If bar{a} = bar{i} - bar{j},bar{b} = bar{j} - | vedclass

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(A) Given $\bar{a} \cdot \bar{d} = 0$,this implies $\bar{d} \perp \bar{a}$.Given $[\bar{b} \bar{c} \bar{d}] = 0$,this implies $\bar{b}, \bar{c}$,and $

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$\pm \frac{1}{\sqrt{6}} (\bar{i} + \bar{j} - 2\bar{k})$

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(A) Given $\bar{a} \cdot \bar{d} = 0$,this implies $\bar{d} \perp \bar{a}$.Given $[\bar{b} \bar{c} \bar{d}] = 0$,this implies $\bar{b}, \bar{c}$,and $\bar{d}$ are coplanar,which means $\bar{d} \perp (\bar{b} 	imes \bar{c})$.Therefore,$\bar{d}$ is parallel to $\bar{a} 	imes (\bar{b} 	imes \bar{c})$.Since $\bar{d}$ is a unit vector,$\bar{d} = \pm \frac{\bar{a} 	imes (\bar{b} 	imes \bar{c})}{|\bar{a} 	imes (\bar{b} 	imes \bar{c})|}$.Here,$\bar{a} = (1, -1, 0)$,$\bar{b} = (0, 1, -1)$,and $\bar{c} = (-1, 0, 1)$.Calculating the cross product $\bar{b} 	imes \bar{c} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 0 & 1 & -1 \ -1 & 0 & 1 \end{vmatrix} = \bar{i}(1-0) - \bar{j}(0-1) + \bar{k}(0+1) = (1, 1, 1)$.Now,$\bar{a} 	imes (\bar{b} 	imes \bar{c}) = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \bar{i}(-1-0) - \bar{j}(1-0) + \bar{k}(1+1) = (-1, -1, 2)$.The magnitude is $|\bar{a} 	imes (\bar{b} 	imes \bar{c})| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.Thus,$\bar{d} = \pm \frac{1}{\sqrt{6}} (-\bar{i} - \bar{j} + 2\bar{k}) = \pm \frac{1}{\sqrt{6}} (\bar{i} + \bar{j} - 2\bar{k})$.

If $\bar{a} = \bar{i} - \bar{j}$,$\bar{b} = \bar{j} - \bar{k}$,$\bar{c} = \bar{k} - \bar{i}$ and $\bar{d}$ is a unit vector such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$,then the vector $\bar{d} = ....$

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