For non-zero vectors bar{a}, bar{b}, bar{c},i | vedclass

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(D) Given $\bar{a} 	imes \bar{b} = \bar{c}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Substituting $\bar{c} = \bar{a} 	imes \bar{b}$ into the second eq

Vedclass · Accepted Answer

$|\bar{b}| = 1, |\bar{c}| = |\bar{a}|$

Vedclass · Answer

(D) Given $\bar{a} 	imes \bar{b} = \bar{c}$ and $\bar{b} 	imes \bar{c} = \bar{a}$.Substituting $\bar{c} = \bar{a} 	imes \bar{b}$ into the second equation: $\bar{b} 	imes (\bar{a} 	imes \bar{b}) = \bar{a}$.Using the vector triple product formula $\bar{b} 	imes (\bar{a} 	imes \bar{b}) = (\bar{b} \cdot \bar{b})\bar{a} - (\bar{b} \cdot \bar{a})\bar{b} = |\bar{b}|^2 \bar{a} - (\bar{a} \cdot \bar{b})\bar{b}$.Thus,$|\bar{b}|^2 \bar{a} - (\bar{a} \cdot \bar{b})\bar{b} = \bar{a}$.Since $\bar{b} 	imes \bar{c} = \bar{a}$,$\bar{b}$ is perpendicular to $\bar{a}$,so $\bar{a} \cdot \bar{b} = 0$.Substituting this,we get $(|\bar{b}|^2 - 1)\bar{a} = 0$. Since $\bar{a} 
eq 0$,$|\bar{b}|^2 = 1$,so $|\bar{b}| = 1$.From $\bar{a} 	imes \bar{b} = \bar{c}$,we have $|\bar{a} 	imes \bar{b}| = |\bar{c}|$.Since $\bar{a} \perp \bar{b}$,$|\bar{a} 	imes \bar{b}| = |\bar{a}||\bar{b}| \sin(90^\circ) = |\bar{a}|(1)(1) = |\bar{a}|$.Therefore,$|\bar{c}| = |\bar{a}|$. Thus,the correct option is $D$.

For non-zero vectors $\bar{a}, \bar{b}, \bar{c}$,if $\bar{a} \times \bar{b} = \bar{c}$ and $\bar{b} \times \bar{c} = \bar{a}$,then:

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