The shortest distance between the lines frac{ | vedclass

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(D) The given lines are $L_1: \frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $L_2: \frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$.Th

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$3\sqrt{30}$

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(D) The given lines are $L_1: \frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $L_2: \frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$.The points on the lines are $A(3, 8, 3)$ and $B(-3, -7, 6)$.The direction vectors are $\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.Calculate the cross product $\vec{b_1} 	imes \vec{b_2}$:$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 1 \ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4 - 2) - \hat{j}(12 - (-3)) + \hat{k}(6 - 3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.The vector $\vec{AB} = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k} = -6\hat{i} - 15\hat{j} + 3\hat{k}$.The shortest distance $d$ is given by $d = \frac{|\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.$\vec{AB} \cdot (\vec{b_1} 	imes \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.$d = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = \frac{90\sqrt{30}}{30} = 3\sqrt{30}$.

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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