C ⃗ ⋅ ( A ⃗ × B ⃗ ) = ∣ A ⃗ ∣ ∣ B ⃗ ∣ ∣ C ⃗ ∣ ⋅ s i n ( θ ) ⋅ c o s ( ϕ ) = A ⃗ ⋅ ( B ⃗ × C ⃗ ) = B ⃗ ⋅ ( C ⃗ × A ⃗ ) = − C ⃗ ⋅ ( B ⃗ × A ⃗ ) \vec{C} \cdot ( \vec{A} \times \vec{B}) = | \vec{A} | | \vec{B} | | \vec{C} | \cdot sin(\theta) \cdot cos(\phi) = \vec{A} \cdot ( \vec{B} \times \vec{C}) = \vec{B} \cdot ( \vec{C} \times \vec{A}) = - \vec{C} \cdot ( \vec{B} \times \vec{A}) C⋅(A×B)=∣A∣∣B∣∣C∣⋅sin(θ)⋅cos(ϕ)=A⋅(B×C)=B⋅(C×A)=−C⋅(B×A)
C ⃗ × ( A ⃗ × B ⃗ ) ≠ ( C ⃗ × A ⃗ ) × B ⃗ A ⃗ × ( B ⃗ × C ⃗ ) = ( A ⃗ ⋅ C ⃗ ) ⋅ B ⃗ − ( A ⃗ ⋅ B ⃗ ) ⋅ C ⃗ \vec{C} \times ( \vec{A} \times \vec{B}) \ne ( \vec{C} \times \vec{A} ) \times \vec{B} \\\\\ \\\ \vec{A} \times ( \vec{B} \times \vec{C}) = ( \vec{A} \cdot \vec{C} ) \cdot \vec{B} - ( \vec{A} \cdot \vec{B} ) \cdot \vec{C} C×(A×B)=(C×A)×B A×(B×C)=(A⋅C)⋅B−(A⋅B)⋅C
{ x = ρ ⋅ c o s ϕ y = ρ ⋅ s i n ϕ z = z { ρ = x 2 + y 2 ϕ = a r c t a n ( y x ) z = z d S = ρ ⋅ d ϕ ⋅ d z d V = ρ ⋅ d ρ ⋅ d ϕ ⋅ d z \left \{ \begin{array}{c} x=\rho \cdot cos\phi \\ y=\rho \cdot sin\phi \\ z=z \end{array} \right . \\\\\ \\\ \left \{ \begin{array}{c} \rho= \sqrt{x^2+y^2} \\ \phi = arctan(\frac{y}{x}) \\ z=z \end{array} \right. \\\\\ \\\ dS= \rho\cdot d\phi \cdot dz \\\\\ \\\ dV = \rho \cdot d\rho \cdot d\phi \cdot dz ⎩⎨⎧x=ρ⋅cosϕy=ρ⋅sinϕz=z ⎩⎨⎧ρ=x2+y2ϕ=arctan(xy)z=z dS=ρ⋅dϕ⋅dz dV=ρ⋅dρ⋅dϕ⋅dz
{ x = r ⋅ s i n θ ⋅ c o s ϕ y = r ⋅ s i n θ ⋅ s i n ϕ z = r ⋅ c o s θ { r = x 2 + y 2 + z 2 θ = a r c t a n ( x 2 + y 2 z ) ϕ = a r c t a n ( y x ) d S = r 2 ⋅ s i n θ ⋅ d θ ⋅ d ϕ d V = r 2 ⋅ s i n θ ⋅ d r ⋅ d θ ⋅ d ϕ \left \{ \begin{array}{c} x=r \cdot sin\theta \cdot cos\phi \\ y=r \cdot sin\theta \cdot sin\phi \\ z=r \cdot cos\theta \end{array} \right . \\\\\ \\\ \left \{ \begin{array}{c} r= \sqrt{x^2+y^2+z^2} \\ \theta = arctan(\frac{ \sqrt{x^2+y^2} }{z}) \\ \phi = arctan(\frac{y}{x}) \end{array} \right. \\\\\ \\\ dS= r^2\cdot sin\theta \cdot d\theta \cdot d\phi \\\\\ \\\ dV = r^2\cdot sin\theta \cdot dr \cdot d\theta \cdot d\phi ⎩⎨⎧x=r⋅sinθ⋅cosϕy=r⋅sinθ⋅sinϕz=r⋅cosθ ⎩⎪⎨⎪⎧r=x2+y2+z2θ=arctan(zx2+y2)ϕ=arctan(xy) dS=r2⋅sinθ⋅dθ⋅dϕ dV=r2⋅sinθ⋅dr⋅dθ⋅dϕ
[ A ρ A ϕ A z ] = [ c o s ϕ s i n ϕ 0 − s i n ϕ c o s ϕ 0 0 0 1 ] [ A x A y A z ] [ A r A θ A ϕ ] = [ s i n θ c o s ϕ s i n θ s i n ϕ c o s θ c o s θ c o s ϕ c o s θ s i n ϕ − s i n θ − s i n ϕ c o s ϕ 0 ] [ A x A y A z ] [ A r A θ A ϕ ] = [ s i n θ 0 c o s θ c o s θ 0 − s i n ϕ 0 1 0 ] [ A ρ A ϕ A z ] \begin{bmatrix}A_\rho\\A_\phi\\A_z \end{bmatrix} =\begin{bmatrix}cos\phi & sin\phi & 0 \\ -sin\phi & cos\phi & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}A_x\\A_y\\A_z \end{bmatrix} \\\\\ \\\ \begin{bmatrix}A_r\\A_\theta\\A_\phi \end{bmatrix} =\begin{bmatrix}sin\theta cos\phi &sin\theta sin\phi & cos\theta \\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta\\ -sin\phi & cos\phi & 0 \end{bmatrix} \begin{bmatrix}A_x\\A_y\\A_z \end{bmatrix} \\\\\ \\\ \begin{bmatrix}A_r\\A_\theta\\A_\phi \end{bmatrix} =\begin{bmatrix}sin\theta & 0 & cos\theta \\ cos\theta & 0 & -sin\phi\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix}A_\rho\\A_\phi\\A_z \end{bmatrix} ⎣⎡AρAϕAz⎦⎤=⎣⎡cosϕ−sinϕ0sinϕcosϕ0001⎦⎤⎣⎡AxAyAz⎦⎤ ⎣⎡ArAθAϕ⎦⎤=⎣⎡sinθcosϕcosθcosϕ−sinϕsinθsinϕcosθsinϕcosϕcosθ−sinθ0⎦⎤⎣⎡AxAyAz⎦⎤ ⎣⎡ArAθAϕ⎦⎤=⎣⎡sinθcosθ0001cosθ−sinϕ0⎦⎤⎣⎡AρAϕAz⎦⎤
∇ f = ∂ f ∂ x e x ⃗ + ∂ f ∂ y e y ⃗ + ∂ f ∂ z e z ⃗ = ∂ f ∂ ρ e ρ ⃗ + 1 ρ ∂ f ∂ ϕ e ϕ ⃗ + ∂ f ∂ z e z ⃗ = ∂ f ∂ r e r ⃗ + 1 r ∂ f ∂ θ e θ ⃗ + 1 r s i n θ ∂ f ∂ ϕ e ϕ ⃗ 梯 度 的 模 为 最 大 变 化 率 \nabla f = \frac{\partial f}{\partial x}\vec{e_x}+\frac{\partial f}{\partial y}\vec{e_y}+\frac{\partial f}{\partial z}\vec{e_z} \\\\\ \\\ = \frac{\partial f}{\partial \rho}\vec{e_\rho}+ \color{red} \frac 1 \rho \color{black}\frac{\partial f}{\partial \phi}\vec{e_\phi}+\frac{\partial f}{\partial z}\vec{e_z} \\\\\ \\\ = \frac{\partial f}{\partial r}\vec{e_r}+\color{red}\frac 1r \color{black} \frac{\partial f}{\partial \theta}\vec{e_\theta}+ \color{red}\frac{1}{rsin\theta } \color{black} \frac{\partial f}{\partial \phi}\vec{e_\phi}\\\ \\ 梯度的模为最大变化率 ∇f=∂x∂fex+∂y∂fey+∂z∂fez =∂ρ∂feρ+ρ1∂ϕ∂feϕ+∂z∂fez =∂r∂fer+r1∂θ∂feθ+rsinθ1∂ϕ∂feϕ 梯度的模为最大变化率
∇ ⋅ A ⃗ = ∂ A x ∂ x + ∂ A y ∂ y + ∂ A z ∂ z = 1 ρ ∂ ( ρ A ρ ) ∂ ρ + 1 ρ ∂ ( A ϕ ) ∂ ϕ + ∂ ( A z ) ∂ z = 1 r 2 ∂ ( r 2 A r ) ∂ r + 1 r s i n θ ∂ ( s i n θ A θ ) ∂ θ + 1 r s i n θ ∂ ( A ϕ ) ∂ ϕ \nabla \cdot \bm{\vec{A}} = \frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z} \\\\\ \\\ =\color{red} \frac 1 \rho \color{black} \frac{\partial (\color{red}\rho \color{black} A_\rho)}{\partial \rho}+ \color{red} \frac 1 \rho \color{black}\frac{\partial (A_\phi)}{\partial \phi}+\frac{\partial (A_z)}{\partial z} \\\\\ \\\ =\color{red} \frac{1}{ r^2} \color{black} \frac{\partial (\color{red} r^2\color{black} A_r)}{\partial r}+ \color{red}\frac{1}{rsin\theta } \color{black} \color{black} \frac{\partial (\color{red} sin\theta\color{black} A_\theta)}{\partial \theta}+ \color{red}\frac{1}{rsin\theta } \color{black} \frac{\partial (A_\phi)}{\partial \phi} ∇⋅A=∂x∂Ax+∂y∂Ay+∂z∂Az =ρ1∂ρ∂(ρAρ)+ρ1∂ϕ∂(Aϕ)+∂z∂(Az) =r21∂r∂(r2Ar)+rsinθ1∂θ∂(sinθAθ)+rsinθ1∂ϕ∂(Aϕ)
∇ × B ⃗ = ∣ e x ⃗ e y ⃗ e z ⃗ ∂ ∂ x ∂ ∂ y ∂ ∂ z B x ⃗ B y ⃗ B z ⃗ ∣ = 1 ρ ∣ e ρ ⃗ ρ e ϕ ⃗ e z ⃗ ∂ ∂ ρ ∂ ∂ ϕ ∂ ∂ z B ρ ⃗ ρ B ϕ ⃗ B z ⃗ ∣ = 1 r 2 s i n θ ∣ e r ⃗ r e θ ⃗ r s i n θ e ϕ ⃗ ∂ ∂ r ∂ ∂ θ ∂ ∂ ϕ B r ⃗ r B θ ⃗ r s i n θ B ϕ ⃗ ∣ \nabla \times \bm{\vec{B}}= \begin{vmatrix}\vec{e_x} & \vec{e_y} & \vec{e_z} \\\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\\\ \vec{B_x} & \vec{B_y} & \vec{B_z} \end{vmatrix} \\\\\ \\\ =\color{red} \frac 1 \rho \color{black} \begin{vmatrix}\vec{e_\rho} & \color{red} \rho \color{black}\vec{e_\phi} & \vec{e_z} \\\\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z}\\\\ \vec{B_\rho} & \color{red} \rho \color{black}\vec{B_\phi} & \vec{B_z} \end{vmatrix} \\\\\ \\\ =\color{red}\frac{1}{r^2sin\theta } \color{black} \begin{vmatrix}\vec{e_r} & \color{red} r \color{black}\vec{e_\theta} & \color{red} rsin\theta \color{black}\vec{e_\phi} \\\\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi}\\\\ \vec{B_r} & \color{red} r \color{black}\vec{B_\theta} & \color{red} rsin\theta \color{black}\vec{B_\phi} \end{vmatrix} ∇×B=∣∣∣∣∣∣∣∣∣∣ex∂x∂Bxey∂y∂Byez∂z∂Bz∣∣∣∣∣∣∣∣∣∣ =ρ1∣∣∣∣∣∣∣∣∣∣eρ∂ρ∂Bρρeϕ∂ϕ∂ρBϕez∂z∂Bz∣∣∣∣∣∣∣∣∣∣ =r2sinθ1∣∣∣∣∣∣∣∣∣∣er∂r∂Brreθ∂θ∂rBθrsinθeϕ∂ϕ∂rsinθBϕ∣∣∣∣∣∣∣∣∣∣
∇ 2 f = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 = 1 ρ ∂ ∂ ρ ( ρ ∂ f ∂ ρ ) + 1 ρ 2 ∂ 2 f ∂ ϕ 2 + ∂ 2 f ∂ z 2 = 1 r 2 ∂ ∂ r ( r 2 ∂ f ∂ r ) + 1 r 2 s i n θ ∂ ∂ θ ( s i n θ ∂ f ∂ θ ) + 1 r 2 s i n 2 θ ∂ 2 f ∂ ϕ 2 \nabla^2f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2} \\\\\ \\\ = \color{red} \frac 1 \rho\color{black} \frac{\partial} {\partial \rho}( \color{red} \rho\color{black} \frac{\partial f}{\partial \rho})+\color{red}\frac{1}{\rho^2 } \color{black} \frac{\partial^2f}{\partial \phi^2}+\frac{\partial^2f}{\partial z^2} \\\\\ \\\ =\color{red}\frac{1}{r^2 } \color{black}\frac{\partial} {\partial r}( \color{red} r^2\color{black} \frac{\partial f}{\partial r})+\color{red}\frac{1}{r^2sin\theta } \color{black}\frac{\partial}{\partial \theta}(\color{red} sin\theta\color{black} \frac{\partial f}{\partial \theta})+\color{red}\frac{1}{r^2sin^2\theta } \color{black} \frac{\partial^2f}{\partial \phi^2} ∇2f=∂x2∂2f+∂y2∂2f+∂z2∂2f =ρ1∂ρ∂(ρ∂ρ∂f)+ρ21∂ϕ2∂2f+∂z2∂2f =r21∂r∂(r2∂r∂f)+r2sinθ1∂θ∂(sinθ∂θ∂f)+r2sin2θ1∂ϕ2∂2f
1 . 梯 度 无 旋 ∇ × ( ∇ f ) ≡ 0 \bm1.梯度无旋 \\ \nabla \times (\nabla f) \equiv 0 1.梯度无旋∇×(∇f)≡0
2 . 旋 度 无 散 ∇ ⋅ ( ∇ × F ⃗ ) ≡ 0 \bm2.旋度无散 \\ \nabla \cdot (\nabla \times \bm{\vec{F}}) \equiv 0 2.旋度无散∇⋅(∇×F)≡0
3 . 旋 度 的 旋 度 ∇ × ( ∇ × F ⃗ ) = ∇ ( ∇ ⋅ F ⃗ ) − ∇ 2 F ⃗ 散 度 的 梯 度 − 并 积 的 散 度 散 度 的 梯 度 指 向 源 密 集 的 地 方 \bm3.旋度的旋度 \\ \nabla \times (\nabla \times \bm{\vec{F}}) = \nabla (\nabla \cdot \bm{\vec{F}}) -\nabla^2 \bm{\vec{F}} \\\\\ \\\ 散度的梯度-并积的散度\\ 散度的梯度指向源密集的地方 3.旋度的旋度∇×(∇×F)=∇(∇⋅F)−∇2F 散度的梯度−并积的散度散度的梯度指向源密集的地方
4 . 点 乘 的 梯 度 ∇ ( a ⃗ ⋅ b ⃗ ) = ( a ⃗ ⋅ ∇ ) b ⃗ + ( b ⃗ ⋅ ∇ ) a ⃗ + a ⃗ × ( ∇ × b ⃗ ) + b ⃗ × ( ∇ × a ⃗ ) 体 现 ∇ 的 矢 量 性 与 微 分 性 \bm4.点乘的梯度 \\ \nabla (\vec{a} \cdot \vec{b} ) =(\vec{a}\cdot\nabla)\vec{b}+(\vec{b}\cdot\nabla)\vec{a}+\vec{a}\times(\nabla\times\vec{b})+\vec{b}\times(\nabla\times\vec{a}) \\\\\ \\\ 体现\nabla的矢量性与微分性 4.点乘的梯度∇(a⋅b)=(a⋅∇)b+(b⋅∇)a+a×(∇×b)+b×(∇×a) 体现∇的矢量性与微分性
5 . 叉 乘 的 散 度 ∇ ⋅ ( a ⃗ × b ⃗ ) = b ⃗ ⋅ ( ∇ × a ⃗ ) − a ⃗ ⋅ ( ∇ × b ⃗ ) \bm5.叉乘的散度\\ \nabla\cdot(\vec{a}\times\vec{b})=\vec{b}\cdot(\nabla\times\vec{a})-\vec{a}\cdot(\nabla\times\vec{b}) 5.叉乘的散度∇⋅(a×b)=b⋅(∇×a)−a⋅(∇×b)
6 . 叉 乘 的 旋 度 ∇ × ( a ⃗ × b ⃗ ) = a ⃗ ( ∇ ⋅ b ⃗ ) − b ⃗ ( ∇ ⋅ a ⃗ ) + ( b ⃗ ⋅ ∇ ) a ⃗ − ( a ⃗ ⋅ ∇ ) b ⃗ \bm6.叉乘的旋度\\ \nabla\times(\vec{a}\times\vec{b})=\vec{a}(\nabla\cdot\vec{b})-\vec{b}(\nabla\cdot\vec{a})+(\vec{b}\cdot\nabla)\vec{a}-(\vec{a}\cdot\nabla)\vec{b} 6.叉乘的旋度∇×(a×b)=a(∇⋅b)−b(∇⋅a)+(b⋅∇)a−(a⋅∇)b
1. 高 斯 散 度 定 理 ∫ V ∇ ⋅ F ⃗ d V = ∮ S F ⃗ ⋅ d S ⃗ \bm{1.高斯散度定理} \\ \int_V^\ \nabla \cdot \bm{\vec{F}} dV = \oint_S \bm{\vec{F}} \cdot d\vec{S} 1.高斯散度定理∫V ∇⋅FdV=∮SF⋅dS
2 . 斯 托 克 斯 定 理 ∫ S ( ∇ × F ⃗ ) ⋅ d S = ∮ C F ⃗ ⋅ d l ⃗ \bm2.斯托克斯定理\\ \int_S^\ (\nabla \times \bm{\vec{F}} )\cdot dS = \oint_C \bm{\vec{F}} \cdot d\vec{l} 2.斯托克斯定理∫S (∇×F)⋅dS=∮CF⋅dl
3 . 标 量 格 林 定 理 ∮ V ∇ ⋅ ( Ψ ∇ Φ ) d V = ∮ S ( Ψ ∇ Φ ) ⋅ d S 对 ( Ψ ∇ Φ ) 用 高 斯 散 度 定 理 \bm3.标量格林定理\\ \oint_V \nabla \cdot (\Psi \nabla \Phi) dV = \oint_S (\Psi \nabla \Phi) \cdot dS \\\\\ \\\ 对 (\Psi \nabla \Phi) 用高斯散度定理 3.标量格林定理∮V∇⋅(Ψ∇Φ)dV=∮S(Ψ∇Φ)⋅dS 对(Ψ∇Φ)用高斯散度定理
4 . 矢 量 格 林 定 理 ∮ V ∇ ⋅ ( P ⃗ × ∇ × Q ⃗ ) d V = ∮ S ( P ⃗ × ∇ × Q ⃗ ) ⋅ d S 对 ( P ⃗ × ∇ × Q ⃗ ) 用 高 斯 散 度 定 理 \bm4.矢量格林定理\\ \oint_V \nabla \cdot (\vec{P}\times \nabla \times \vec{Q}) dV = \oint_S (\vec{P}\times \nabla \times \vec{Q}) \cdot dS \\\\\ \\\ 对 (\vec{P}\times \nabla \times \vec{Q})用高斯散度定理 4.矢量格林定理∮V∇⋅(P×∇×Q)dV=∮S(P×∇×Q)⋅dS 对(P×∇×Q)用高斯散度定理
{ 不 写 ∑ 重 复 下 标 自 动 求 和 和 式 相 乘 下 标 不 能 相 同 a ⃗ ⋅ b ⃗ = a 1 b 1 + a 2 b 2 + a 3 b 3 = ∑ i a i b i = a i b i \left \{ \begin{array}{c} 不写\sum \\ 重复下标自动求和 \\ 和式相乘下标不能相同 \end{array} \right. \\\\\ \\\ \vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3=\sum_i a_ib_i \color{red}= a_ib_i ⎩⎨⎧不写∑重复下标自动求和和式相乘下标不能相同 a⋅b=a1b1+a2b2+a3b3=i∑aibi=aibi
δ i j = { 0 , i ≠ j 1 , i = j 提 取 a ⃗ ⋅ b ⃗ = a i b i = δ i j a i b j ( 1 ) δ i j = e i ⃗ ⋅ e j ⃗ ( 2 ) I → → = [ δ 11 δ 12 δ 13 δ 21 δ 22 δ 23 δ 31 δ 32 δ 33 ] ( 3 ) δ i m ⋅ δ m j = δ i j ( 4 ) e i ⃗ = [ δ i 1 δ i 2 δ i 3 ] \delta_{ij} = \begin{cases} 0, & i \ne j \\ 1, & i=j \end{cases} \\\\\ \\\ 提取 \\\\\ \\\ \vec{a}\cdot\vec{b}=a_ib_i \color{red}=\delta_{ij}a_ib_j\color{black} \\\\\ \\\ (1)\delta_{ij}=\vec{e_i}\cdot\vec{e_j}\\\\\ \\\ (2)\overrightarrow {\overrightarrow {I}}=\begin{bmatrix}\delta_{11} & \delta_{12} &\delta_{13}\\ \delta_{21} &\delta_{22}&\delta_{23}\\ \delta_{31} & \delta_{32} &\delta_{33}\end{bmatrix}\\\\\ \\\ (3)\delta_{im}\cdot\delta_{mj}=\delta_{ij}\\\\\ \\\ (4)\vec{e_i}=\begin{bmatrix}\delta_{i1} \\ \delta_{i2}\\\delta_{i3} \end{bmatrix} δij={ 0,1,i=ji=j 提取 a⋅b=aibi=δijaibj (1)δij=ei⋅ej (2)I=⎣⎡δ11δ21δ31δ12δ22δ32δ13δ23δ33⎦⎤ (3)δim⋅δmj=δij (4)ei=⎣⎡δi1δi2δi3⎦⎤
ε i j k = { 1 , i j k = 123 , 231 , 312. − 1 , i j k = 321 , 213 , 132. 0 , o t h e r w i s e a ⃗ × b ⃗ = ε i j k a i b j e k ⃗ ( 1 ) ε i j k = e i ⃗ ⋅ ( e j ⃗ × e k ⃗ ) ( 2 ) ε i j k = − ε j i k ( 3 ) ε i j k ε l m n = ∣ δ i l δ i m δ i n δ j l δ j m δ j n δ k l δ k m δ k n ∣ ① ε i j k ε m n k = δ i m δ j n − δ i n δ j m ② ε i j k ε m j k = 2 δ i m ③ ε i j k ε i j k = 6 ④ e i ⃗ × e j ⃗ = ε i j k e k ⃗ \varepsilon_{ijk} = \begin{cases} 1, & i jk=123, 231, 312. \\ -1, & ijk=321, 213, 132. \\ 0,& otherwise \end{cases} \\\\\ \\\ \vec{a}\times\vec{b}\color{red}=\varepsilon_{ijk}a_ib_j\vec{e_k}\color{black}\\\\\ \\\ (1) \varepsilon_{ijk}=\vec{e_i}\cdot(\vec{e_j}\times\vec{e_k})\\\\\ \\\ (2)\varepsilon_{ijk}=-\varepsilon_{jik}\\\\\ \\\ (3)\varepsilon_{ijk}\varepsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} &\delta_{in}\\ \delta_{jl} &\delta_{jm}&\delta_{jn}\\ \delta_{kl} & \delta_{km} &\delta_{kn}\end{vmatrix}\\\\\ \\ ① \varepsilon_{ijk}\varepsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}\\\\\ \\ ② \varepsilon_{ijk}\varepsilon_{mjk}=2\delta_{im}\\\\\ \\ ③ \varepsilon_{ijk}\varepsilon_{ijk}=6\\\ \\ ④ \vec{e_i}\times\vec{e_j} = \varepsilon_{ijk}\vec{e_k} εijk=⎩⎪⎨⎪⎧1,−1,0,ijk=123,231,312.ijk=321,213,132.otherwise a×b=εijkaibjek (1)εijk=ei⋅(ej×ek) (2)εijk=−εjik (3)εijkεlmn=∣∣∣∣∣∣δilδjlδklδimδjmδkmδinδjnδkn∣∣∣∣∣∣ ①εijkεmnk=δimδjn−δinδjm ②εijkεmjk=2δim ③εijkεijk=6 ④ei×ej=εijkek