# Vector Cross Product is Anticommutative

## Theorem

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

### Complex Plane

The same result holds for complex numbers:

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

## Proof 1

 $\ds \mathbf b \times \mathbf a$ $=$ $\ds \begin {bmatrix} b_i \\ b_j \\ b_k \end {bmatrix} \times \begin {bmatrix} a_i \\ a_j \\ a_k \end {bmatrix}$ $\ds$ $=$ $\ds \begin {bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end {bmatrix}$ $\ds \mathbf a \times \mathbf b$ $=$ $\ds \begin {bmatrix} a_i \\ a_j \\ a_k \end {bmatrix} \times \begin {bmatrix} b_i \\ b_j \\ b_k \end {bmatrix}$ $\ds$ $=$ $\ds \begin {bmatrix} a_j b_k - a_k b_j \\ a_k b_i - a_i b_k \\ a_i b_j - a_j b_i \end {bmatrix}$ $\ds$ $=$ $\ds \begin {bmatrix} -\paren {a_k b_j - a_j b_k} \\ -\paren {a_i b_k - a_k b_i} \\ -\paren {a_j b_i - a_i b_j} \end {bmatrix}$ $\ds$ $=$ $\ds -1 \begin {bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end {bmatrix}$ $\ds$ $=$ $\ds -\paren {\mathbf b \times \mathbf a}$

$\blacksquare$

## Proof 2

 $\ds \paren {\mathbf a + \mathbf b} \times \paren {\mathbf a + \mathbf b}$ $=$ $\ds \mathbf a \times \mathbf a + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + \mathbf b \times \mathbf b$ Vector Cross Product Operator is Bilinear $\ds 0$ $=$ $\ds 0 + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + 0$ Cross Product of Vector with Itself is Zero $\ds \mathbf a \times \mathbf b$ $=$ $\ds -\paren {\mathbf b \times \mathbf a}$ simplifying

$\blacksquare$

## Proof 3

 $\ds \mathbf a \times \mathbf b$ $=$ $\ds \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end {vmatrix}$ Definition of Vector Cross Product $\ds$ $=$ $\ds -\begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ b_i & b_j & b_k \\ a_i & a_j & a_k \end {vmatrix}$ Determinant with Rows Transposed $\ds$ $=$ $\ds -\paren {\mathbf b \times \mathbf a}$ Definition of Vector Cross Product

$\blacksquare$

## Proof 4

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:

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