Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k

For a, b, c with [a b c] ≠ 0, the reciprocal system is a′ = (b × c)/[a b c], b′ = (c × a)/[a b c], c′ = (a × b)/[a b c].

Here a = (2, −3, 1), b = (1, 2, −1), c = (3, −1, 2).

Cross products

b × c = (2·2 − (−1)(−1), (−1)(3) − (1)(2), (1)(−1) − (2)(3))
      = (4 − 1, −3 − 2, −1 − 6) = (3, −5, −7)
c × a = ((−1)(1) − (2)(−3), (2)(2) − (3)(1), (3)(−3) − (−1)(2))
      = (−1 + 6, 4 − 3, −9 + 2) = (5, 1, −7)
a × b = ((−3)(−1) − (1)(2), (1)(1) − (2)(−1), (2)(2) − (−3)(1))
      = (3 − 2, 1 + 2, 4 + 3) = (1, 3, 7)

Triple product

[a b c] = a · (b × c) = 2(3) + (−3)(−5) + (1)(−7) = 6 + 15 − 7 = 14.

Reciprocal vectors

a′ = (3i − 5j − 7k)/14
b′ = (5i + j − 7k)/14
c′ = (i + 3j + 7k)/14

Check: a · a′ = (6 + 15 − 7)/14 = 1 ✓; a · b′ = (10 − 3 − 7)/14 = 0 ✓; c · c′ = (3 − 3 + 14)/14 = 1 ✓.

Answer: a′ = (3i − 5j − 7k)/14, b′ = (5i + j − 7k)/14, c′ = (i + 3j + 7k)/14.