Let vec{a}=9 hat{i}-13 hat{j}+25 hat{k},vec{b | vedclass

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(D) Given $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$.From $\vec{r}

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$569$

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(D) Given $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$.From $\vec{r} 	imes \vec{a}=(\vec{b}+\vec{c}) 	imes \vec{a}$,we have $(\vec{r}-(\vec{b}+\vec{c})) 	imes \vec{a} = 0$.This implies $\vec{r}-(\vec{b}+\vec{c}) = \lambda \vec{a}$ for some scalar $\lambda$,so $\vec{r} = \lambda \vec{a} + \vec{b} + \vec{c}$.Given $\vec{r} \cdot (\vec{b}-\vec{c}) = 0$,substitute $\vec{r}$:$(\lambda \vec{a} + \vec{b} + \vec{c}) \cdot (\vec{b}-\vec{c}) = 0$.$\lambda (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}) + (\vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) = 0$.$\lambda (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}) + |\vec{b}|^2 - |\vec{c}|^2 = 0$.Calculate dot products: $\vec{a} \cdot \vec{b} = (9)(3) + (-13)(7) + (25)(-13) = 27 - 91 - 325 = -389$.$\vec{a} \cdot \vec{c} = (9)(17) + (-13)(-2) + (25)(1) = 153 + 26 + 25 = 204$.$|\vec{b}|^2 = 3^2 + 7^2 + (-13)^2 = 9 + 49 + 169 = 227$.$|\vec{c}|^2 = 17^2 + (-2)^2 + 1^2 = 289 + 4 + 1 = 294$.$\lambda (-389 - 204) + 227 - 294 = 0 \Rightarrow -593 \lambda - 67 = 0 \Rightarrow \lambda = -\frac{67}{593}$.Thus,$\vec{r} = -\frac{67}{593} \vec{a} + \vec{b} + \vec{c}$.$593 \vec{r} + 67 \vec{a} = 593(\vec{b} + \vec{c})$.$\frac{|593 \vec{r} + 67 \vec{a}|^2}{(593)^2} = |\vec{b} + \vec{c}|^2$.$\vec{b} + \vec{c} = (3+17)\hat{i} + (7-2)\hat{j} + (-13+1)\hat{k} = 20\hat{i} + 5\hat{j} - 12\hat{k}$.$|\vec{b} + \vec{c}|^2 = 20^2 + 5^2 + (-12)^2 = 400 + 25 + 144 = 569$.

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