Vector Cross Product is Anticommutative/Proof 4

Theorem

The vector cross product is anticommutative:

$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\left({\mathbf b \times \mathbf a}\right)$

Proof

From the definition of the vector cross product:

The vector cross product, denoted $\mathbf a \times \mathbf b$, is defined as:

$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$

where:

$\norm {\mathbf a}$ denotes the length of $\mathbf a$

$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction

$\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.

Hence we have that:

$\mathbf a \times \mathbf b$ is a vector whose direction is specified according to the right-hand rule

while:

$\mathbf b \times \mathbf a$ is a vector whose direction, also specified according to the right-hand rule, is exactly the opposite of that for $\mathbf a \times \mathbf b$.

That is:

$\mathbf a \times \mathbf b = -\mathbf b \times \mathbf a$

$\blacksquare$

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $4$. The Vector Product: $(2.13)$