Why does b · b' also equal 1?

[b c a] = [a b c] by cyclic permutation

Reciprocal system: a·a'+b·b'+c·c'=3

Short answer

By definition, a·a' = a·(b×c)/[abc] = 1, and similarly b·b' = 1, c·c' = 1, so the sum is 3.

If a, b, c and a', b', c' are reciprocal system of vectors, then show that a·a' + b·b' + c·c' = 3.

Let V = [a b c] ≠ 0. The reciprocal system is a′ = (b × c)/V, b′ = (c × a)/V, c′ = (a × b)/V.

a · a′ = a · (b × c) / V = [a b c]/V = 1
b · b′ = b · (c × a) / V = [b c a]/V = [a b c]/V = 1
c · c′ = c · (a × b) / V = [c a b]/V = [a b c]/V = 1

Here [b c a] = [c a b] = [a b c] by cyclic permutation of the triple product.

a · a′ + b · b′ + c · c′ = 1 + 1 + 1.

Hence a · a′ + b · b′ + c · c′ = 3.

✓ Answer checked by Claude Opus 5.5 (high effort)