Bilinearity and algebraic determinant template
Cross product is bilinear: linear in each argument when the other is fixed. Distribution rules include $\mathbf{a}\times(\mathbf{b}+\mathbf{c})=\mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}$ and $(c\mathbf{a})\times\mathbf{b}=c(\mathbf{a}\times\mathbf{b})$ .
The symbolic determinant
$$\mathbf{a}\times\mathbf{b}=\det\begin{bmatrix}\hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{bmatrix}$$
abbreviates the cofactor expansion. Treat $\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}$ as placeholders in the top row, not real scalars.
Expanding the template for $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$ yields the component formulas you can verify by hand once, then trust symbolically afterward.

Cross product is not associative. Compare $\mathbf{i}\times(\mathbf{i}\times\mathbf{j})$ with $(\mathbf{i}\times\mathbf{i})\times\mathbf{j}$ to see different outputs. The BAC-CAB vector identity packages the non-associativity in a useful form.

Jacobi-type cyclic sums appear in Lie algebras; for cross products the takeaway is algebraic manipulation requires care, unlike dot products where reordering is safer.
When teaching, emphasize bilinearity first and treat the determinant layout as a compressed bookkeeping device rather than a definition pulled from nowhere.
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- Topic: Mathematics
- Difficulty: Intermediate
- Completed: 0 users