# The proof of $\hat{b}(\hat{a}\cdot\hat{c})-\hat{c}(\hat{a}\cdot\hat{b})=\hat{a}\times(\hat{b}\times\hat{c})$

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:

$r=(\hat{a}\cdot\hat{c})$

$s=-(\hat{a}\cdot\hat{b})$

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

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+1: for showing what you have done... –  Fabian Jan 11 '12 at 10:41
+1 : another 1 for interesting questions –  dato datuashvili Jan 11 '12 at 10:58

Try with multiplying with $\hat{b}$ and $\hat{c}$ and use the cyclic property $\hat{a} \cdot (\hat{b} \times \hat{c}) = \hat{b} \cdot (\hat{c} \times \hat{a})$ of the triple product.