Let vec{x}, vec{y} and vec{z} be three vector | vedclass

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(A, B, C) Given $|\vec{x}| = |\vec{y}| = |\vec{z}| = \sqrt{2}$ and the angle between each pair is $\frac{\pi}{3}$.Thus,$\vec{x} \cdot \vec{y} = \vec{y

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$(A, B, C)$

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(A, B, C) Given $|\vec{x}| = |\vec{y}| = |\vec{z}| = \sqrt{2}$ and the angle between each pair is $\frac{\pi}{3}$.Thus,$\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{z} = \vec{z} \cdot \vec{x} = |\vec{x}||\vec{y}| \cos(\frac{\pi}{3}) = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1$.Since $\vec{a}$ is perpendicular to $\vec{x}$ and $\vec{y} 	imes \vec{z}$,$\vec{a}$ is parallel to $\vec{x} 	imes (\vec{y} 	imes \vec{z})$.Using the vector triple product formula,$\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} = 1\vec{y} - 1\vec{z} = \vec{y} - \vec{z}$.So,$\vec{a} = \lambda(\vec{y} - \vec{z})$.Then $\vec{a} \cdot \vec{y} = \lambda(\vec{y} \cdot \vec{y} - \vec{z} \cdot \vec{y}) = \lambda(2 - 1) = \lambda$. Thus,$\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})$,which is $(B)$.Similarly,$\vec{b}$ is perpendicular to $\vec{y}$ and $\vec{z} 	imes \vec{x}$,so $\vec{b}$ is parallel to $\vec{y} 	imes (\vec{z} 	imes \vec{x}) = (\vec{y} \cdot \vec{x})\vec{z} - (\vec{y} \cdot \vec{z})\vec{x} = 1\vec{z} - 1\vec{x} = \vec{z} - \vec{x}$.So,$\vec{b} = \mu(\vec{z} - \vec{x})$.Then $\vec{b} \cdot \vec{z} = \mu(\vec{z} \cdot \vec{z} - \vec{x} \cdot \vec{z}) = \mu(2 - 1) = \mu$. Thus,$\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})$,which is $(A)$.Now,$\vec{a} \cdot \vec{b} = \lambda \mu (\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x}) = \lambda \mu (\vec{y} \cdot \vec{z} - \vec{y} \cdot \vec{x} - \vec{z} \cdot \vec{z} + \vec{z} \cdot \vec{x}) = \lambda \mu (1 - 1 - 2 + 1) = -\lambda \mu = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$,which is $(C)$.

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