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No. In n dimensions a cross product of n-1 n-vectors is the determinant with the top row being the basis elements (e_1, e_2, ..., e_n) and the next n-1 rows bei
by mjcohen 3y ago
No. In n dimensions a cross product of n-1 n-vectors is the determinant with the top row being the basis elements (e_1, e_2, ..., e_n) and the next n-1 rows being the n-1 vectors. The result is a n-vector orthogonal to the n-1 vectors.
In 2 dimensions it is
i j
x y
In 3 dimensions it is
i j k
x1 y1 z1
x2 y2 z2
And so on.
- itishappy 3y agoThat's not a particularly fruitful generalization. The traditional definition begins: > The cross product on a Euclidean space V is a bilinear map from V × V to V. This is required for some of the cross product's useful properties such as: |a x b| = |a| |b| sin(theta) This only works in 0, 1, 3, and 7 dimensions.