If $x, y, z$ are three unit vectors in threedimensional space, then the minimum value
of ${|x - y|}^{2} + {|y + z|}^{2} + {|z + x|}^{2}$ is

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$3 \sqrt{3}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$|x| = 1 = |y| = |z| {|x + y|}^{2} + {|y + z|}^{2} + {|z + x|}^{2} = 1 + 1 + 2 x . y + 1 + 1 + 2 y . z + 1 + 1 + 2 z . x = 6 + 2 (x . y + y . z + z . x) \leq 6 + 2 (\frac{- 3}{2}) \leq 3$

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