Let bar{a}=hat{i}+hat{j}+hat{k},bar{b} and ba | vedclass

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(B) Given $\bar{a} = \hat{i} + \hat{j} + \hat{k}$ and $\bar{c} = \hat{j} - \hat{k}$.We are given $\bar{a} 	imes \bar{b} = \bar{c}$.Taking the cross p

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(B) Given $\bar{a} = \hat{i} + \hat{j} + \hat{k}$ and $\bar{c} = \hat{j} - \hat{k}$.We are given $\bar{a} 	imes \bar{b} = \bar{c}$.Taking the cross product with $\bar{a}$ on both sides: $\bar{a} 	imes (\bar{a} 	imes \bar{b}) = \bar{a} 	imes \bar{c}$.Using the vector triple product formula $\bar{a} 	imes (\bar{a} 	imes \bar{b}) = (\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}$.Since $\bar{a} \cdot \bar{a} = 1^2 + 1^2 + 1^2 = 3$,we have $(\bar{a} \cdot \bar{b})\bar{a} - 3\bar{b} = \bar{a} 	imes \bar{c}$.Let $\bar{v} = \bar{a} 	imes \bar{c} = (\hat{i} + \hat{j} + \hat{k}) 	imes (\hat{j} - \hat{k}) = -2\hat{i} + \hat{j} + \hat{k}$.The projection of $\bar{b}$ on $\bar{v}$ is $l = \frac{|\bar{b} \cdot \bar{v}|}{|\bar{v}|}$.From $(\bar{a} \cdot \bar{b})\bar{a} - 3\bar{b} = \bar{v}$,we take the dot product with $\bar{v}$:$(\bar{a} \cdot \bar{b})(\bar{a} \cdot \bar{v}) - 3(\bar{b} \cdot \bar{v}) = \bar{v} \cdot \bar{v} = |\bar{v}|^2$.Since $\bar{a} \cdot \bar{v} = \bar{a} \cdot (\bar{a} 	imes \bar{c}) = 0$,we get $-3(\bar{b} \cdot \bar{v}) = |\bar{v}|^2$.Thus,$|\bar{b} \cdot \bar{v}| = \frac{|\bar{v}|^2}{3}$.Then $l = \frac{|\bar{b} \cdot \bar{v}|}{|\bar{v}|} = \frac{|\bar{v}|^2}{3|\bar{v}|} = \frac{|\bar{v}|}{3}$.$|\bar{v}|^2 = (-2)^2 + 1^2 + 1^2 = 4 + 1 + 1 = 6$.So $l = \frac{\sqrt{6}}{3}$,which means $l^2 = \frac{6}{9} = \frac{2}{3}$.Therefore,$3l^2 = 3 	imes \frac{2}{3} = 2$.

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