Let vec{a}=hat{i}+hat{j}+hat{k},vec{b}=2hat{i | vedclass

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(C) Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$.$\vec{d}=\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j

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$6$

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(C) Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$.$\vec{d}=\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 2 & 1 \end{vmatrix} = -\hat{i}+\hat{j}$.$|\vec{d}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.Given $|\vec{c}-2\vec{a}|^2=8 \implies |\vec{c}|^2 + 4|\vec{a}|^2 - 4(\vec{a} \cdot \vec{c}) = 8$.Since $|\vec{a}|^2 = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$,let $|\vec{c}| = x$.$x^2 + 4(3) - 4x = 8 \implies x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies |\vec{c}| = 2$.Given the angle between $\vec{d}$ and $\vec{c}$ is $\frac{\pi}{4}$,$|\vec{d} 	imes \vec{c}| = |\vec{d}||\vec{c}| \sin(\frac{\pi}{4}) = \sqrt{2} \cdot 2 \cdot \frac{1}{\sqrt{2}} = 2$.Thus,$|\vec{d} 	imes \vec{c}|^2 = 4$.Using the vector triple product property or projection,we find $\vec{b} \cdot \vec{c}$.Since $\vec{d} = \vec{a} 	imes \vec{b}$,$|\vec{d} 	imes \vec{c}|^2 = |(\vec{a} 	imes \vec{b}) 	imes \vec{c}|^2 = |(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}|^2 = 4$.$|2\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}|^2 = 4 \implies 4|\vec{b}|^2 + (\vec{b} \cdot \vec{c})^2 |\vec{a}|^2 - 4(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{b}) = 4$.$|\vec{b}|^2 = 9$,$\vec{a} \cdot \vec{b} = 5$,$|\vec{a}|^2 = 3$.$4(9) + 3(\vec{b} \cdot \vec{c})^2 - 4(\vec{b} \cdot \vec{c})(5) = 4 \implies 3(\vec{b} \cdot \vec{c})^2 - 20(\vec{b} \cdot \vec{c}) + 32 = 0$.Solving for $\vec{b} \cdot \vec{c}$,we get $\vec{b} \cdot \vec{c} = \frac{20 \pm \sqrt{400 - 384}}{6} = \frac{20 \pm 4}{6} = 4, \frac{8}{3}$.For $\vec{b} \cdot \vec{c} = \frac{8}{3}$,$|10 - 3(\frac{8}{3})| + 4 = |10-8| + 4 = 2+4 = 6$.

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