The shortest distance between the lines frac{ | vedclass

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(B) The lines are given by $\vec{r_1} = (3, 8, 3) + \lambda(3, -1, 1)$ and $\vec{r_2} = (-3, -7, 6) + \mu(-3, 2, 4)$.Let $\vec{a_1} = (3, 8, 3)$,$\vec

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$3\sqrt{30}$

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(B) The lines are given by $\vec{r_1} = (3, 8, 3) + \lambda(3, -1, 1)$ and $\vec{r_2} = (-3, -7, 6) + \mu(-3, 2, 4)$.Let $\vec{a_1} = (3, 8, 3)$,$\vec{a_2} = (-3, -7, 6)$,$\vec{b_1} = (3, -1, 1)$,and $\vec{b_2} = (-3, 2, 4)$.The shortest distance $d$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$.First,$\vec{a_2} - \vec{a_1} = (-3-3, -7-8, 6-3) = (-6, -15, 3)$.Next,$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12 - (-3)) + \hat{k}(6-3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.Thus,$d = \frac{|270|}{\sqrt{270}} = \sqrt{270} = 3\sqrt{30}$.

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

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