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Let V be the set of eight vectors of the form {a*i_hat + b*j_hat + c*k_hat: a, b, c each in {-1, 1}}. If the number of ways to choose three non-coplanar vectors from V is 2^p, find p.

1. 5
2. 6
3. 7
4. 8

Correct answer: 5

Solution

V has 8 vectors: (+/-1, +/-1, +/-1) — the vertices of a unit cube. Each vector has its negative in V. C(8,3) = 56 total triplets. Three vectors are coplanar (through origin) iff they are linearly dependent. Cases of linear dependence: (1) Two are negatives of each other: choose a pair {v, -v}: C(4,1) = 4 pairs; third vector can be any of remaining 6: 4*6 = 24 triplets. (2) Three are linearly dependent without any pair being negatives: this happens when one vector is in the span of the other two. For vectors from the cube vertices, this requires one to be a linear combination of two others — with all components ±1. Check: can (1,1,1) = a*(1,1,-1) + b*(1,-1,1)? a+b=1, a-b=1, -a+b=1 => a=1, b=0 but then -a+b = -1 ≠ 1. Checking systematically, no three such vectors (without a -(vector) pair) are linearly dependent since the determinant of any 3 cube-vertex vectors (none being negatives) is always ±4 ≠ 0. So coplanar count = 24. Non-coplanar = 56 - 24 = 32 = 2⁵. So p = 5.

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