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
\(\displaystyle \mathbf b \times \mathbf a\) | \(=\) | \(\displaystyle \begin{bmatrix} b_i \\ b_j \\ b_k \end{bmatrix} \times \begin{bmatrix} a_i \\ a_j \\ a_k \end{bmatrix}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \begin{bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end{bmatrix}\) | |||||||||||
\(\displaystyle \mathbf a \times \mathbf b\) | \(=\) | \(\displaystyle \begin{bmatrix} a_i \\ a_j \\ a_k \end{bmatrix} \times \begin{bmatrix} b_i \\ b_j \\ b_k \end{bmatrix}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \begin{bmatrix} a_j b_k - a_k b_j \\ a_k b_i - a_i b_k \\ a_i b_j - a_j b_i \end{bmatrix}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \begin{bmatrix} -\left({a_k b_j - a_j b_k}\right) \\ -\left({a_i b_k - a_k b_i}\right) \\ -\left({a_j b_i - a_i b_j}\right)\end{bmatrix}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle -1 \begin{bmatrix} b_j a_k - a_j b_k \\ b_k a_i - b_i a_k \\ b_i a_j - a_i b_j \end{bmatrix}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle -\left({\mathbf b \times \mathbf a}\right)\) |
$\blacksquare$
Proof 2
\(\displaystyle \left({\mathbf a + \mathbf b}\right) \times \left({\mathbf a + \mathbf b}\right)\) | \(=\) | \(\displaystyle \mathbf a \times \mathbf a + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + \mathbf b \times \mathbf b\) | Vector Cross Product Operator is Bilinear | ||||||||||
\(\displaystyle 0\) | \(=\) | \(\displaystyle 0 + \mathbf a \times \mathbf b + \mathbf b \times \mathbf a + 0\) | Cross Product of Vector with Itself is Zero | ||||||||||
\(\displaystyle \mathbf a \times \mathbf b\) | \(=\) | \(\displaystyle -\left({\mathbf b \times \mathbf a}\right)\) | simplifying |
$\blacksquare$
Proof 3
\(\displaystyle \mathbf a \times \mathbf b\) | \(=\) | \(\displaystyle \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ a_i & a_j & a_k \\ b_i & b_j & b_k \\ \end{vmatrix}\) | Definition of Vector Cross Product | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle -\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ b_i & b_j & b_k \\ a_i & a_j & a_k \\ \end{vmatrix}\) | Determinant with Rows Transposed | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle -\left({\mathbf b \times \mathbf a}\right)\) | Definition of Vector Cross Product |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 22$: Cross or Vector Product: $22.13$