Q6P - Introduction to Electrodynamics

Question

Prove that. $\mathbf{[}\mathbf{A}\mathbf{\times }\left(B\times C\right)\mathbf{]}\mathbf{+}\mathbf{ }\mathbf{[}\mathbf{B}\mathbf{\times }\left(C\times A\right)\mathbf{]}\mathbf{+}\mathbf{[}\mathbf{C}\mathbf{\times }\left(A\times B\right)\mathbf{]}\mathbf{=}\mathbf{0}$ Under what conditions does $\mathit{A}\mathbf{\times }\left(B\times C\right)\mathbf{=}\left(A\times B\right)\mathbf{\times }\mathit{C}$ ?

Step-by-Step Solution

Answer

The value of $A\times \left(B\times C\right)+B\times \left(C\times A\right)+C\times \left(A\times B\right)$ proved to be equal to 0 . The given condition is possible only when vector  is either parallel or anti parallel to $\left(C\times A\right)$ .

Step 1: Apply the BAC-CAB rule to left side of the given equation

To prove, $\mathrm{A}\times (\mathrm{B}\times \mathrm{C})+ \mathrm{B}\times (\mathrm{C}\times \mathrm{A})+\mathrm{C}\times (\mathrm{A}\times \mathrm{B})=0$ , firstly evaluate its left side, ,and compare the result with 0.

 

The left side of the equation is $\mathrm{A}\times (\mathrm{B}\times \mathrm{C})+ \mathrm{B}\times (\mathrm{C}\times \mathrm{A})+\mathrm{C}\times (\mathrm{A}\times \mathrm{B})$, on using BAC-CAB rule, it is obtained as,

 

$$\begin{gathered} \mathrm{A}\times (\mathrm{B}\times \mathrm{C})+ \mathrm{B}\times (\mathrm{C}\times \mathrm{A})+\mathrm{C}\times (\mathrm{A}\times \mathrm{B})=\left[\begin{array}{c}B\left(A·C\right)-C\left(A·B\right)+C\left(B·A\right)\\ -A\left(B·C\right)+A\left(B·C\right)-B\left(A·C\right)\end{array}\right] \\ =0 \end{gathered}$$ 

Thus, it is proved that LHS=RHS.   

Step 2: Find whether cross product A × ( B × C ) follow commutative law or not

The cross product $\mathrm{A}\times \mathrm{B}$ is equal to $-\left(\mathrm{B}\times \mathrm{A}\right)$ . Thus, the given condition can rewritten as $\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=-\mathrm{C}\left(\mathrm{A}\times \mathrm{B}\right)$ . On using BAC-CAB rule, in $\mathrm{A}\times \left(\mathrm{B}\times \mathrm{C}\right)=-\mathrm{C}\left(\mathrm{A}\times \mathrm{B}\right)$ , the result is obtained as follows,

 

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

 

Rewrite as, 

 

$$\begin{gathered} -\left[\mathrm{C}\left(\mathrm{A}·\mathrm{B}\right)-\mathrm{A}\left(\mathrm{B}·\mathrm{C}\right)\right]=0 \\ -\mathrm{B}\times \left(\mathrm{C}\times \mathrm{A}\right)=0 \end{gathered}$$                       

 

If the two vectors are parallel or anti parallel with each other , then their cross product is zero. Thus the given condition is possible only when B is either parallel or anti parallel to $\left(\mathrm{C}\times \mathrm{A}\right)$ .