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Reciprocal system of vectors

Short answer

Calculate [a b c] = -35. Reciprocal vectors: a' = (b×c)/[a b c] = (8i+7j+2k)/35, b' = (c×a)/[a b c] = (2i -7j -17k)/35, c' = (a×b)/[a b c] = (i - j - k)/5.

Find the set of reciprocal system to the set of vectors 2i+3j−k, −i+2j−3k and 3i−4j+2k.

For vectors a, b, c with [a b c] = 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],

so that a · a′ = b · b′ = c · c′ = 1 and a · b′ = a · c′ = … = 0.

Here a = (2, 3, −1), b = (−1, 2, −3), c = (3, −4, 2).

Cross products

b × c = (2·2 − (−3)(−4), (−3)(3) − (−1)(2), (−1)(−4) − 2·3)
      = (4 − 12, −9 + 2, 4 − 6) = (−8, −7, −2)
c × a = ((−4)(−1) − 2·3, 2·2 − 3(−1), 3·3 − (−4)(2))
      = (4 − 6, 4 + 3, 9 + 8) = (−2, 7, 17)
a × b = (3(−3) − (−1)(2), (−1)(−1) − 2(−3), 2·2 − 3(−1))
      = (−9 + 2, 1 + 6, 4 + 3) = (−7, 7, 7)

Scalar triple product

[a b c] = a · (b × c) = 2(−8) + 3(−7) + (−1)(−2) = −16 − 21 + 2 = −35.

Reciprocal vectors

a′ = (−8i − 7j − 2k)/(−35) = (8i + 7j + 2k)/35
b′ = (−2i + 7j + 17k)/(−35) = (2i − 7j − 17k)/35
c′ = (−7i + 7j + 7k)/(−35) = (i − j − k)/5

Check: a · a′ = (16 + 21 − 2)/35 = 1 ✓; a · b′ = (4 − 21 + 17)/35 = 0 ✓; b · c′ = (−1 − 2 + 3)/5 = 0 ✓.

Answer: a′ = (8i + 7j + 2k)/35, b′ = (2i − 7j − 17k)/35, c′ = (i − j − k)/5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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