Vector Algebra
Non-Coplanar Vector Selection
Grade 12
Question:
<p>Consider the set of eight vectors
\(V=\{a\hat{i}+b\hat{j}+c\hat{k}:a,b,c\in\{-1,1\}\}\).
Three non-coplanar vectors can be chosen from \(V\) in \(2^p\) ways.
Find \(p\).</p>
Step-by-Step Solution
Key Concept: Count the total number of 3-element subsets of V and subtract the coplanar triples. Three vectors from V (vertices of a cube centred at origin) are coplanar iff their scalar triple product is 0.
The 8 vectors form the vertices of a cube $\{(\pm1,\pm1,\pm1)\}$.
Total 3-element subsets: $\binom{8}{3}=56$.
Coplanar triples (scalar triple product = 0): each face of the cube contributes
$\binom{4}{3}=4$ triples; 6 faces → 24 coplanar triples.
Non-coplanar triples: $56-24=32=2^5\Rightarrow p=\boxed{5}$.
Correct Answer: 5