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(A) Let $b = b_1 i + b_2 j + b_3 k$.Given $a 	imes b = j - k$,we have:$j - k = \begin{vmatrix} i & j & k \ 1 & 1 & 1 \ b_1 & b_2 & b_3 \end{vmatrix

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$i$

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(A) Let $b = b_1 i + b_2 j + b_3 k$.Given $a 	imes b = j - k$,we have:$j - k = \begin{vmatrix} i & j & k \ 1 & 1 & 1 \ b_1 & b_2 & b_3 \end{vmatrix}$Expanding the determinant:$j - k = i(b_3 - b_2) - j(b_3 - b_1) + k(b_2 - b_1)$Comparing coefficients:$b_3 - b_2 = 0 \Rightarrow b_3 = b_2$$-(b_3 - b_1) = 1 \Rightarrow b_1 - b_3 = 1 \Rightarrow b_1 = b_3 + 1$$b_2 - b_1 = -1$Substituting $b_3 = b_2$ into $b_1 = b_3 + 1$,we get $b_1 = b_2 + 1$.Given $a \cdot b = 1$,we have:$(i + j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1$$b_1 + b_2 + b_3 = 1$Substituting $b_1 = b_2 + 1$ and $b_3 = b_2$:$(b_2 + 1) + b_2 + b_2 = 1$$3b_2 + 1 = 1$$3b_2 = 0 \Rightarrow b_2 = 0$Thus,$b_3 = 0$ and $b_1 = 0 + 1 = 1$.Therefore,$b = 1i + 0j + 0k = i$.

If $a = i + j + k$,$a \cdot b = 1$ and $a \times b = j - k$,then $b = \dots$

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