Question 71873
2018
If a, b and c are unit vectors, then |a − b|² + |b − c|² + |c − a|² does not exceed:
Answer: B. 9 — Let a, b and c be unit vectors, so |a|² = |b|² = |c|² = 1. Expand each squared magnitude using |x − y|² = |x|² + |y|² − 2(x·y): |a − b|² + |b − c|² + |c − a|²…
- A.
4
- B.
9
- C.
8
- D.
6
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Correct answer: B
Let a, b and c be unit vectors, so |a|² = |b|² = |c|² = 1.
Expand each squared magnitude using |x − y|² = |x|² + |y|² − 2(x·y):
|a − b|² + |b − c|² + |c − a|² = 2(|a|² + |b|² + |c|²) − 2(a·b + b·c + c·a) = 6 − 2(a·b + b·c + c·a).
Now use the fact that the squared magnitude of any vector is never negative:
|a + b + c|² = |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 3 + 2(a·b + b·c + c·a) ≥ 0.
Hence a·b + b·c + c·a ≥ −3/2. Substituting this lower bound into the expression:
|a − b|² + |b − c|² + |c − a|² = 6 − 2(a·b + b·c + c·a) ≤ 6 − 2(−3/2) = 9.
The maximum value 9 is attained when a + b + c = 0, e.g. three unit vectors at 120° to one another. Therefore the expression does not exceed 9.