Let vec{a} = hat{i} + hat{j} + hat{k},vec{b}, | vedclass

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$,$\vec{a} 	imes \vec{b} = \vec{c}$,and $\vec{a} \cdot \vec{b} = 1$.Fir

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$2$

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$,$\vec{a} 	imes \vec{b} = \vec{c}$,and $\vec{a} \cdot \vec{b} = 1$.First,calculate $\vec{a} 	imes \vec{c} = (\hat{i} + \hat{j} + \hat{k}) 	imes (\hat{j} - \hat{k}) = \hat{i}(-1-1) - \hat{j}(-1-0) + \hat{k}(1-0) = -2\hat{i} + \hat{j} + \hat{k}$.The magnitude $|\vec{a} 	imes \vec{c}| = \sqrt{(-2)^{2} + 1^{2} + 1^{2}} = \sqrt{4 + 1 + 1} = \sqrt{6}$.The projection length $l$ of $\vec{b}$ on $\vec{a} 	imes \vec{c}$ is given by $l = \frac{|\vec{b} \cdot (\vec{a} 	imes \vec{c})|}{|\vec{a} 	imes \vec{c}|}$.Using the scalar triple product property,$\vec{b} \cdot (\vec{a} 	imes \vec{c}) = -\vec{a} \cdot (\vec{b} 	imes \vec{c})$. Alternatively,since $\vec{a} 	imes \vec{b} = \vec{c}$,we have $\vec{b} \cdot (\vec{a} 	imes \vec{c}) = \vec{b} \cdot (\vec{a} 	imes (\vec{a} 	imes \vec{b})) = \vec{b} \cdot ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) - |\vec{a}|^{2}|\vec{b}|^{2}$.Easier approach: $(\vec{a} 	imes \vec{b}) \cdot \vec{c} = \vec{c} \cdot \vec{c} = |\vec{c}|^{2} = 0^{2} + 1^{2} + (-1)^{2} = 2$.Since $\vec{b} \cdot (\vec{a} 	imes \vec{c}) = -\vec{c} \cdot (\vec{a} 	imes \vec{b}) = -|\vec{c}|^{2} = -2$,we have $l = \frac{|-2|}{\sqrt{6}} = \frac{2}{\sqrt{6}}$.Thus,$l^{2} = \frac{4}{6} = \frac{2}{3}$.Therefore,$3l^{2} = 3 	imes \frac{2}{3} = 2$.

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