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

[Ask4material]

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

formula: \hat{b}(\hat{a}\cdot\hat{c})-\hat{c}(\hat{a}\cdot\hat{b})=\hat{a}\times(\hat{b}\times\hat{c})

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

\hat{b}r+\hat{c}s=\hat{a}\times(\hat{b}\times\hat{c})

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

\hat{a}\cdot both sides:

\hat{a}\cdot(\hat{b}r+\hat{c}s)=\hat{a}\cdot[\hat{a}\times(\hat{b}\times\hat{c})]

(\hat{a}\cdot\hat{b})r+(\hat{a}\cdot\hat{c})s=0

It seems needing another condition to distinguish \hat{a}\times(\hat{b}\times\hat{c}) and (\hat{b}\times\hat{c})\times\hat{a}

 

[CompuChip]

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

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 a = \hat i, b = \hat j, c = \hat k).