
x
1
y
2
−
x
2
y
1
)
×
(
x
3
,
y
3
,
z
3
)
=
(
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
)
=
(
x
2
(
x
1
x
3
+
y
1
y
3
+
z
1
z
3
)
,
y
2
(
x
1
x
3
+
y
1
y
3
+
z
1
z
3
)
,
z
2
(
x
1
x
3
+
y
1
y
3
+
z
1
z
3
)
)
−
(
x
1
(
x
2
x
3
+
y
2
y
3
+
z
2
z
3
)
,
y
1
(
x
2
x
3
+
y
2
y
3
+
z
2
z
3
)
,
z
1
(
x
2
x
3
+
y
2
y
3
+
z
2
z
3
)
)
=
(
x
2
(
a
⃗
⋅
c
⃗
)
,
y
2
(
a
⃗
⋅
c
⃗
)
,
z
2
(
a
⃗
⋅
c
⃗
)
)
−
(
x
1
(
b
⃗
⋅
c
⃗
)
,
y
1
(
b
⃗
⋅
c
⃗
)
,
z
1
(
b
⃗
⋅
c
⃗
)
)
=
b
⃗
(
a
⃗
⋅
c
⃗
)
−
a
⃗
(
b
⃗
⋅
c
⃗
)
\begin{aligned} &(\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) \\ &=\left(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}\right. \\ &\left.-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}\right) \\ &=\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{aligned}
(a
×b
)×c
=(y1z2−z1y2,z1x2−x1z2,x1y2−x2y1)×(x3,y3,z3)=(z1z3x2−x1z2z3−x1y2y3+x2y1y3,x1x3y2−x2x3y1−y1z2z3+z1z3y2,y1y3z2−y2y3z1−x2x3z1+x1x3z2)=(x2(x1x3+y1y3+z1z3),y2(x1x3+y1y3+z1z3),z2(x1x3+y1y3+z1z3))−(x1(x2x3+y2y3+z2z3),y1(x2x3+y2y3+z2z3),z1(x2x3+y2y3+z2z3))=(x2(a
⋅c
),y2(a
⋅c
),z2(a
⋅c
))−(x1(b
⋅c
),y1(b
⋅c
),z1(b
⋅c
))=b
(a
⋅c
)−a
(b
⋅c
) 这就是二重向量叉乘化简公式
(
a
⃗
×
b
⃗
)
×
c
⃗
=
b
⃗
(
a
⃗
⋅
c
⃗
)
−
a
⃗
(
b
⃗
⋅
c
⃗
)
(\vec{a} \times \vec{b}) \times \vec{c}=\vec{b}(\vec{a} \cdot \vec{c})-\vec{a}(\vec{b} \cdot \vec{c})
(a
×b
)×c
=b
(a
⋅c
)−a
(b
⋅c
)
转化一下，得
a
⃗
×
(
b
⃗
×
c
⃗
)
=
b
⃗
(
a
⃗
⋅
c
⃗
)
−
c
⃗
(
a
⃗
⋅
b
⃗
)
\vec{a} \times (\vec{b} \times \vec{c})=\vec{b}(\vec{a} \cdot \vec{c})-\vec{c}(\vec{a} \cdot \vec{b})
a
×(b
×c
)=b
(a
⋅c
)−c
(a
⋅b
)
向量积（矢积）与数量积（标积）的区别