Question

(a) Write the triple scalar productin $\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})$tensor form and show that it is equal to the determinant in Chapter 6, equation.$\mathbf{(}\mathbf{3}\mathbf{.2}\mathbf{)}$ Hint: See.$\mathbf{(}\mathbf{5}\mathbf{.5}\mathbf{)}$

(b) Write equation$\mathbf{(}\mathbf{3}\mathbf{.2}\mathbf{)}$of Chapter 6 in tensor form to show the equivalence of the various expressions for the triple scalar product. Hint: Change the dummy indices as needed.

Short answer

1. The equation, $\left(3.2\right)$i.e.$\det \text{A}=a_{1i}{a}_{2j}{a}_{3k}{ϵ}_{ijk}$,
2. The triple scalar product as

$\begin{array}{c}\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})\\ =\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})\end{array}$

Step-by-step solution

Given Information

Equation$\left(5.5\right)$, i.e.,$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})={ϵ}_{ijk}{A}_i{B}_j{C}_k.$

Definition of Triple Scalar Product.

The dot product of one of the vectors with the cross product of the other two is known as the triple scalar product.

Step 3(a):Show thattriple scalar product is equal to the determinant

$\begin{array}{c}\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=A_i{(\mathbf{B}\times \mathbf{C})}_i\\ ={ϵ}_{ijk}{A}_i{B}_j{C}_k\\ ={ϵ}_{ijk}{a}_{1i}{a}_{2j}{a}_{3k}\\ =\det A\end{array}$$a_{1i}\equiv A_i,a_{2j}\equiv B_j,a_{3k}\equiv C_k$

$=\det \left(\begin{array}{c}A\\ B\\ C\end{array}\right)$

Hence, the equation $\left(3.2\right)$is obtained, i.e$\det \text{A}=a_{1i}{a}_{2j}{a}_{3k}{ϵ}_{ijk}$.,

Step 4(b):Write equation (3.2) of Chapter 6 in tensor form

Use above result to solve for (b).

$\begin{array}{c}{ϵ}_{ijk}{A}_i{B}_j{C}_k=\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})\\ ={ϵ}_{jki}{A}_i{B}_j{C}_k\\ ={ϵ}_{jki}{B}_j{C}_k{A}_i\\ =\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})\end{array}$

Simplify further,

$\begin{array}{c}{ϵ}_{ijk}{A}_i{B}_j{C}_k={ϵ}_{kij}{A}_i{B}_j{C}_k\\ ={ϵ}_{kij}{C}_k{A}_i{B}_j\\ =\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})\end{array}$

Hence the triple scalar product is:

$\begin{array}{c}\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})\\ =\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})\end{array}$.