Chapter 10, Problem 32P

To determine

To Compute: $\overrightarrow{A}.\left(\overrightarrow{B}\times \overrightarrow{C}\right)=\overrightarrow{C}.\left(\overrightarrow{A}\times \overrightarrow{B}\right)=\overrightarrow{B}.\left(\overrightarrow{C}\times \overrightarrow{A}\right)$

Explanation of Solution

Given:

  $\begin{array}{c}\overrightarrow{A}=a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\\ \overrightarrow{B}=b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\\ \overrightarrow{C}=c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\end{array}$

Formula used:

The property of vector cross product:

  $\begin{array}{l}·\widehat{i}\times \widehat{j}=\widehat{k},\widehat{j}\times \widehat{k}=\widehat{i},\widehat{k}\times \widehat{i}=\widehat{j}\\ ·\widehat{j}\times \widehat{i}=-\widehat{k},\widehat{k}\times \widehat{j}=-\widehat{i},\widehat{i}\times \widehat{k}=-\widehat{j}\\ ·\widehat{i}\times \widehat{i}=0,\widehat{j}\times \widehat{j}=0,\widehat{k}\times \widehat{k}=0\end{array}$

The properties of vector dot product:

  $\begin{array}{l}·\widehat{i}\text{.}\widehat{j}=0,\widehat{j}\text{.}\widehat{k}=0,\widehat{k}\text{.}\widehat{i}=0,\\ ·\widehat{j}\text{.}\widehat{i}=0,\widehat{k}\text{.}\widehat{j}=0,\widehat{i}\text{.}\widehat{k}=0\\ ·\widehat{i}\text{. }\widehat{i}=1,\widehat{j}\text{. }\widehat{j}=1,\widehat{k}\text{. }\widehat{k}=1\end{array}$

Calculation:

Calculation of $\overrightarrow{A}.\left(\overrightarrow{B}\times \overrightarrow{C}\right)$ :

$\begin{array}{c}\overrightarrow{A}.\left(\overrightarrow{B}\times \overrightarrow{C}\right)=\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right).\left[\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right)\times \left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right)\right]\\ =\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right).\left[\begin{array}{l}{b}_x{c}_x\left(\widehat{i}\times \widehat{i}\right)+b_x{c}_y\left(\widehat{i}\times \widehat{j}\right)+b_x{c}_z\left(\widehat{i}\times \widehat{k}\right)+b_y{c}_x\left(\widehat{j}\times \widehat{i}\right)+b_y{c}_y\left(\widehat{j}\times \widehat{j}\right)+b_y{c}_z\left(\widehat{j}\times \widehat{k}\right)\\ +b_z{c}_x\left(\widehat{k}\times \widehat{i}\right)+b_z{c}_y\left(\widehat{k}\times \widehat{j}\right)+b_z{c}_z\left(\widehat{k}\times \widehat{k}\right)\end{array}\right]\end{array}$

  $\begin{array}{c}=\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right).\left[\begin{array}{l}{b}_x{c}_x\left(0\right)+b_x{c}_y\left(\widehat{k}\right)+b_x{c}_z\left(-\widehat{j}\right)+b_y{c}_x\left(-\widehat{k}\right)+b_y{c}_y\left(0\right)+b_y{c}_z\left(\widehat{i}\right)\\ +b_z{c}_x\left(-\widehat{j}\right)+b_z{c}_y\left(-\widehat{i}\right)+b_z{c}_z\left(0\right)\end{array}\right]\\ =\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right).\left[b_x{c}_y\left(\widehat{k}\right)+b_x{c}_z\left(-\widehat{j}\right)+b_y{c}_x\left(-\widehat{k}\right)+b_y{c}_z\left(\widehat{i}\right)+b_z{c}_x\left(-\widehat{j}\right)+b_z{c}_y\left(-\widehat{i}\right)\right]\end{array}$

  $\begin{array}{c}=\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right).\left[\left(b_y{c}_z-b_z{c}_y\right)\left(\widehat{i}\right)-\left(b_x{c}_z+b_z{c}_x\right)\left(\widehat{j}\right)+\left(b_x{c}_y-b_y{c}_x\right)\left(\widehat{k}\right)\right]\\ =a_x\left(b_y{c}_z-b_z{c}_y\right)\left(\widehat{i}.\widehat{i}\right)-a_y\left(b_x{c}_z+b_z{c}_x\right)\left(\widehat{j}.\widehat{j}\right)+a_z\left(b_x{c}_y-b_y{c}_x\right)\left(\widehat{k}.\widehat{k}\right)\end{array}$

  $\begin{array}{c}=a_x\left(b_y{c}_z-b_z{c}_y\right)-a_y\left(b_x{c}_z+b_z{c}_x\right)+a_z\left(b_x{c}_y-b_y{c}_x\right)\\ =a_x{b}_y{c}_z-a_x{b}_z{c}_y-a_y{b}_x{c}_z-a_y{b}_z{c}_x+a_z{b}_x{c}_y-a_z{b}_y{c}_x\mathrm{...............}\text{(i)}\end{array}$

  $\begin{array}{c}\overrightarrow{C}.\left(\overrightarrow{A}\times \overrightarrow{B}\right)=\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right).\left[\left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right)\times \left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right)\right]\\ =\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right).\left[\begin{array}{l}{a}_x{b}_x\left(\widehat{i}\times \widehat{i}\right)+a_x{b}_y\left(\widehat{i}\times \widehat{j}\right)+a_x{b}_z\left(\widehat{i}\times \widehat{k}\right)+a_y{b}_x\left(\widehat{j}\times \widehat{i}\right)+a_y{b}_y\left(\widehat{j}\times \widehat{j}\right)+a_y{b}_z\left(\widehat{j}\times \widehat{k}\right)\\ +a_z{b}_x\left(\widehat{k}\times \widehat{i}\right)+a_z{b}_y\left(\widehat{k}\times \widehat{j}\right)+a_z{b}_z\left(\widehat{k}\times \widehat{k}\right)\end{array}\right]\end{array}$

  $\begin{array}{c}=\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right).\left[\begin{array}{l}{a}_x{b}_x\left(0\right)+a_x{b}_y\left(\widehat{k}\right)+a_x{b}_z\left(-\widehat{j}\right)+a_y{b}_x\left(-\widehat{k}\right)+a_y{b}_y\left(0\right)+a_y{b}_z\left(\widehat{i}\right)\\ +a_z{b}_x\left(-\widehat{j}\right)+a_z{b}_y\left(-\widehat{i}\right)+a_z{b}_z\left(0\right)\end{array}\right]\\ =\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right).\left[a_x{b}_y\left(\widehat{k}\right)+a_x{b}_z\left(-\widehat{j}\right)+a_y{b}_x\left(-\widehat{k}\right)+a_y{b}_z\left(\widehat{i}\right)+a_z{b}_x\left(-\widehat{j}\right)+a_z{b}_y\left(-\widehat{i}\right)\right]\end{array}$

  $\begin{array}{c}=\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right).\left[\left(a_y{b}_z-a_z{b}_y\right)\left(\widehat{i}\right)-\left(a_x{b}_z+a_z{b}_x\right)\left(\widehat{j}\right)+\left(a_x{b}_y-a_y{b}_x\right)\left(\widehat{k}\right)\right]\\ =c_x\left(a_y{b}_z-a_z{b}_y\right)\left(\widehat{i}.\widehat{i}\right)-c_y\left(a_x{b}_z+a_z{b}_x\right)\left(\widehat{j}.\widehat{j}\right)+c_z\left(a_x{b}_y-a_y{b}_x\right)\left(\widehat{k}.\widehat{k}\right)\\ =c_x\left(a_y{b}_z-a_z{b}_y\right)-c_y\left(a_x{b}_z+a_z{b}_x\right)+c_z\left(a_x{b}_y-a_y{b}_x\right)\end{array}$

  $\begin{array}{c}=c_x{a}_y{b}_z-c_x{a}_z{b}_y-c_y{a}_x{b}_z-c_y{a}_z{b}_x+c_z{a}_x{b}_y-c_z{a}_y{b}_x\\ =a_x{b}_y{c}_z-a_x{b}_z{c}_y-a_y{b}_x{c}_z-a_y{b}_z{c}_x+a_z{b}_x{c}_y-a_z{b}_y{c}_x\mathrm{...............}\text{(ii)}\end{array}$

  $\begin{array}{c}\overrightarrow{B}.\left(\overrightarrow{C}\times \overrightarrow{A}\right)=\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right).\left[\left(c_x\widehat{i}+c_y\widehat{j}+c_z\widehat{k}\right)\times \left(a_x\widehat{i}+a_y\widehat{j}+a_z\widehat{k}\right)\right]\\ =\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right).\left[\begin{array}{l}{c}_x{a}_x\left(\widehat{i}\times \widehat{i}\right)+c_x{a}_y\left(\widehat{i}\times \widehat{j}\right)+c_x{a}_z\left(\widehat{i}\times \widehat{k}\right)+c_y{a}_x\left(\widehat{j}\times \widehat{i}\right)+c_y{a}_y\left(\widehat{j}\times \widehat{j}\right)+c_y{a}_z\left(\widehat{j}\times \widehat{k}\right)\\ +c_z{a}_x\left(\widehat{k}\times \widehat{i}\right)+c_z{a}_y\left(\widehat{k}\times \widehat{j}\right)+c_z{a}_z\left(\widehat{k}\times \widehat{k}\right)\end{array}\right]\end{array}$

$=\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right).\left[\begin{array}{l}{c}_x{a}_x\left(0\right)+c_x{a}_y\left(\widehat{k}\right)+c_x{a}_z\left(-\widehat{j}\right)+c_y{a}_x\left(-\widehat{k}\right)+c_y{a}_y\left(0\right)+c_y{a}_z\left(\widehat{i}\right)\\ +c_z{a}_x\left(-\widehat{j}\right)+c_z{a}_y\left(-\widehat{i}\right)+c_z{a}_z\left(0\right)\end{array}\right]$

  $\begin{array}{c}=\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right).\left[c_x{a}_y\left(\widehat{k}\right)+c_x{a}_z\left(-\widehat{j}\right)+c_y{a}_x\left(-\widehat{k}\right)+c_y{a}_z\left(\widehat{i}\right)+c_z{a}_x\left(-\widehat{j}\right)+c_z{a}_y\left(-\widehat{i}\right)\right]\\ =\left(b_x\widehat{i}+b_y\widehat{j}+b_z\widehat{k}\right).\left[\left(c_y{a}_z-c_z{a}_y\right)\left(\widehat{i}\right)-\left(c_x{a}_z+c_z{a}_x\right)\left(\widehat{j}\right)+\left(c_x{a}_y-c_y{a}_x\right)\left(\widehat{k}\right)\right]\end{array}$

  $\begin{array}{c}=b_x\left(c_y{a}_z-c_z{a}_y\right)\left(\widehat{i}.\widehat{i}\right)-b_y\left(c_x{a}_z+c_z{a}_x\right)\left(\widehat{j}.\widehat{j}\right)+b_z\left(c_x{a}_y-c_y{a}_x\right)\left(\widehat{k}.\widehat{k}\right)\\ =b_x\left(c_y{a}_z-c_z{a}_y\right)-b_y\left(c_x{a}_z+c_z{a}_x\right)+b_z\left(c_x{a}_y-c_y{a}_x\right)\\ =b_x{c}_y{a}_z-b_x{c}_z{a}_y-b_y{c}_x{a}_z-b_y{c}_z{a}_x+b_z{c}_x{a}_y-b_z{c}_y{a}_x\\ =a_x{b}_y{c}_z-a_x{b}_z{c}_y-a_y{b}_x{c}_z-a_y{b}_z{c}_x+a_z{b}_x{c}_y-a_z{b}_y{c}_x\mathrm{...............}\text{(iii)}\end{array}$

Equations (i), (ii) and (iii) are equal.

Therefore,

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

Conclusion:

Hence, $\overrightarrow{A}.\left(\overrightarrow{B}\times \overrightarrow{C}\right)=\overrightarrow{C}.\left(\overrightarrow{A}\times \overrightarrow{B}\right)=\overrightarrow{B}.\left(\overrightarrow{C}\times \overrightarrow{A}\right)$