Unit 11: Cross products in the light of linear transformations
About this unit
Fixing u and v, the map x ↦ det[x u v] is linear, so it is a dot product with one vector p. The determinant is a box volume, base area times height, so p must be perpendicular to u and v with length equal to their area: p is u × v.
By the end, you can: See the cross product u × v as the dual vector of the linear map x ↦ det[x u v], and use that to explain its formula and its geometry.
Modules
4 modulesLearn
- A determinant with a gapArticle · Fix u and v in 3D; f(x) = det[x u v] is a linear map from 3D to the number line, so by duality there is exactly one vector p with f(x) = p · x, and its entries are the cross-product formula.Not started
- Volume meets shadowArticle · det[x u v] is the signed volume of the box on x, u, v: base area times x's height; p · x is |p| times x's shadow on p. They agree only if p is perpendicular to u and v with length equal to the area: p = u × v.Not started
- Why the formula looks like a determinantReview · The map x ↦ det[x u v], its dual vector, the geometric match, and the triple product.Not started