(adsbygoogle = window.adsbygoogle || []).push({}); [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:

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

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:

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

[itex](\hat{a}\cdot\hat{b})r+(\hat{a}\cdot\hat{c})s=0[/itex]

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]

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

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