vec{a}, vec{b}, vec{c} are three vectors each | vedclass

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(A) Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$ and the angle between any two vectors is $\frac{\pi}{3}$.$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}

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$|\vec{x}|=|\vec{y}|$

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(A) Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{2}$ and the angle between any two vectors is $\frac{\pi}{3}$.$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{3}) = \sqrt{2} 	imes \sqrt{2} 	imes \frac{1}{2} = 1$.Similarly,$\vec{b} \cdot \vec{c} = 1$ and $\vec{c} \cdot \vec{a} = 1$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$:$\vec{x} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = 1 \cdot \vec{b} - 1 \cdot \vec{c} = \vec{b} - \vec{c}$.Now,$|\vec{x}|^2 = |\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c}) = 2 + 2 - 2(1) = 2$.Similarly,$\vec{y} = \vec{b} 	imes (\vec{c} 	imes \vec{a}) = (\vec{b} \cdot \vec{a})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a} = 1 \cdot \vec{c} - 1 \cdot \vec{a} = \vec{c} - \vec{a}$.Now,$|\vec{y}|^2 = |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 2 + 2 - 2(1) = 2$.Since $|\vec{x}|^2 = 2$ and $|\vec{y}|^2 = 2$,we have $|\vec{x}| = |\vec{y}| = \sqrt{2}$.

$\vec{a}, \vec{b}, \vec{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\vec{a}, \vec{b})=(\vec{b}, \vec{c})=(\vec{c}, \vec{a})=\frac{\pi}{3}$. If $\vec{x}=\vec{a} \times(\vec{b} \times \vec{c})$ and $\vec{y}=\vec{b} \times(\vec{c} \times \vec{a})$,then

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