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I have this exercise that is difficult for me. $a, b, c$ are three $3$-vectors and $$(a+b-c)\cdot(a-b+c) \wedge (-a+b+c) = -4a\cdot b\wedge c$$ where $\cdot$ is the dot product, $\wedge$ is cross product. Could you explain this result?

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    $\begingroup$ @DanielRust I think only one way makes sense (cross products first). $\endgroup$ – Jonathan Y. Sep 30 '13 at 15:00
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Using multiple times the bilinearity of these products and simple identities like $x\cdot x\wedge y=0$ and $x\wedge x=0$ we get $$(a+b-c)\cdot(a-b+c) \wedge (-a+b+c) = (a+b-c)\cdot[2(a-b) \wedge c] = (a+b)\cdot[2(a-b) \wedge c]= (2a-(a-b))\cdot[2(a-b) \wedge c] = 2a\cdot[2(a-b) \wedge c] = -4a\cdot b\wedge c.$$

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  • $\begingroup$ Thanks but why (a−b+c)∧(−a+b+c)= 2(a−b)∧c? $\endgroup$ – user40267 Sep 30 '13 at 15:57
  • $\begingroup$ $((a-b)+c) \wedge (c-(a-b)) = (a-b) \wedge c - c \wedge (a-b) -(a-b) \wedge (a-b) + c \wedge c = 2(a-b) \wedge c$. $\endgroup$ – njguliyev Sep 30 '13 at 17:03

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