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

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$3 \sqrt{3}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

Given $| \bar{x} | = | \bar{y} | = | \bar{z} | = 1,$

now $| \bar{x} + \bar{y} + \bar{z} |^{2} \geq 0 \Rightarrow 2 \sum (\bar{x} . \bar{y}) \geq - 3$

Now $\sum {(\bar{x} + \bar{y})}^{2} \geq 6 - 3 \geq 3$

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