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If $\hat{x}, \hat{y}$ and $\hat{z}$ are three unit vectors in threedimensional space, then the minimum value of $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$
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Verified Answer
The correct answer is:
3
3
$(\hat{x}+\hat{y}+\hat{z})^2 \geq 0$
$$
\begin{aligned}
&\Rightarrow 3+2 \Sigma \hat{x} \cdot \hat{y} \geq 0 \\
&\Rightarrow 2 \Sigma \hat{x} \cdot \hat{y} \geq-3
\end{aligned}
$$
Now, $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ $=6+2 \sum \hat{x} . \hat{y} \geq 6+(-3)$
$$
\Rightarrow|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2 \geq 3
$$
$$
\begin{aligned}
&\Rightarrow 3+2 \Sigma \hat{x} \cdot \hat{y} \geq 0 \\
&\Rightarrow 2 \Sigma \hat{x} \cdot \hat{y} \geq-3
\end{aligned}
$$
Now, $|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ $=6+2 \sum \hat{x} . \hat{y} \geq 6+(-3)$
$$
\Rightarrow|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2 \geq 3
$$
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