# Test of handedness

I'm reading a book on linear algebra, where the author gives a method to test the handedness or chirality of a given set of 3 basis vectors.

if (v1 x v2) . v3 > 0 then it's right-handed, while if it's less than 0, it's left handed.

While what beats me is that numbers are just numbers, left or right handedness of a system depends on the viewer and how he interprets the given data.

Taking the canonical i, j, k basis vectors, in both left and right handed systems i x j = k, thereby k.k = ||k||^2 > 0 (always), then how does this test hold true?

Orientation (handedness) is not about a set of vectors, it is about an ordered list of vectors. That is, a certain ordering, $(i,j,k)$ is agreed to as right handed. Then $(j,i,k)$ is left handed. This may or may not agree with some notion you have from physics, hard to predict.
• @legends2k, $j \times i = -k,$ so $(j \times i) \cdot k < 0,$ so the ordered basis $(j,i,k)$ has the opposite orientation compared with $(i,j,k).$ Looking at your question again, I do not know what you mean by a "left or right handed system," so I would say the use of the word "system" and the phrase "depends on the viewer" indicate thinking from some view of this that is not that of linear algebra or differential geometry texts. Perhaps computer graphics? There is no viewer in a vector space. – Will Jagy Mar 11 '13 at 23:41