Vector Cross Product is Anticommutative/Complex

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Theorem

The complex cross product is anticommutative:

$\forall z_1, z_2 \in \C: z_1 \times z_2 = -\paren {z_2 \times z_1}$


Proof

Let:

$z_1 := x_1 + i y_1, z_2 = x_2 + i y_2$

Then:

\(\displaystyle z_1 \times z_2\) \(=\) \(\displaystyle x_1 y_2 - y_1 x_2\) Definition 1 of Complex Cross Product
\(\displaystyle \) \(=\) \(\displaystyle -\left({x_2 y_1 - y_2 x_1}\right)\) Real Addition is Commutative and Real Multiplication is Commutative
\(\displaystyle \) \(=\) \(\displaystyle -\left({z_2 \times z_1}\right)\) Definition 1 of Complex Cross Product

$\blacksquare$


Examples

Example: $\paren {2 + 5 i} \times \paren {3 - i} = -\paren {\paren {3 - i} \times \paren {2 + 5 i} }$

Example: $\paren {2 + 5 i} \times \paren {3 - i}$

Let:

$z_1 = 2 + 5 i$
$z_2 = 3 - i$

Then:

$z_1 \times z_2 = -17$

where $\times$ denotes (complex) cross product.


Example: $\paren {3 - i} \times \paren {2 + 5 i}$

Let:

$z_1 = 3 - i$
$z_2 = 2 + 5 i$

Then:

$z_1 \times z_2 = 17$

where $\times$ denotes (complex) cross product.


As can be seen:

$\paren {2 + 5 i} \times \paren {3 - i} = -\paren {\paren {3 - i} \times \paren {2 + 5 i} }$

$\blacksquare$


Sources