If a = (1, 1, 1) and c = (0, 1, -1) are two v | vedclass

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(D) Let $b = b_1 i + b_2 j + b_3 k$. Given $a \cdot b = 3$,we have $b_1 + b_2 + b_3 = 3$ ......$(i)$. Given $a 	imes b = c$,where $a = (1, 1, 1)$ and

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$\left( \frac{5}{3}, \frac{2}{3}, \frac{2}{3} \right)$

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(D) Let $b = b_1 i + b_2 j + b_3 k$. Given $a \cdot b = 3$,we have $b_1 + b_2 + b_3 = 3$ ......$(i)$. Given $a 	imes b = c$,where $a = (1, 1, 1)$ and $c = (0, 1, -1)$,we calculate the cross product: $a 	imes b = \begin{vmatrix} i & j & k \ 1 & 1 & 1 \ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2)i + (b_1 - b_3)j + (b_2 - b_1)k$. Comparing this with $c = (0, 1, -1)$,we get: $b_3 - b_2 = 0$ ......$(ii)$ $b_1 - b_3 = 1$ ......$(iii)$ $b_2 - b_1 = -1$ ......$(iv)$ From $(ii)$,$b_2 = b_3$. Substituting into $(i)$,$b_1 + 2b_2 = 3$. From $(iii)$,$b_1 = 1 + b_3 = 1 + b_2$. Substituting $b_1$ into $b_1 + 2b_2 = 3$,we get $(1 + b_2) + 2b_2 = 3$,so $3b_2 = 2$,which means $b_2 = \frac{2}{3}$. Then $b_3 = \frac{2}{3}$ and $b_1 = 1 + \frac{2}{3} = \frac{5}{3}$. Thus,$b = \left( \frac{5}{3}, \frac{2}{3}, \frac{2}{3} ight)$.

If $a = (1, 1, 1)$ and $c = (0, 1, -1)$ are two vectors and $b$ is a vector such that $a \times b = c$ and $a \cdot b = 3$,then $b$ is equal to

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