Question

Write out the twelve triple scalar products involving A, B, and C and verify the facts stated just above (3.3)

Short answer

The twelve triple scalar products involving A, B and C are as follows.

$$\begin{gathered} \mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)=\left(\mathrm{B}\times \mathrm{C}\right).\mathrm{A}=\left(\mathrm{A}\times \mathrm{B}\right).\mathrm{C}=\mathrm{C}.\left(\mathrm{A}\times \mathrm{B}\right)=\left(\mathrm{C}\times \mathrm{A}\right).\mathrm{B}=\mathrm{B}.\left(\mathrm{C}\times \mathrm{A}\right) \\ \mathrm{B}.\left(\mathrm{A}\times \mathrm{C}\right)=\left(\mathrm{A}\times \mathrm{C}\right).\mathrm{B}=\left(\mathrm{B}\times \mathrm{A}\right).\mathrm{C}=\left(\mathrm{C}\times \mathrm{B}\right).\mathrm{A}=\mathrm{A}.\left(\mathrm{C}\times \mathrm{B}\right)=\mathrm{C}.\left(\mathrm{B}\times \mathrm{A}\right)=‐\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right) \end{gathered}$$

Step-by-step solution

Given Condition

The vectors $\mathrm{A}$, B and C.

Concept of triple scalar product

The triple scalar product for three vectors is written as $\mathbf{A}$.$\left(B\times C\right)$

The determinant is as follows.

$\mathbf{A}\mathbf{.}\left(B\times C\right)\mathbf{=}\left|\begin{array}{ccc}{A}_x& A_y& A_z\\ B_x& B_y& B_z\\ C_x& C_y& C_z\end{array}\right|$

Verify the interchange of rows of determinant

The triple scalar products involving A, B and C is given as follows.

$$\begin{gathered} \mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)={\mathrm{A}}_{\mathrm{x}}.{\left(\mathrm{B}\times \mathrm{C}\right)}_{\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}.{\left(\mathrm{B}\times \mathrm{C}\right)}_{\mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}.{\left(\mathrm{B}\times \mathrm{C}\right)}_{\mathrm{z}}\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right) \\ =\left|\begin{array}{ccc}{\mathrm{A}}_{\mathrm{x}}& {\mathrm{A}}_{\mathrm{y}}& {\mathrm{A}}_{\mathrm{z}}\\ {\mathrm{B}}_{\mathrm{x}}& {\mathrm{B}}_{\mathrm{y}}& {\mathrm{B}}_{\mathrm{z}}\\ {\mathrm{C}}_{\mathrm{x}}& {\mathrm{C}}_{\mathrm{y}}& {\mathrm{C}}_{\mathrm{z}}\end{array}\right| \end{gathered}$$

On interchanging the rows of determinants, there will be a minus sign.

Verification of the below determinant is s follows.

$$\begin{gathered} \left(\mathrm{A}\times \mathrm{B}\right).\mathrm{C}=\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right) \\ =\mathrm{C}.\left(\mathrm{A}\times \mathrm{B}\right) \\ =‐\left(\mathrm{A}\times \mathrm{C}\right).\mathrm{B} \end{gathered}$$

Interchange rows of determinant

The determinant equivalent to$\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)$is obtained when two rows are interchanged.

$\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)=\left(\mathrm{B}\times \mathrm{C}\right).\mathrm{A}=\left(\mathrm{A}\times \mathrm{B}\right).\mathrm{C}=\mathrm{C}.\left(\mathrm{A}\times \mathrm{B}\right)=\left(\mathrm{C}\times \mathrm{A}\right).\mathrm{B}=\mathrm{B}.\left(\mathrm{C}\times \mathrm{A}\right)$

The determinant equivalent to$\left(‐\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)\right)$is obtained when one row is interchanged.

$\mathrm{B}.\left(\mathrm{A}\times \mathrm{C}\right)=\left(\mathrm{A}\times \mathrm{C}\right).\mathrm{B}=\left(\mathrm{B}\times \mathrm{A}\right).\mathrm{C}=\left(\mathrm{C}\times \mathrm{B}\right).\mathrm{A}=\mathrm{A}.\left(\mathrm{C}\times \mathrm{B}\right)=\mathrm{C}.\left(\mathrm{B}\times \mathrm{A}\right)=‐\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)$

Hence, the twelve triple scalar products involving A B and C are as follows.

$\mathrm{B}.\left(\mathrm{A}\times \mathrm{C}\right)=\left(\mathrm{A}\times \mathrm{C}\right).\mathrm{B}=\left(\mathrm{B}\times \mathrm{A}\right).\mathrm{C}=\left(\mathrm{C}\times \mathrm{B}\right).\mathrm{A}=\mathrm{A}.\left(\mathrm{C}\times \mathrm{B}\right)=\mathrm{C}.\left(\mathrm{B}\times \mathrm{A}\right)=‐\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)$

$\mathrm{A}.\left(\mathrm{B}\times \mathrm{C}\right)=\left(\mathrm{B}\times \mathrm{C}\right).\mathrm{A}=\left(\mathrm{A}\times \mathrm{B}\right).\mathrm{C}=\mathrm{C}.\left(\mathrm{A}\times \mathrm{B}\right)=\left(\mathrm{C}\times \mathrm{A}\right).\mathrm{B}=\mathrm{B}.\left(\mathrm{C}\times \mathrm{A}\right)$