Let overrightarrow{a} = hat{i} + 2hat{j} - 3h | vedclass

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(B) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{b} 	imes \overrightarrow{r}$,we can write $\overrightarrow{r} 	imes \overr

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$15$

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(B) Given $\overrightarrow{r} 	imes \overrightarrow{a} = \overrightarrow{b} 	imes \overrightarrow{r}$,we can write $\overrightarrow{r} 	imes \overrightarrow{a} + \overrightarrow{r} 	imes \overrightarrow{b} = 0$,which implies $\overrightarrow{r} 	imes (\overrightarrow{a} + \overrightarrow{b}) = 0$.This means $\overrightarrow{r}$ is parallel to $(\overrightarrow{a} + \overrightarrow{b})$.$\overrightarrow{a} + \overrightarrow{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (2\hat{i} - 3\hat{j} + 5\hat{k}) = 3\hat{i} - \hat{j} + 2\hat{k}$.So,$\overrightarrow{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k})$ for some scalar $\lambda$.Substituting $\overrightarrow{r}$ into $\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$:$\lambda(3\alpha - 2 + 2) = 3 \Rightarrow 3\lambda\alpha = 3 \Rightarrow \lambda\alpha = 1$.Substituting $\overrightarrow{r}$ into $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$:$\lambda(6 - 5 - 2\alpha) = -1 \Rightarrow \lambda(1 - 2\alpha) = -1 \Rightarrow \lambda - 2\lambda\alpha = -1$.Since $\lambda\alpha = 1$,we have $\lambda - 2(1) = -1 \Rightarrow \lambda = 1$.Then $\alpha = 1/\lambda = 1$.Thus,$\overrightarrow{r} = 3\hat{i} - \hat{j} + 2\hat{k}$.$|\overrightarrow{r}|^{2} = 3^{2} + (-1)^{2} + 2^{2} = 9 + 1 + 4 = 14$.Finally,$\alpha + |\overrightarrow{r}|^{2} = 1 + 14 = 15$.

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