In a question I am reading, the following question appears.
What if $\vec{A},\vec{B}$, and $\vec{C}$ are mutually perpendicular and form a right-handed set?
What exactly does "form a right-handed set" mean? That $\hat{A}\times\hat{B}=\hat{C}$? Or does it mean the stronger equality $\vec{A}\times \vec{B}=\vec{C}$?
We are accustomed to embed our geometrical figures into the world around us, and in this environment the idea of "righhandedness" makes sense. But mathematically relevant is only "samehandedness", as follows:
Call two triples $(a_0,b_0,c_0)$, $\>(a_1,b_1,c_1)$ of linearly independent vectors in ${\mathbb R}^3$ (or any three-dimensional real vector space) equivalent if the first triple can be continuously deformed into the second, such that at all times $t\in[0,1]$ the three vectors $a(t)$, $b(t)$, $c(t)$ are linearly independent. It turns out that there are exactly two equivalence classes, and that two triples are equivalent (or equally oriented) iff the matrix representing the one in terms of the other has positive determinant.
Usually one considers the standard basis of ${\mathbb R}^3$ as positively oriented and draws these three vectors in such a way that they seem "righthanded". Now for the cross product: Two linearly independent vectors $a$ and $b$ determine a two-dimensional plane whose orthogonal complement is a one dimensional vector space $L$, spanned by some vector $n\ne0$. Exactly one of $n$ and $-n$ complements $(a,b)$ to a positively oriented triple, and it is this choice of normal that is used in the definition of $a\times b$.