Let vec{a}=hat{i}+hat{j}+hat{k}, vec{b}=3 hat | vedclass

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(C) Given $\vec{a} 	imes \hat{d} = \vec{b} 	imes \hat{d}$,we have $(\vec{a} - \vec{b}) 	imes \hat{d} = 0$. This implies $\hat{d}$ is parallel to $\

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$5$

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(C) Given $\vec{a} 	imes \hat{d} = \vec{b} 	imes \hat{d}$,we have $(\vec{a} - \vec{b}) 	imes \hat{d} = 0$. This implies $\hat{d}$ is parallel to $\vec{a} - \vec{b}$. $\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1-(-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$. Since $\hat{d}$ is a unit vector,$\hat{d} = \pm \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3}$. Given $\vec{c} \cdot \vec{a} = 0$,we have $(\lambda \hat{j} + \mu \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$,so $\lambda + \mu = 0$,which means $\mu = -\lambda$. Thus,$\vec{c} = \lambda(\hat{j} - \hat{k})$. Given $\vec{c} \cdot \hat{d} = 1$,we have $\lambda(\hat{j} - \hat{k}) \cdot \pm \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) = 1$. $\pm \frac{\lambda}{3}(-1 - 2) = 1 \Rightarrow \mp \lambda = 1 \Rightarrow \lambda = \mp 1$. In both cases,$\lambda^2 = 1$ and $\mu^2 = \lambda^2 = 1$. We need to calculate $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 9 \lambda^2 |\hat{d}|^2 + \mu^2 |\vec{c}|^2 + 6 \lambda \mu (\hat{d} \cdot \vec{c})$. Since $|\hat{d}|^2 = 1$,$|\vec{c}|^2 = \lambda^2 + \mu^2 = 2\lambda^2 = 2$,and $\hat{d} \cdot \vec{c} = 1$: $|3 \lambda \hat{d} + \mu \vec{c}|^2 = 9(1)(1) + (1)(2) + 6(\lambda)(-\lambda)(1) = 9 + 2 - 6\lambda^2 = 11 - 6(1) = 5$.

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