If overrightarrow{a} = hat{i} + 2hat{k},overr | vedclass

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(A) Given $\overrightarrow{r} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{c} = \overrightarrow{0}$.This can be written as $\

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$34$

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(A) Given $\overrightarrow{r} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{c} = \overrightarrow{0}$.This can be written as $\overrightarrow{r} 	imes \overrightarrow{b} - \overrightarrow{c} 	imes \overrightarrow{b} = \overrightarrow{0}$,which implies $(\overrightarrow{r} - \overrightarrow{c}) 	imes \overrightarrow{b} = \overrightarrow{0}$.This means $\overrightarrow{r} - \overrightarrow{c} = \lambda \overrightarrow{b}$ for some scalar $\lambda$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{b}$.Given $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,we substitute $\overrightarrow{r}$:$(\overrightarrow{c} + \lambda \overrightarrow{b}) \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{a} + \lambda (\overrightarrow{b} \cdot \overrightarrow{a}) = 0$.Calculating the dot products:$\overrightarrow{c} \cdot \overrightarrow{a} = (7)(1) + (-3)(0) + (4)(2) = 7 + 0 + 8 = 15$.$\overrightarrow{b} \cdot \overrightarrow{a} = (1)(1) + (1)(0) + (1)(2) = 1 + 0 + 2 = 3$.Thus,$15 + \lambda(3) = 0 \Rightarrow \lambda = -5$.Now,$\overrightarrow{r} = \overrightarrow{c} - 5\overrightarrow{b} = (7\hat{i} - 3\hat{j} + 4\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = 2\hat{i} - 8\hat{j} - \hat{k}$.Finally,$\overrightarrow{r} \cdot \overrightarrow{c} = (2\hat{i} - 8\hat{j} - \hat{k}) \cdot (7\hat{i} - 3\hat{j} + 4\hat{k}) = (2)(7) + (-8)(-3) + (-1)(4) = 14 + 24 - 4 = 34$.

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