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[Help]The proof of (a.c)b-(a.b)c=aX(bXc)?

  1. Jan 11, 2012 #1
    [Help]The proof of (a.c)b-(a.b)c=aX(bXc)

    formula: [itex]\hat{b}(\hat{a}\cdot\hat{c})-\hat{c}(\hat{a}\cdot\hat{b})=\hat{a}\times(\hat{b}\times\hat{c})[/itex]

    [itex]\hat{a}\times(\hat{b}\times\hat{c})[/itex] is on the [itex]\hat{b}[/itex], [itex]\hat{c}[/itex] plane, so:


    want to proof:[itex]\begin{array}{l}r=(\hat{a}\cdot\hat{c})\\s=-(\hat{a}\cdot\hat{b})\end{array}[/itex]

    [itex]\hat{a}\cdot[/itex] both sides:



    It seems needing another condition to distinguish [itex]\hat{a}\times(\hat{b}\times\hat{c})[/itex] and [itex](\hat{b}\times\hat{c})\times\hat{a}[/itex]
    Last edited: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2


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    All you have now is that
    [tex]r = - s \frac{\hat a \cdot \hat c}{\hat a \cdot \hat b}[/tex]

    One obvious choice is to set s = -(a . b) and then r = (a . c), but you can also set s = (a . b) or s = 1, or any other value. If you look at this geometrically, you will find the one-dimensional subspace spanned by a x (b x c).

    You can cut it down to two options by preserving the length, which will give you r2 + s2 = 1. And you should be able to discern between the two of those by also preserving the direction (e.g. take [itex]a = \hat i, b = \hat j, c = \hat k[/itex]).
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