Reciprocal system: sum of cross products

Short answer

Using reciprocal vectors a'=(b×c)/[abc] etc., compute a'×b' = c/[abc], b'×c' = a/[abc], c'×a' = b/[abc], and sum to obtain (a+b+c)/[abc].

If a, b, c and a', b', c' are the reciprocal system of vectors then prove that a'×b' + b'×c' + c'×a' = (a+b+c)/[a b c], [a b c] ≠ 0.

Let [a b c] = V ≠ 0. The reciprocal system is

a′ = (b × c)/V, b′ = (c × a)/V, c′ = (a × b)/V.

a′ × b′ = (b × c) × (c × a) / V².

Using P × (Q × R) = Q(P · R) − R(P · Q) with P = b × c, Q = c, R = a:

(b × c) × (c × a) = c[(b × c) · a] − a[(b × c) · c] = V c − 0 = V c.

So a′ × b′ = V c / V² = c/V.

b′ × c′ = (c × a) × (a × b) / V² = V a / V² = a/V
c′ × a′ = (a × b) × (b × c) / V² = V b / V² = b/V

a′ × b′ + b′ × c′ + c′ × a′ = (c + a + b)/V.

Hence a′×b′ + b′×c′ + c′×a′ = (a + b + c)/[a b c].

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions