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    I need to prove (a\times b)\times c=(a.c)b-(b.c)a

    I have these in my notes:
    starting from

    (a\times b)\times c=\lambda a+\mu b

    (a\times b)\times c=(\lambda')(b.c)a+\mu'(a.c)b
    wherea,b,c are vectors
    I dont follow any of these from the begining ..any help ..
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    We proved this using summation convention - if you're familiar with that, the proof mainly relies on the following identity:

     \epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}

    where  \epsilon_{abc} is the Levi-Civita symbol and  \delta_{ab} is the Kronecker delta symbol.
    (Original post by rbnphlp)
    I need to prove (a\times b)\times c=(a.c)b-(b.c)a

    I have these in my notes:
    starting from

    (a\times b)\times c=\lambda a+\mu b

    (a\times b)\times c=(\lambda')(b.c)a+\mu'(a.c)b
    wherea,b,c are vectors
    I dont follow any of these from the begining ..any help ..
    The statements you're starting with just say that the vector on the LHS lies in the plane containing a and b; this is true since the vector is perp. to axb. Then the next statement just rescales the factors. Considering dotting the second statement with c.
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    That's the best way.

    However there are no intuitive proofs of this - you MUST calculate in the end (reasonably) well e.g. careful algebra.
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    (Original post by Unbounded)
    We proved this using summation convention - if you're familiar with that, the proof mainly relies on the following identity:

     \epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}

    where  \epsilon_{abc} is the Levi-Civita symbol and  \delta_{ab} is the Kronecker delta symbol.

    The statements you're starting with just say that the vector on the LHS lies in the plane containing a and b; this is true since the vector is perp. to axb. Then the next statement just rescales the factors. Considering dotting the second statement with c.
    thanks ..what do you mean by rescaling the factors?I dont understand how (b.c)lambda'=lambda
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    (Original post by Unbounded)
    We proved this using summation convention - if you're familiar with that, the proof mainly relies on the following identity:

     \epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}

    where  \epsilon_{abc} is the Levi-Civita symbol and  \delta_{ab} is the Kronecker delta symbol.

    The statements you're starting with just say that the vector on the LHS lies in the plane containing a and b; this is true since the vector is perp. to axb. Then the next statement just rescales the factors. Considering dotting the second statement with c.
    I think this may be assuming a little much, no offence intended to OP. Are you proving it in 3 dimensions? If so, probably the most straightforward way (despite being a little time-consuming) is to expand out the left- and right-hand sides as vectors, and see what comes out. For example, assuming the definition of vector product:

    \left( \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times  \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\right) \times \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} =  \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix} \times \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} = ...

    and continue until you have one vector. Expand the right-hand side out until you have one vector, and compare the two; they will be equal.
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    (Original post by rbnphlp)
    thanks ..what do you mean by rescaling the factors?I dont understand how (b.c)lambda'=lambda
    a.b is a scalar so lamda/(a.b) = lamda' is a new scalar you should consider.

    Make sure (trivial) that you consider when a or b or c is zero.
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    (Original post by DeanK22)
    a.b is a scalar so lamda/(a.b) = lamda' is a new scalar you should consider.

    Make sure (trivial) that you consider when a or b or c is zero.
    thanks dean..I think I get the idea!

    (Original post by dreameater)
    thanks for the alternative
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      (Original post by Dream Eater)
      Are you proving it in 3 dimensions?
      For completeness, this identity doesn't hold for the seven-dimensional cross product.
     
     
     
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