If hat{x}, hat{y}, and hat{z} are three unit | vedclass

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(B) Given that $\hat{x}, \hat{y}, \hat{z}$ are unit vectors,so $|\hat{x}| = |\hat{y}| = |\hat{z}| = 1$.We know that for any vectors,$|\hat{x} + \hat{y

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$3$

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(B) Given that $\hat{x}, \hat{y}, \hat{z}$ are unit vectors,so $|\hat{x}| = |\hat{y}| = |\hat{z}| = 1$.We know that for any vectors,$|\hat{x} + \hat{y} + \hat{z}|^2 \geq 0$.Expanding this,we get $|\hat{x}|^2 + |\hat{y}|^2 + |\hat{z}|^2 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$.Substituting the magnitudes,$1 + 1 + 1 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$,which implies $3 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq 0$.Thus,$2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq -3$.Now,consider the expression $S = |\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$.$S = (\hat{x} \cdot \hat{x} + \hat{y} \cdot \hat{y} + 2\hat{x} \cdot \hat{y}) + (\hat{y} \cdot \hat{y} + \hat{z} \cdot \hat{z} + 2\hat{y} \cdot \hat{z}) + (\hat{z} \cdot \hat{z} + \hat{x} \cdot \hat{x} + 2\hat{z} \cdot \hat{x})$.$S = 2(|\hat{x}|^2 + |\hat{y}|^2 + |\hat{z}|^2) + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x})$.$S = 2(1 + 1 + 1) + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) = 6 + 2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x})$.Since $2(\hat{x} \cdot \hat{y} + \hat{y} \cdot \hat{z} + \hat{z} \cdot \hat{x}) \geq -3$,the minimum value of $S$ is $6 - 3 = 3$.

If $\hat{x}, \hat{y},$ and $\hat{z}$ are three unit vectors in three-dimensional space,then find the minimum value of $|\hat{x} + \hat{y}|^2 + |\hat{y} + \hat{z}|^2 + |\hat{z} + \hat{x}|^2$.

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