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Triple product times (a×b) as determinant

Short answer

Write a×b = xl+ym+zn; dotting with m×n etc. gives x = [(m·a)(n·b)−(n·a)(m·b)]/[l m n], and similarly for y, z. Then [l m n](a×b) equals the expansion of the determinant with rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n) along the third column.

Prove if l, m, n be three non-coplanar vectors then l m n = | l·a l·b l ; m·a m·b m ; n·a n·b n |. [3×3 determinant: rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n)]

[Reading: [l m n](a × b) equals the 3 × 3 determinant with rows (l·a, l·b, l), (m·a, m·b, m), (n·a, n·b, n), whose third column holds the vectors l, m, n, expanded along that column.]

l, m, n are non-coplanar, so V = [l m n] ≠ 0 and they form a basis. Write

a × b = x l + y m + z n.

Find x, y, z

Dot with m × n. Since m · (m × n) = 0 = n · (m × n),

(a × b) · (m × n) = x l · (m × n) = x V.

By Lagrange's identity, (a × b) · (m × n) = (a·m)(b·n) − (a·n)(b·m). So

x V = (m·a)(n·b) − (n·a)(m·b).

Dotting with n × l and with l × m in the same way:

y V = (n·a)(l·b) − (l·a)(n·b)
z V = (l·a)(m·b) − (m·a)(l·b)

Combine

V (a × b) = [(m·a)(n·b) − (n·a)(m·b)] l
          + [(n·a)(l·b) − (l·a)(n·b)] m
          + [(l·a)(m·b) − (m·a)(l·b)] n

The determinant

Expand along the third column (l, m, n):

| l·a  l·b  l |
| m·a  m·b  m | = l[(m·a)(n·b) − (m·b)(n·a)]
| n·a  n·b  n |   − m[(l·a)(n·b) − (l·b)(n·a)]
                  + n[(l·a)(m·b) − (l·b)(m·a)]

Compare term by term with V(a × b): the coefficients of l, m, n are identical (for m: −[(l·a)(n·b) − (l·b)(n·a)] = (n·a)(l·b) − (l·a)(n·b)).

Hence [l m n](a × b) = | l·a l·b l ; m·a m·b m ; n·a n·b n |.

✓ Answer checked by Claude Opus 5.5 (high effort)

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