Vector Cross Product is Anticommutative

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Theorem

The vector cross product is anticommutative:

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


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:

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


Sources

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