If a = (1, -1, 1) and c = (-1, -1, 0), then t | vedclass

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(D) Let $b = b_1 i + b_2 j + b_3 k$.Given $a \cdot b = 1$,we have $(i - j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1 \Rightarrow b_1 - b_2 + b_3 = 1$ ...

Vedclass · Accepted Answer

None of these

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(D) Let $b = b_1 i + b_2 j + b_3 k$.Given $a \cdot b = 1$,we have $(i - j + k) \cdot (b_1 i + b_2 j + b_3 k) = 1 \Rightarrow b_1 - b_2 + b_3 = 1$ ... $(i)$Given $a 	imes b = c$,we calculate the cross product:$a 	imes b = \begin{vmatrix} i & j & k \ 1 & -1 & 1 \ b_1 & b_2 & b_3 \end{vmatrix} = i(-b_3 - b_2) - j(b_3 - b_1) + k(b_2 + b_1) = (-b_2 - b_3)i + (b_1 - b_3)j + (b_1 + b_2)k$.Equating this to $c = -i - j + 0k$,we get:$-b_2 - b_3 = -1 \Rightarrow b_2 + b_3 = 1$ ... $(ii)$$b_1 - b_3 = -1$ ... $(iii)$$b_1 + b_2 = 0$ ... $(iv)$From $(iv)$,$b_2 = -b_1$. Substituting into $(ii)$,$-b_1 + b_3 = 1 \Rightarrow b_3 - b_1 = 1$.Adding this to $(iii)$,$(b_3 - b_1) + (b_1 - b_3) = 1 - 1 \Rightarrow 0 = 0$.Using $(i)$,$b_1 - (-b_1) + b_3 = 1 \Rightarrow 2b_1 + b_3 = 1$.From $(iii)$,$b_3 = b_1 - 1$.Substituting into $2b_1 + b_3 = 1$: $2b_1 + b_1 - 1 = 1 \Rightarrow 3b_1 = 2 \Rightarrow b_1 = 2/3$.This leads to $b_2 = -2/3$ and $b_3 = -1/3$.Checking the options,none match $(2/3, -2/3, -1/3)$. Thus,the correct answer is None of these.

If $a = (1, -1, 1)$ and $c = (-1, -1, 0),$ then the vector $b$ satisfying $a \times b = c$ and $a \cdot b = 1$ is

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