I am reading Lovelock and Rund's book on Tensors and they make a statement that I wanted to validate about normal vectors in high-dimensional spaces.
It should be remarked that the above geometrical definition of $A \times B$ is meaningful solely for the case of a three-dimensional space, because in a higher-dimensional space the plane $C$ does not possess a unique normal.
I was just wondering why planes in high dimensional spaces do not possess unique normals?
The (simple) bivector describing a plane in $\mathbb{R}^n$ belongs to a $n(n-1)/2$ dimensional space, which number equals 3 for n=3. In higher dimensions we cannot associate it with a unique n-dimensional normal vector.