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

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(D) Given $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$.Since $[\bar{b} \bar{c} \bar{d}] = 0$,the vect

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$\pm \left( \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}} \right)$

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(D) Given $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$.Since $[\bar{b} \bar{c} \bar{d}] = 0$,the vector $\bar{d}$ must be coplanar with $\bar{b}$ and $\bar{c}$.Thus,$\bar{d} = x\bar{b} + y\bar{c} = x(\hat{j} - \hat{k}) + y(\hat{k} - \hat{i}) = -y\hat{i} + x\hat{j} + (y - x)\hat{k}$.Given $\bar{a} \cdot \bar{d} = 0$,we have $(\hat{i} - \hat{j}) \cdot (-y\hat{i} + x\hat{j} + (y - x)\hat{k}) = 0$.$-y - x = 0 \implies y = -x$.Substituting $y = -x$ into $\bar{d}$,we get $\bar{d} = x\hat{i} + x\hat{j} - 2x\hat{k} = x(\hat{i} + \hat{j} - 2\hat{k})$.Since $\bar{d}$ is a unit vector,$|\bar{d}| = 1 \implies |x| \sqrt{1^2 + 1^2 + (-2)^2} = 1 \implies |x| \sqrt{6} = 1 \implies x = \pm \frac{1}{\sqrt{6}}$.Therefore,$\bar{d} = \pm \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})$.

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

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