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(A) Given $\vec{r} 	imes \vec{a} = \vec{b}$.Since $\vec{a}$ and $\vec{b}$ are perpendicular,$\vec{a} \cdot \vec{b} = 0$.We can express $\vec{r}$ as a

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$\vec{r} = x\vec{a} + \frac{1}{\vec{a} \cdot \vec{a}} (\vec{a} 	imes \vec{b})$

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(A) Given $\vec{r} 	imes \vec{a} = \vec{b}$.Since $\vec{a}$ and $\vec{b}$ are perpendicular,$\vec{a} \cdot \vec{b} = 0$.We can express $\vec{r}$ as a linear combination of $\vec{a}$,$\vec{b}$,and $\vec{a} 	imes \vec{b}$:$\vec{r} = x\vec{a} + y\vec{b} + z(\vec{a} 	imes \vec{b})$ for some scalars $x, y, z$.Taking the cross product with $\vec{a}$ on both sides:$(\vec{r} 	imes \vec{a}) = (x\vec{a} + y\vec{b} + z(\vec{a} 	imes \vec{b})) 	imes \vec{a}$$\vec{b} = x(\vec{a} 	imes \vec{a}) + y(\vec{b} 	imes \vec{a}) + z((\vec{a} 	imes \vec{b}) 	imes \vec{a})$Since $\vec{a} 	imes \vec{a} = 0$ and $(\vec{a} 	imes \vec{b}) 	imes \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{a})\vec{a}$:$\vec{b} = y(\vec{b} 	imes \vec{a}) + z((\vec{a} \cdot \vec{a})\vec{b} - 0)$$\vec{b} = -y(\vec{a} 	imes \vec{b}) + z(\vec{a} \cdot \vec{a})\vec{b}$Comparing coefficients of $\vec{b}$ and $(\vec{a} 	imes \vec{b})$:$-y = 0 \implies y = 0$$z(\vec{a} \cdot \vec{a}) = 1 \implies z = \frac{1}{\vec{a} \cdot \vec{a}}$Substituting these into the expression for $\vec{r}$:$\vec{r} = x\vec{a} + \frac{1}{\vec{a} \cdot \vec{a}} (\vec{a} 	imes \vec{b})$.

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is:

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If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=\hat{j}-\hat{k}$,find a vector $\bar{c}$ such that $\bar{a} \times \bar{c}=\bar{b}$ and $\bar{a} \cdot \bar{c}=3$.

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