I know the vector identity A → × ( B → × C → ) = ( A → ⋅ C → ) B → − ( A → ⋅ B → ) C → Now, is there a succinct way of obtaining | A → × ( B → × C → ) | 2 using vector algebra? I know we can expand, multiply and group the terms back, but is there a neater way of obtaining the result?Using Levi-Civita symbols, would this be easier?

Question

I know the vector identity             A      →        ×  (            B      →        ×            C      →        )  =  (            A      →        ⋅            C      →        )            B      →        −  (            A      →        ⋅            B      →        )            C      →      Now, is there a succinct way of obtaining       |              A      →        ×  (            B      →        ×            C      →        )            |        2   using vector algebra? I know we can expand, multiply and group the terms back, but is there a neater way of obtaining the result?Using Levi-Civita symbols, would this be easier?

Szulikto · Accepted Answer

Use triple product square equation or exterior algebra inner product to get                    (1)                    (                              A            →                          ⋅        (                              B            →                          ×                              C            →                          )                  )          2                =                  |                                                                                          A                    →                                                  ⋅                                                      A                    →                                                                                                                    A                    →                                                  ⋅                                                      B                    →                                                                                                                    A                    →                                                  ⋅                                                      C                    →                                                                                                                                            B                    →                                                  ⋅                                                      A                    →                                                                                                                    B                    →                                                  ⋅                                                      B                    →                                                                                                                    B                    →                                                  ⋅                                                      C                    →                                                                                                                                            C                    →                                                  ⋅                                                      A                    →                                                                                                                    C                    →                                                  ⋅                                                      B                    →                                                                                                                    C                    →                                                  ⋅                                                      C                    →                                                                                |                .            Now we also have the equation                    (2)                              |                                      A            →                          ×        (                              B            →                          ×                              C            →                          )                              |                    2                =                  |                                      A            →                                                |                    2                                   |                                      B            →                          ×                              C            →                                                |                    2                −        (                              A            →                          ⋅        (                              B            →                          ×                              C            →                          )                  )          2                    and a little bit of inspection leads to the result                    (3)                              |                                      A            →                          ×        (                              B            →                          ×                              C            →                          )                              |                    2                =        −                  |                                                    0                                                                                  A                    →                                                  ⋅                                                      B                    →                                                                                                                    A                    →                                                  ⋅                                                      C                    →                                                                                                                                            B                    →                                                  ⋅                                                      A                    →                                                                                                                    B                    →                                                  ⋅                                                      B                    →                                                                                                                    B                    →                                                  ⋅                                                      C                    →                                                                                                                                            C                    →                                                  ⋅                                                      A                    →                                                                                                                    C                    →                                                  ⋅                                                      B                    →                                                                                                                    C                    →                                                  ⋅                                                      C                    →                                                                                |                .            The advantage of exterior algebra is that it works in                 R        n       while the usual cross product is restricted to                 R        3    .

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