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If $a, b, c$ are unit vectors, then ${|a - b|}^{2} + | b - c |^{2} + | c - a |^{2}$ does not exceed

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answer is B.

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Detailed Solution

$|a| = |b| = |c| = 1 {|a - b|}^{2} + {|b - c|}^{2} + {|c - a|}^{2} = 1 + 1 - 2 a . b + 1 + 1 - 2 b . c + 1 + 1 - 2 c . a = 6 - 2 (a . b + b . c + c . a) \leq 6 - 2 (\frac{- 3}{2}) \leq 9$

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