If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that a + b + c = 0 => a x b = b x c = c x a

Question

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that $ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \Rightarrow \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a} $

So taking the first part with cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively

$$ \left\{\begin{array}{test} \overrightarrow{a} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = 0 \\ \overrightarrow{b} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = 0 \end{array}\right\} \Rightarrow \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a} $$

Is this way of thinking right, or the two equations have infinite solutions? I mean, no matter which vector we multiply with $( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} )$ it's going to be zero anyway. So is this a gap in our reasoning?

Answer 1

It looks from the comments that you did not understand the details
that are missing from your proof, so here is a start.

$\overrightarrow{a} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = \overrightarrow0$, hence $\overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{c} = \overrightarrow0$.
Use that $\overrightarrow{a} \times \overrightarrow{a}=\overrightarrow0$, and that $\overrightarrow{a} \times \overrightarrow{c}= -\overrightarrow{c} \times \overrightarrow{a}$ to obtain
$\overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{c} \times \overrightarrow{a} = \overrightarrow0$, and finally $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{a} $.
Then do a similar argument at least once more.

Answer 2

It's the right way of thinking. $\vec a+\vec b+\vec c=0$ can be understood geometrically as that the three vectors are coplanar. in that case, the cross product is a vector perpendicular to this plane, so at least $\vec a \times \vec b$, $ \vec b \times \vec c$, $\vec c \times \vec a$ point in the same direction. The magnitude of $\vec a \times \vec b$ is the area of parralelogram with sides $\vec a$ and $\vec b$. You have a similar interpretation for the other cross products. If $\vec c=-(\vec a+\vec b)$, you can also prove geometrically that the three areas are equal.

In conclusion, yes, you can multiply any vector with $\vec a+\vec b+\vec c$ and you still get $0$, but in general it's just a meaningless statement.