Question

prove the BAC-CAB identity using these equations

Tmn eimm 26 jmn eijk6. ikevh ijk

Answer 1

The BAC - CAB identity is
  $\vec{A}\times (\vec{B}\times \vec{C})=\vec{B} (\vec{A}. \vec{C})-\vec{C} (\vec{A}. \vec{B})$

We use the fact that
   $\vec{A}=A_i\hat{e}_i=A_i\hat{e}_{j}\delta_{ij}$
And
  $\vec{A}.\vec{B}=A_iB_i=A_iB_j\delta_{ij}$
And
   $\hat{e}_i.\hat{e}_j=\delta_{ij}$
And
  $\hat{e}_i\times \hat{e}_j=\epsilon_{ijk}\hat{e}_k$
And so,
   $\vec{A}\times (\vec{B}\times \vec{C})=A_i\hat{e}_i\times (B_j\hat{e}_j\times C_k\hat{e}_k)$
  $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=A_iB_jC_k\hat{e}_i\times (\hat{e}_j\times \hat{e}_k)$
  $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=A_iB_jC_k\hat{e}_i\times \epsilon_{jkl}\hat{e}_l$
  $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=A_iB_jC_k\epsilon_{jkl}\hat{e}_i\times \hat{e}_l$
   $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=A_iB_jC_k\epsilon_{jkl}\epsilon_{ilm}\hat{e}_m$
   $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=A_iB_jC_k(-\delta_{ij}\delta_{km}+\delta_{ik}\delta_{jm})\hat{e}_m$
   $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=-(A_iB_j\delta_{ij})(C_k\delta_{km}\hat{e}_m)+(A_{i}C_k\delta_{ik})(\delta_{jm}B_j\hat{e}_m)$
  $\Rightarrow \vec{A}\times (\vec{B}\times \vec{C})=-(\vec{A}.\vec{B})\vec{C}+ (\vec{A}.\vec{C})\vec{B}$