Vector Cross Product is Anticommutative/Proof 1

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Theorem

The vector cross product is anticommutative:

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


Proof

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

$\blacksquare$