Believe it or not, there are even more property in the cross product. This time our goal is the BAC-CAB identity
$ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) $.
To prove this, we need this result first, the so-called contraction identity.
$ \epsilon_{ijk}\epsilon_{ist} = \delta_{js}\delta_{kt} - \delta_{jt}\delta_{ks} $.
To be honest, I don't really understand how does one come up with the identity, but it is relatively easy to verify that it is correct, and under what condition (that $ i $ does not equal any of $ j, k, s, t $
With that, we express the right hand side as follow:
$ \begin{eqnarray*} \vec{a} \times (\vec{b} \times \vec{c}) &=& \vec{a} \times (\epsilon_{pqr}b_pc_qe_r) \\ &=& \epsilon_{srt}a_s\epsilon_{pqr}b_pc_qe_t \\ &=& \epsilon_{srt}\epsilon_{pqr}a_sb_pc_qe_t \\ &=& \epsilon_{rts}\epsilon_{rpq}a_sb_pc_qe_t \\ &=& (\delta_{tp}\delta_{sq} - \delta_{tq}\delta_{sp})a_sb_pc_qe_t \\ &=& \delta_{tp}\delta_{sq}a_sb_pc_qe_t - \delta_{tq}\delta_{sp}a_sb_pc_qe_t \\ &=& a_sb_tc_se_t - a_sb_sc_te_t \\ &=& \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) \end{eqnarray*} $
$ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) $.
To prove this, we need this result first, the so-called contraction identity.
$ \epsilon_{ijk}\epsilon_{ist} = \delta_{js}\delta_{kt} - \delta_{jt}\delta_{ks} $.
To be honest, I don't really understand how does one come up with the identity, but it is relatively easy to verify that it is correct, and under what condition (that $ i $ does not equal any of $ j, k, s, t $
With that, we express the right hand side as follow:
$ \begin{eqnarray*} \vec{a} \times (\vec{b} \times \vec{c}) &=& \vec{a} \times (\epsilon_{pqr}b_pc_qe_r) \\ &=& \epsilon_{srt}a_s\epsilon_{pqr}b_pc_qe_t \\ &=& \epsilon_{srt}\epsilon_{pqr}a_sb_pc_qe_t \\ &=& \epsilon_{rts}\epsilon_{rpq}a_sb_pc_qe_t \\ &=& (\delta_{tp}\delta_{sq} - \delta_{tq}\delta_{sp})a_sb_pc_qe_t \\ &=& \delta_{tp}\delta_{sq}a_sb_pc_qe_t - \delta_{tq}\delta_{sp}a_sb_pc_qe_t \\ &=& a_sb_tc_se_t - a_sb_sc_te_t \\ &=& \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) \end{eqnarray*} $
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