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

For a, b, c with [a b c] ≠ 0: a′ = (b × c)/[a b c], b′ = (c × a)/[a b c], c′ = (a × b)/[a b c].

a = (−1, 1, 1), b = (1, −1, 1), c = (1, 1, −1).

Cross products

b × c = ((−1)(−1) − (1)(1), (1)(1) − (1)(−1), (1)(1) − (−1)(1))
      = (1 − 1, 1 + 1, 1 + 1) = (0, 2, 2)
c × a = ((1)(1) − (−1)(1), (−1)(−1) − (1)(1), (1)(1) − (1)(−1))
      = (1 + 1, 1 − 1, 1 + 1) = (2, 0, 2)
a × b = ((1)(1) − (1)(−1), (1)(1) − (−1)(1), (−1)(−1) − (1)(1))
      = (1 + 1, 1 + 1, 1 − 1) = (2, 2, 0)

Triple product

[a b c] = a · (b × c) = (−1)(0) + (1)(2) + (1)(2) = 4.

Reciprocal vectors

a′ = (0, 2, 2)/4 = (j + k)/2
b′ = (2, 0, 2)/4 = (i + k)/2
c′ = (2, 2, 0)/4 = (i + j)/2

Check: a · a′ = (0 + 1 + 1)/2 = 1 ✓; b · b′ = (1 + 0 + 1)/2 = 1 ✓; c · c′ = (1 + 1 + 0)/2 = 1 ✓; a · b′ = (−1 + 0 + 1)/2 = 0 ✓.

Answer: a′ = (j + k)/2, b′ = (k + i)/2, c′ = (i + j)/2.