# How these are valid if the coordinate is a right-handed coordinate system?

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

(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.
• 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.