---
tag: 07ZS685JJC
kind: statement
role: proposition
name: Anticommutativity of the cross product
msc: [15A72]
uses: [0N5N15HWZ7]
sources:
  - url: https://en.wikipedia.org/wiki/Cross_product
    locator: Algebraic properties
---

## Statement

Let $a$ and $b$ be vectors in $\mathbb{R}^3$, and let $\times$ be the [[cross product|0N5N15HWZ7]]. Then
$$b \times a = -(a \times b).$$

## Remarks

With $b = a$ the statement gives $a \times a = -(a \times a)$, thus $a \times a = 0$ for each vector $a$.
The proof is a direct computation: each component of $b \times a$ is the component of $a \times b$ with the two products in the other order.
The cross product is thus not commutative. For example, $e_1 \times e_2 = e_3$ but $e_2 \times e_1 = -e_3$.

## Explanations

### L1 human

If you swap the two factors of a cross product, the result points the other way. Its length stays the same. The right-hand rule shows this: swapping the factors turns your hand over.