Let vec{a} = 2hat{i} + hat{k}, vec{b} = hat{i | vedclass

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(B) Given $\vec{r} 	imes \vec{b} = \vec{c} 	imes \vec{b}$, we can write $(\vec{r} - \vec{c}) 	imes \vec{b} = 0$. This implies $\vec{r} - \vec{c} =

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$34$

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(B) Given $\vec{r} 	imes \vec{b} = \vec{c} 	imes \vec{b}$, we can write $(\vec{r} - \vec{c}) 	imes \vec{b} = 0$. This implies $\vec{r} - \vec{c} = t\vec{b}$ for some scalar $t$, so $\vec{r} = \vec{c} + t\vec{b}$.Substituting $\vec{r}$ into $\vec{r} \cdot \vec{a} = 0$, we get $(\vec{c} + t\vec{b}) \cdot \vec{a} = 0$, which means $\vec{c} \cdot \vec{a} + t(\vec{b} \cdot \vec{a}) = 0$.Calculate $\vec{c} \cdot \vec{a} = (4)(2) + (-3)(0) + (7)(1) = 8 + 0 + 7 = 15$.Calculate $\vec{b} \cdot \vec{a} = (1)(2) + (1)(0) + (1)(1) = 2 + 0 + 1 = 3$.Thus, $15 + 3t = 0$, which gives $t = -5$.Now, $\vec{r} = \vec{c} - 5\vec{b} = (4\hat{i} - 3\hat{j} + 7\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - 8\hat{j} + 2\hat{k}$.Finally, $\vec{r} \cdot \vec{c} = (-1)(4) + (-8)(-3) + (2)(7) = -4 + 24 + 14 = 34$.

Let $\vec{a} = 2\hat{i} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$, then $\vec{r} \cdot \vec{c} = $

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