Question

Let ${\mathit{e}}_{\mathbf{1}}\mathbf{,}{\mathit{e}}_{\mathbf{2}}\mathbf{,}{\mathit{e}}_{\mathbf{3}}$bea set of orthogonal unit vectors forming a right-handed system if taken in cyclic order. Show that the triple scalarproduct .${\mathit{e}}_{\mathbf{i}}\mathbf{.}\mathbf{(}{\mathit{e}}_{\mathbf{j}}\mathbf{\times }{\mathit{e}}_{\mathbf{k}}\mathbf{)}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{i}\mathbf{j}\mathbf{k}}$

Short answer

It has been shown that the scalar triple product is .$e_i\cdot (e_j\times e_k)={\epsilon }_{ijk}$

Step-by-step solution

Given Information

The given set of orthogonal unit vectors is .$e_1,e_2,e_3$

Definition of a cartesian tensor 

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Find the value 

Let$i^{th}$component is given by.$i^{th}{\mathit{(}\mathit{a}\mathit{\times }\mathit{b}\mathit{)}}_{\mathit{i}}\mathit{=}\sum _{lm}{\epsilon }_{ilm}{a}_{\mathit{l}}{b}_{\mathit{m}}$

The scalar triple product is given below.

role="math" localid="1664357511514" $\begin{array}{l}{\mathit{e}}_{\mathit{i}}\mathit{\cdot }\mathit{(}{\mathit{e}}_{\mathit{j}}\mathit{\times }{\mathit{e}}_{\mathit{k}}\mathit{)}\mathit{=}{\mathit{(}{\mathit{e}}_{\mathit{j}}\mathit{\times }{\mathit{e}}_{\mathit{k}}\mathit{)}}_{\mathit{i}}\\ {\mathit{e}}_{\mathit{i}}\mathit{\cdot }\mathit{(}{\mathit{e}}_{\mathit{j}}\mathit{\times }{\mathit{e}}_{\mathit{k}}\mathit{)}\mathit{=}\sum _{lm}{\epsilon }_{ilm}{\mathit{\delta }}_{\mathit{j}\mathit{l}}{\mathit{\delta }}_{\mathit{k}\mathit{m}}\\ {\mathit{e}}_{\mathit{i}}\mathit{\cdot }\mathit{(}{\mathit{e}}_{\mathit{j}}\mathit{\times }{\mathit{e}}_{\mathit{k}}\mathit{)}\mathit{=}{\mathit{\epsilon }}_{\mathit{i}\mathit{j}\mathit{k}}\end{array}$

Hence,thescalar triple product is .$e_i\cdot (e_j\times e_k)={\epsilon }_{ijk}\begin{array}{l}\\ \\ \end{array}$