91Ó°ÊÓ

The equation $\left(5·8\right)$, i.e.,role="math" localid="1659336734341" ${\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{i}\mathbf{j}\mathbf{k}}{\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{i}\mathbf{m}\mathbf{n}}\mathbf{=}{\mathcal{S}}_{\mathbf{j}\mathbf{m}}{\mathcal{S}}_{\mathbf{k}\mathbf{n}}\mathbf{-}{\mathcal{S}}_{\mathbf{j}\mathbf{n}}\mathcal{S}\mathit{k}\mathit{m}$

The equation ,$\left(5.9\right)$ i.e., .role="math" localid="1659337040545" ${\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{a}\mathbf{b}\mathbf{c}}{\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{p}\mathbf{q}\mathbf{b}}\mathbf{=}{\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{b}\mathbf{c}\mathbf{a}}{\stackrel{\mathbf{,}}{\mathbf{o}}}_{\mathbf{b}\mathbf{p}\mathbf{q}}\mathbf{=}{\mathcal{S}}_{\mathbf{c}\mathbf{p}}{\mathcal{S}}_{\mathbf{a}\mathbf{q}}\mathbf{-}{\mathcal{S}}_{\mathbf{c}\mathbf{q}}{\mathcal{S}}_{\mathbf{a}\mathbf{p}}\mathbf{.}$

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

Solve for (a).

$$\begin{gathered} \left(A×B\right)·\left(\left(B×C\right)×\left(C×A\right)\right)={\stackrel{,}{o}}_{ijk}{\left(A×B\right)}_i{\left(B×C\right)}_j{\left(C×A\right)}_K \\ ={\stackrel{,}{o}}_{ijk}{\stackrel{,}{o}}_{imun}{\stackrel{,}{o}}_{j{\overline{)+ok}}_jË\copyright }{A}_m{B}_n{B}_I{C}_r{C}_p{A}_q \\ =\left({\mathcal{S}}_{jm}{\mathcal{S}}_{km}-{\mathcal{S}}_{jn}{\mathcal{S}}_{km}\right){\stackrel{,}{o}}_{jIr}{\stackrel{,}{o}}_{kpq}{A}_m{B}_n{B}_I{C}_r{C}_p{A}_q \\ =\left({\stackrel{,}{o}}_{mIr}{\stackrel{,}{o}}_{nqq}\stackrel{}{{\stackrel{,}{o}}_{nl}+}{\stackrel{,}{o}}_{mwp}\right)A_m{B}_n{B}_I{C}_r{C}_p{A}_q \end{gathered}$$

Simplify further,
$$\begin{gathered} =\left({\stackrel{,}{o}}_{mir}{A}_m{B}_I{C}_r\right)\left({\stackrel{,}{o}}_{mep}{A}_v{B}_n{C}_p\right)-\left({\stackrel{,}{o}}_{nir}{A}_n{B}_I{C}_r\right)\left({\stackrel{,}{o}}_{mup}{A}_{re}{B}_p{C}_q\right) \\ ={\left(A·\left(B×C\right)\right)}^2 \end{gathered}$$

The equation ,$\left(3.13\right)$ i.e.,$\left(A×B\right)·\left(\left(B×C\right)×\left(C×A\right)×\left(C×A\right)\right)={\left(A·\left(B×C\right)\right)}^2$

Solve for (b).

$$\begin{gathered} A×\left(B×C\right)+B×\left(C×A\right)+C×\left(A×B\right)=\left(A·C\right)B-\left(A·B\right)C+\left(B·A\right)C-\left(B·C\right)A+\left(C·B\right)-\left(C·A\right)B \\ =0 \end{gathered}$$

The equation ,$\left(3.14\right)$i.e., $A×\left(B×C\right)+B×\left(C×A\right)+C×\left(A×B\right)=0$

Solve for (c).

$$\begin{gathered} \left(A×B\right)·\left(C×D\right)={\left(A×B\right)}_i{\left(c×d\right)}_i \\ ={\stackrel{,}{o}}_{ijk}{A}_j{B}_k{C}_n{D}_m-{\mathcal{S}}_{kn}{\mathcal{S}}_{jm}{A}_j{B}_k{C}_n{D}_m \\ ={\left({\stackrel{,}{o}}_{mir}{A}_m{B}_{I}C\right)}_r\left({\stackrel{,}{o}}_{mep}{A}_v{B}_n{C}_p\right)-\left({\stackrel{,}{o}}_{nir}{B}_n{B}_I{C}_r\right)\left({\stackrel{,}{o}}_{mup}{A}_{re}{B}_p{C}_q\right) \end{gathered}$$

On further simplification,

$$\begin{gathered} \left(A×B\right)·\left(C×D\right)=A_j{B}_k{C}_j{D}_K-A_j{B}_k{C}_k{D}_j \\ =\left(A·C\right)\left(B·D\right)-\left(A·D\right)\left(B·C\right) \end{gathered}$$

Lagrange’s identity:$\left(A×B\right)·\left(C×D\right)=\left(A·C\right)\left(B·D\right)-\left(A·D\right)\left(B·C\right)$

Solve for (d).

$$\begin{gathered} {\left(\left(A×B\right)\left(C×D\right)\right)}_i={\stackrel{,}{o}}_{ojk}{\left(A×B\right)}_j{\left(C×D\right)}_k \\ ={\stackrel{,}{o}}_{ijk}{\stackrel{,}{o}}_{jnm}{\stackrel{,}{o}}_{kpq}{A}_n{B}_m{C}_p{D}_q \\ ={\stackrel{,}{o}}_{nmj}{A}_n{B}_m{C}_j{D}_i-{\stackrel{,}{o}}_{nmj}{A}_n{B}_m{C}_j{D}_i \end{gathered}$$

Simplify further,

$$\begin{gathered} =\left(ABD\right)C_i-\left(ABC\right)D_i \\ =\left(ABD\right)C-\left(ABC\right)D_i \\ \left(\left(A×B\right)×\left(C×D\right)\right)=\left(ABD\right)C-\left(ABC\right)D \end{gathered}$$

Hence, the equation required, i.e., $\left(\left(A×B\right)×\left(C×D\right)\right)=\left(ABD\right)C-\left(ABC\right)D$.