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

For a, b, c with [a b c] ≠ 0, the reciprocal set 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, −1, −2), c = (−1, 2, 2).

Cross products

b × c = ((−1)(2) − (−2)(2), (−2)(−1) − (1)(2), (1)(2) − (−1)(−1))
      = (−2 + 4, 2 − 2, 2 − 1) = (2, 0, 1)
c × a = ((2)(−1) − (2)(3), (2)(2) − (−1)(−1), (−1)(3) − (2)(2))
      = (−2 − 6, 4 − 1, −3 − 4) = (−8, 3, −7)
a × b = ((3)(−2) − (−1)(−1), (−1)(1) − (2)(−2), (2)(−1) − (3)(1))
      = (−6 − 1, −1 + 4, −2 − 3) = (−7, 3, −5)

Triple product

[a b c] = a · (b × c) = 2(2) + 3(0) + (−1)(1) = 3.

Reciprocal vectors

a′ = (2i + k)/3
b′ = (−8i + 3j − 7k)/3
c′ = (−7i + 3j − 5k)/3

Check: a · a′ = (4 − 1)/3 = 1 ✓; b · b′ = (−8 − 3 + 14)/3 = 1 ✓; c · c′ = (7 + 6 − 10)/3 = 1 ✓; a · b′ = (−16 + 9 + 7)/3 = 0 ✓.

Answer: a′ = (2i + k)/3, b′ = (−8i + 3j − 7k)/3, c′ = (−7i + 3j − 5k)/3.