How about this.
Warning: I'm relying on the identity . This is proved by considering the identity .
First assume that , and are linearly dependent. WLOG assume there are scalars , with .
Then and and so
, proving the result in this case.
Now assume that , and are linearly independent, thus forming a basis for .
We have and .
Hence .
Similarly and .
Since , and form a basis and is orthogonal to all of them, the result follows.
Here is a proof. Consider the special orthogonal group as a Lie group. Its Lie algebra is the set of skew-symmetric matrices, and the Lie bracket is the commutator . A skew-symmetric matrix can be associated in a natural way with a vector in , and it is easy to check that the vector associated to the bracket is . By the Jacobi identity for the Lie bracket, the result follows.
this is a clever solution with lots of details! in simple words: first we note that if where is an associative ring, and if we define for all then the
Jacobi identity holds, i.e. now let be the set of skew-symmetric matrices with real entries, i.e. consists of all matrices
in the form define the map by clearly is an isomorphism of vector spaces and it's easy to see that call this result since is an associative ring, we have
taking of both sides and using the linearity of and gives us: which completes the proof.