\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Vector cross-product

Definition of $\times$ : It is a bilinear, anti-symmetric operator that maps $\mathbb{E}^3 \times \mathbb{E}^3 \rightarrow \mathbb{E}^3 $, such that

\[\begin{align} \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_1&:=0,\\ \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_2&:=0,\\ \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_3&:=0,\\ \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_2 &:= \hat{\boldsymbol{\mathsf{E}}}_3, & \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_1 &:= -\hat{\boldsymbol{\mathsf{E}}}_3, \\ \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_3 &:= \hat{\boldsymbol{\mathsf{E}}}_1, & \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_2 &:= -\hat{\boldsymbol{\mathsf{E}}}_1, \\ \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_1 &:= \hat{\boldsymbol{\mathsf{E}}}_2, & \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_3 &:= -\hat{\boldsymbol{\mathsf{E}}}_2. \\ \end{align}\] \[\begin{align} (a_i \hat{\boldsymbol{\mathsf{E}}}_i) \times (b_j \hat{\boldsymbol{\mathsf{E}}}_j) = & a_1 b_1 \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_1+ a_1 b_2 \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_2+ a_1 b_3 \hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_3+\\ & a_2 b_1 \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_1+ a_2 b_2 \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_2+ a_2 b_3 \hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_3+ \\ & a_3 b_1 \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_1+ a_3 b_2 \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_2+ a_3 b_3 \hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_3 \\ =& a_1 b_2 (\hat{\boldsymbol{\mathsf{E}}}_3)+ a_1 b_3 (-\hat{\boldsymbol{\mathsf{E}}}_2)+ a_2 b_1 (-\hat{\boldsymbol{\mathsf{E}}}_3)+ a_2 b_3 (\hat{\boldsymbol{\mathsf{E}}}_1)+ a_3 b_1 (\hat{\boldsymbol{\mathsf{E}}}_2)+ a_3 b_2 (-\hat{\boldsymbol{\mathsf{E}}}_1) \\ =& (a_2b_3-a_3b_2)\hat{\boldsymbol{\mathsf{E}}}_1+ (a_3b_1-a_1b_3)\hat{\boldsymbol{\mathsf{E}}}_2+ (a_1b_2-a_2b_1)\hat{\boldsymbol{\mathsf{E}}}_3 \end{align}\]

Scalar triple product

\(\begin{align} \left[ \hat{\boldsymbol{\mathsf{E}}}_{i}, \hat{\boldsymbol{\mathsf{E}}}_{j}, \hat{\boldsymbol{\mathsf{E}}}_{k} \right]&= \hat{\boldsymbol{\mathsf{E}}}_{i}\cdot \left( \hat{\boldsymbol{\mathsf{E}}}_{j} \times \hat{\boldsymbol{\mathsf{E}}}_{k} \right) \end{align}\)

Levi-Cevita symbol

\(\begin{align} \epsilon_{ijk} &:= \hat{\boldsymbol{\mathsf{E}}}_{i}\cdot \left( \hat{\boldsymbol{\mathsf{E}}}_{j} \times \hat{\boldsymbol{\mathsf{E}}}_{k} \right) \end{align}\)

Now if $j=k$, then \(\hat{\boldsymbol{\mathsf{E}}}_{i} \times \hat{\boldsymbol{\mathsf{E}}}_{j}=0\), since $\sin(\theta)=0$. Also, the vector \(\hat{\boldsymbol{\mathsf{E}}}_{j} \times\hat{\boldsymbol{\mathsf{E}}}_{k}\) is perpendicular to both $\hat{\boldsymbol{\mathsf{E}}}_j$ and $ \hat{\boldsymbol{\mathsf{E}}}_k$. Thus, if $i$ equals either $j$ or $k$ then the scalar triple product also vanishes. Thus, the scalar triple product is non-zero only when $i$ and $j$ and $k$ are all different. There are six such combinations:

\[\begin{align*} \epsilon_{123}&:=\hat{\boldsymbol{\mathsf{E}}}_1\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_3\right)&=+1\\ \epsilon_{132}&:=\hat{\boldsymbol{\mathsf{E}}}_1\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_2\right)&=-1\\ \epsilon_{231}&:=\hat{\boldsymbol{\mathsf{E}}}_2\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_3 \times \hat{\boldsymbol{\mathsf{E}}}_1\right)&=+1\\ \epsilon_{213}&:=\hat{\boldsymbol{\mathsf{E}}}_2\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_3\right)&=-1\\ \epsilon_{312}&:=\hat{\boldsymbol{\mathsf{E}}}_3\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_1 \times \hat{\boldsymbol{\mathsf{E}}}_2\right)&=+1\\ \epsilon_{321}&:=\hat{\boldsymbol{\mathsf{E}}}_3\cdot\left(\hat{\boldsymbol{\mathsf{E}}}_2 \times \hat{\boldsymbol{\mathsf{E}}}_1\right)&=-1\\ \end{align*}\]

Using the Levi-cevita symbol, we can define the cross-product between two vectors as follows. Let \(\boldsymbol{\mathsf{a}}=a_i\hat{\boldsymbol{\mathsf{E}}}_i\) and \(\boldsymbol{\mathsf{b}}=b_i\hat{\boldsymbol{\mathsf{E}}}_i\) be two vectors.

\[\begin{align} \boldsymbol{\mathsf{a}}\times \boldsymbol{\mathsf{b}}&=\boldsymbol{\mathsf{c}}\\ (a_i\hat{\boldsymbol{\mathsf{E}}}_i)\times (b_j\hat{\boldsymbol{\mathsf{E}}}_j)&=c_p\hat{\boldsymbol{\mathsf{E}}}_p,\\ a_i b_j\hat{\boldsymbol{\mathsf{E}}}_i\times \hat{\boldsymbol{\mathsf{E}}}_j&=c_p\hat{\boldsymbol{\mathsf{E}}}_p,\\ a_i b_j(\hat{\boldsymbol{\mathsf{E}}}_i\times \hat{\boldsymbol{\mathsf{E}}}_j)\cdot \hat{\boldsymbol{\mathsf{E}}}_k&=c_p\hat{\boldsymbol{\mathsf{E}}}_p\cdot \hat{\boldsymbol{\mathsf{E}}}_k,\\ a_i b_j\epsilon_{ijk}&=c_p\delta_{pk},\\ \epsilon_{ijk} a_i b_j&=c_k,\\ \epsilon_{kij} a_i b_j&=c_k.\\ \end{align}\]

The final equation be written as, \(\begin{align} c_i&=\epsilon_{ijk} a_j b_k,\\ \boldsymbol{\mathsf{c}}&=\epsilon_{ijk} a_j b_k\hat{\boldsymbol{\mathsf{E}}}_i. \end{align}\)

Mathematics notes on cross product

computing cross product in mma