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I want to show that this action not depend by A and B vectors, I know that cross product of the same vector by itself is $0$. $$\left((\vec A+\vec B) \times (\vec A\times \vec B)\right)\cdot(\vec A \times \vec B)$$ I can use here in Associative property?
like : $$\left((\vec A \times \vec B) \times (\vec A\times \vec B)\right)\cdot (\vec A+ \vec B)$$ then its zero.
any suggestions? Thanks!

Ofir Attia
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The vector cross product is not associative, but it does distribute over addition:

$$(A+B)\times (A \times B) = A \times (A \times B) + B \times (A \times B).$$

The dot product operation is commutative. So you might have some success in applying the property

$$\alpha\cdot (\beta \times \gamma) = \beta \cdot (\gamma \times \alpha) = \gamma \cdot (\alpha \times \beta).$$

Using this, let $C = A \times B$, and we find

$$\begin{align*} \left((A + B)\times (C)\right) \cdot (C) &= \left(A\times (C) + B \times (C)\right)\cdot (C) \\ &= (C)\cdot (A\times (C)) + (C)\cdot (B\times (C)) \\ &= A \cdot ((C)\times (C)) + B\cdot ((C)\times (C)) \\ &= 0. \end{align*}$$

Emily
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  • from your first line I can say that there is two $0$ there, vector $\times$ the same vector will give $0$, so $\cdot$(dot) it with another vector will give zero. right? – Ofir Attia Jul 29 '13 at 20:02
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    @OfirAttia No, that's not true. You don't know if it's zero yet. You have $A\times C + B \times C$, with $C = A \times B$. You cannot yet conclude that it is zero. – Emily Jul 29 '13 at 20:04
  • Ok, there is another way to show it? short one? I know its zero no matter what, how I can write it? – Ofir Attia Jul 29 '13 at 20:06
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    Never mind, this is nice answer, Thanks! – Ofir Attia Jul 29 '13 at 20:09