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

1. Jan 11, 2012

[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 $(\hat{b}\times\hat{c})\times\hat{a}$

Last edited: Jan 11, 2012
2. Jan 11, 2012

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