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The question I am trying to show under what conditions

$$\vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\times\vec{B})\times\vec{C}.$$

I have found that right-hand side of the above equation is equal to

\begin{align} (\vec{A}\times\vec{B})\times\vec{C} &=-\vec{C}\times(\vec{A}\times\vec{B})\\ &= \vec{C}\times(\vec{B}\times\vec{A}). \end{align}

This is similar to the left-hand side of the original equation.

The conclusion I arrived at was, in order for the equality to be true, either $\vec{A}, \vec{B}$ or $\vec{C}$ have to be zero, or $\vec{A}$ must be equal to $\vec{C}$.

Is this correct? If not, or if I am missing anything, please let me know.

achacttn
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  • What do you think happens when $\vec A = \lambda \vec C$ for some scalar $\lambda$? – Arthur Jul 25 '13 at 14:16
  • I think that since multiplication by scalar would be distributive, it would also be an additional condition for equality. Is that correct? – achacttn Jul 25 '13 at 14:42

1 Answers1

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Your reasoning is correct. The reason is that the cross product is anti-commutative. Therefore, the only situations wherein you will find the desired properties are when one of the vectors is zero, or when $A \propto C$

To prove this, note that $A\times (B\times C) = (A\cdot C)B-(A\cdot B)C$ and likewise $(A\times B)\times C = -(C\cdot B)A+(C\cdot A)B$

Assume the two are equal, and note that the dot product commutes, and we find

$$ (A\cdot B) C = (C\cdot B) A $$

Then, $$\frac{A}{A\cdot B} = \frac{C}{C \cdot B} \implies C = kA.$$

So $C$ is some scalar multiple of $A$.

Community
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Emily
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    Thanks. I have a few quick followup questions.
    1. Is there any way to derive the AX(BXC) = BAC-CAB or is it just a shortcut that cuts down on tedious algebraic manipulation?
    2. Where does the $k$ come from at the very end, that says that another possible solution is that $C$ is a scalar multiple of $A$ and not just equal?
     – achacttn Jul 25 '13 at 14:24
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    @achacttn Since $k = \frac{A\cdot B}{C\cdot B}$. The dot product is just a scalar number, so the ratio of dot products is a scalar number. In this case, we're looking for conditions of equality; $A,B,C$ aren't variables, so there's nothing lost by saying $C = \frac{A\cdot B}{C\cdot B} A$. The triple product relation I used can be found by doing out the algebra; it's called the "triple product expansion", see also http://en.wikipedia.org/wiki/Triple_product – Emily Jul 25 '13 at 14:27
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    @Emily I think there's another possibility when the triple product is associative: when the vector $B$ is orthogonal to both $A$ and $C$. In your prove you assumed that $A\cdot B$ and $C\cdot B$ are both non-zero; which might not be the case. – It'sMe Sep 14 '21 at 18:51