I am trying to understand how $(1),(2) \text{ and } (3)$ are valid if the coordinate is a right-handed coordinate system. The definition of a right-handed coordinate system is a one such that $\overrightarrow{i} \times \overrightarrow{j} = \overrightarrow{k}$ is valid. To me, to prove that I must compute all possibilities to verify that result. Is there any other method?

$$\overrightarrow{i} \times \overrightarrow{j} = \overrightarrow{k} \qquad(1)\\ \overrightarrow{j} \times \overrightarrow{k} = \overrightarrow{i} \qquad(2)\\ \overrightarrow{k} \times \overrightarrow{i} = \overrightarrow{j} \qquad(3)$$

I want to very that $(1),(2) \text{ and } (3)$ is valid as long as it is a right-handed coordinate system. what I mean by possibilities is to try all the different ways that $x,y$ and $z$ can be. For example:

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  • 1
    $\begingroup$ Can you clarify exactly what you're looking for here? What are you trying to prove? What possibilities are you referring to? $\endgroup$
    – J. Murray
    Apr 9, 2020 at 19:17
  • $\begingroup$ @J. Murray I have adjusted it. Please, tell me if you need more clarification. $\endgroup$
    – Ali
    Apr 9, 2020 at 19:54

1 Answer 1


(1) holds in a right-handed coordinate system by definition. To show that (2) and (3) also hold, replace $\vec k$ by $\vec{i}\times\vec{j}$ in them and then expand the double cross product.

  • $\begingroup$ Can you please explain what the double cross product is? And how it can be applied (some derivation)? $\endgroup$
    – Ali
    Apr 9, 2020 at 20:03
  • $\begingroup$ I’ve added a link with the relevant identities. They are a standard part of vector algebra. $\endgroup$
    – G. Smith
    Apr 9, 2020 at 20:17
  • $\begingroup$ In the proof they assume that $\overrightarrow{j} \times \overrightarrow{k} = \overrightarrow{i} \qquad(2) \\ \overrightarrow{k} \times \overrightarrow{i} = \overrightarrow{j} \qquad(3)$ are correct, and this is what I want to prove. $\endgroup$
    – Ali
    Apr 9, 2020 at 21:19
  • $\begingroup$ You can assume any of the three and then prove the other two. You said that (1) was the definition of a right-handed system, but if you want to assume (2) instead of (1) it works similarly. $\endgroup$
    – G. Smith
    Apr 9, 2020 at 21:25
  • $\begingroup$ Physics Chat chat.stackexchange.com/rooms/106530/cross-product $\endgroup$
    – Ali
    Apr 9, 2020 at 21:42

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