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Let a, b, c be unit vectors with a·b = 1/4, b·c = 1/3 and c·a = 1/5.…

Question

Let a, b, c be unit vectors with a·b = 1/4, b·c = 1/3 and c·a = 1/5. Let S denote the square of the scalar triple product [a b c]. Compute floor(1000·S).

✓ Verified answer: 819checked by our engine — not a guess

Step-by-step solution

The square of the scalar triple product equals the Gram determinant: [a b c]^2 = det(G), where G_ij = (vector i)·(vector j).

With unit vectors and the given dot products, G = [[1, 1/4, 1/5],[1/4, 1, 1/3],[1/5, 1/3, 1]].
Expanding: det(G) = 1·(1 - 1/9) - 1/4·(1/4 - 1/15) + 1/5·(1/12 - 1/5).
= 8/9 - 1/4·(11/60) + 1/5·(-7/60) = 8/9 - 11/240 - 7/300.
Common denominator 3600: 8/9 = 3200/3600, 11/240 = 165/3600, 7/300 = 84/3600.
det(G) = (3200 - 165 - 84)/3600 = 2951/3600 ≈ 0.81972.
Then 1000·S ≈ 819.72, so floor = 819.

Final answer819

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