Why can't we have these alternative cross product relationships?

Question

I am currently studying Introduction to Electrodynamics, fourth edition, by David J. Griffiths. Chapter 1.1.2 Vector Algebra: Component Form says the following:

$$\mathbf{\hat{x}} \times \mathbf{\hat{x}} = \mathbf{\hat{y}} \times \mathbf{\hat{y}} = \mathbf{\hat{z}} \times \mathbf{\hat{z}} = \mathbf{0},$$

$$\mathbf{\hat{x}} \times \mathbf{\hat{y}} = - \mathbf{\hat{y}} \times \mathbf{\hat{x}} = \mathbf{\hat{z}},$$

$$\mathbf{\hat{y}} \times \mathbf{\hat{z}} = -\mathbf{\hat{z}} \times \mathbf{\hat{y}} = \mathbf{\hat{x}},$$

$$\mathbf{\hat{z}} \times \mathbf{\hat{x}} = -\mathbf{\hat{x}} \times \mathbf{\hat{z}} = \mathbf{\hat{y}}.$$

These signs pertain to a right-handed coordinate system ($x$-axis out of the page, $y$-axis to the right, $z$-axis up, or any rotated version thereof). In a left-handed system ($z$-axis down), the signs would be reversed: $\mathbf{\hat{x}} \times \mathbf{\hat{y}} = - \mathbf{\hat{z}}$, and so on. We shall use right-handed systems exclusively.

I don't understand why we can't also have that

$$\mathbf{\hat{x}} \times \mathbf{\hat{y}} = \mathbf{\hat{y}} \times - \mathbf{\hat{x}} = \mathbf{\hat{z}},$$

$$\mathbf{\hat{y}} \times \mathbf{\hat{z}} = \mathbf{\hat{z}} \times - \mathbf{\hat{y}} = \mathbf{\hat{x}},$$

$$\mathbf{\hat{z}} \times \mathbf{\hat{x}} = \mathbf{\hat{x}} \times - \mathbf{\hat{z}} = \mathbf{\hat{y}}.$$

I would greatly appreciate it if people would please take the time to explain this.

Answer 1

We do also have those "alternative cross product relationships" as well.

The cross-product is compatible with scalar multiplication

(in particular, multiplication by $-1$),

so $\mathbf{-\hat y}\times\mathbf{\hat x}=\mathbf{\hat y}\times -\mathbf {\hat x}$, for example.

Answer 2

Your suggested equations are equivalent to some of the equations from your textbook, so to avoid redundancy we list one or the other. However, the advantage of writing $-\mathbf{\hat{y}}\times\mathbf{\hat{x}}=\mathbf{\hat{z}}$ instead of $\mathbf{\hat{y}}\times-\mathbf{\hat{x}}=\mathbf{\hat{z}}$ is that the former can be read as $-(\mathbf{\hat{y}}\times\mathbf{\hat{x}})=\mathbf{\hat{z}}$ so that the whole line becomes a concise version of$$\mathbf{\hat{x}}\times\mathbf{\hat{y}}=\mathbf{\hat{z}},\,\mathbf{\hat{y}}\times\mathbf{\hat{x}}=-\mathbf{\hat{z}},$$so that we only ever multiply unsigned unit vectors.