Let three vectors vec{a}, vec{b} and vec{c} b | vedclass

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(B) Given $\vec{a} 	imes \vec{b} = \vec{c}$ and $\vec{b} 	imes \vec{c} = \vec{a}$.Since $\vec{a} 	imes \vec{b} = \vec{c}$,$\vec{c}$ is perpendicula

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$|3\vec{a} + \vec{b} - 2\vec{c}|^2 = 51$

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(B) Given $\vec{a} 	imes \vec{b} = \vec{c}$ and $\vec{b} 	imes \vec{c} = \vec{a}$.Since $\vec{a} 	imes \vec{b} = \vec{c}$,$\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.Since $\vec{b} 	imes \vec{c} = \vec{a}$,$\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.Thus,$\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal.$|\vec{a} 	imes \vec{b}| = |\vec{c}| \implies |\vec{a}||\vec{b}| = |\vec{c}| \implies 2|\vec{b}| = |\vec{c}|$.$|\vec{b} 	imes \vec{c}| = |\vec{a}| \implies |\vec{b}||\vec{c}| = 2$.Substituting $|\vec{c}| = 2|\vec{b}|$,we get $|\vec{b}|(2|\vec{b}|) = 2 \implies |\vec{b}|^2 = 1 \implies |\vec{b}| = 1$.Then $|\vec{c}| = 2(1) = 2$.$(A)$ Projection of $\vec{a}$ on $(\vec{b} 	imes \vec{c}) = \frac{\vec{a} \cdot (\vec{b} 	imes \vec{c})}{|\vec{b} 	imes \vec{c}|} = \frac{\vec{a} \cdot \vec{a}}{|\vec{a}|} = |\vec{a}| = 2$. (True)$(B)$ $|3\vec{a} + \vec{b} - 2\vec{c}|^2 = (3\vec{a} + \vec{b} - 2\vec{c}) \cdot (3\vec{a} + \vec{b} - 2\vec{c}) = 9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2 = 9(4) + 1 + 4(4) = 36 + 1 + 16 = 53 
eq 51$. (Not True)$(C)$ $[\vec{a} \vec{b} \vec{c}] + [\vec{c} \vec{a} \vec{b}] = 2[\vec{a} \vec{b} \vec{c}] = 2(\vec{a} \cdot (\vec{b} 	imes \vec{c})) = 2(\vec{a} \cdot \vec{a}) = 2|\vec{a}|^2 = 2(4) = 8$. (True)$(D)$ $\vec{a} 	imes ((\vec{b} + \vec{c}) 	imes (\vec{b} - \vec{c})) = \vec{a} 	imes (\vec{b} 	imes \vec{b} - \vec{b} 	imes \vec{c} + \vec{c} 	imes \vec{b} - \vec{c} 	imes \vec{c}) = \vec{a} 	imes (\vec{0} - \vec{a} - \vec{a} - \vec{0}) = \vec{a} 	imes (-2\vec{a}) = -2(\vec{a} 	imes \vec{a}) = \vec{0}$. (True)

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{a} \times \vec{b} = \vec{c}$,$\vec{b} \times \vec{c} = \vec{a}$ and $|\vec{a}| = 2$. Then which one of the following is not true?

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