Let a = i - j,b = j - k,c = k - i. If hat{d} | vedclass

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(C) Let $a = i - j$,$b = j - k$,and $c = k - i$.Let $\hat{d} = a_1 i + a_2 j + a_3 k$,where $|\hat{d}| = \sqrt{a_1^2 + a_2^2 + a_3^2} = 1$.This implie

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$\pm \frac{i + j - 2k}{\sqrt{6}}$

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(C) Let $a = i - j$,$b = j - k$,and $c = k - i$.Let $\hat{d} = a_1 i + a_2 j + a_3 k$,where $|\hat{d}| = \sqrt{a_1^2 + a_2^2 + a_3^2} = 1$.This implies $a_1^2 + a_2^2 + a_3^2 = 1$ $(i)$.Given $a \cdot \hat{d} = 0$,we have $(i - j) \cdot (a_1 i + a_2 j + a_3 k) = 0$,which gives $a_1 - a_2 = 0$ $(ii)$.Given $[b, c, \hat{d}] = 0$,the scalar triple product is zero,so $b \cdot (c 	imes \hat{d}) = 0$.This is equivalent to the determinant $\begin{vmatrix} 0 & 1 & -1 \ -1 & 0 & 1 \ a_1 & a_2 & a_3 \end{vmatrix} = 0$.Expanding the determinant: $0(0 - a_2) - 1(-a_3 - a_1) - 1(-a_2 - 0) = 0$,which simplifies to $a_3 + a_1 + a_2 = 0$ $(iii)$.From $(ii)$,$a_1 = a_2$. Substituting into $(iii)$,we get $a_1 + a_1 + a_3 = 0$,so $a_3 = -2a_1$.Substituting $a_2 = a_1$ and $a_3 = -2a_1$ into $(i)$:$a_1^2 + a_1^2 + (-2a_1)^2 = 1 \Rightarrow 6a_1^2 = 1 \Rightarrow a_1 = \pm \frac{1}{\sqrt{6}}$.Thus,$a_1 = \pm \frac{1}{\sqrt{6}}$,$a_2 = \pm \frac{1}{\sqrt{6}}$,and $a_3 = \mp \frac{2}{\sqrt{6}}$.Therefore,$\hat{d} = \pm \frac{i + j - 2k}{\sqrt{6}}$.

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

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