Let bar{a}, bar{b}, bar{c} be vectors such th | vedclass

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(C) Given $\bar{a} 	imes \bar{c} = \bar{b}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Taking the magnitude of the first equation: $|\bar{a} 	imes \bar{

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$|\bar{b}| = |\bar{a}|$

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(C) Given $\bar{a} 	imes \bar{c} = \bar{b}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Taking the magnitude of the first equation: $|\bar{a} 	imes \bar{c}| = |\bar{b}| \implies |\bar{a}| |\bar{c}| \sin 	heta = |\bar{b}|$,where $	heta$ is the angle between $\bar{a}$ and $\bar{c}$.Taking the magnitude of the second equation: $|\bar{b} 	imes \bar{c}| = |\bar{a}| \implies |\bar{b}| |\bar{c}| \sin \phi = |\bar{a}|$,where $\phi$ is the angle between $\bar{b}$ and $\bar{c}$.Substitute $|\bar{b}|$ from the first into the second: $(|\bar{a}| |\bar{c}| \sin 	heta) |\bar{c}| \sin \phi = |\bar{a}|$.Since $\bar{a} 
eq \bar{o}$,we can divide by $|\bar{a}|$: $|\bar{c}|^2 \sin 	heta \sin \phi = 1$.Also,from $\bar{a} 	imes \bar{c} = \bar{b}$,$\bar{b}$ is perpendicular to $\bar{c}$. From $\bar{b} 	imes \bar{c} = \bar{a}$,$\bar{a}$ is perpendicular to $\bar{c}$.Thus,$	heta = 90^\circ$ and $\phi = 90^\circ$,so $\sin 	heta = 1$ and $\sin \phi = 1$.Therefore,$|\bar{c}|^2 = 1 \implies |\bar{c}| = 1$.Now,$|\bar{a}| |\bar{c}| = |\bar{b}|$ and $|\bar{b}| |\bar{c}| = |\bar{a}|$.Since $|\bar{c}| = 1$,we get $|\bar{a}| = |\bar{b}|$. Thus,option $C$ is correct.

Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then:

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