The shortest distance between the lines frac{ | vedclass

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(A) The first line is $L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$.$A$ point on $L_{1}$ is $\vec{a}_{1} = (-7, 6, 0)$ and its direction vector

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$2 \sqrt{29}$

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(A) The first line is $L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$.$A$ point on $L_{1}$ is $\vec{a}_{1} = (-7, 6, 0)$ and its direction vector is $\vec{b}_{1} = (-6, 7, 1)$.The second line is $L_{2}: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}$.$A$ point on $L_{2}$ is $\vec{a}_{2} = (7, 2, 6)$ and its direction vector is $\vec{b}_{2} = (-2, 1, 1)$.The vector $\vec{a}_{2} - \vec{a}_{1} = (7 - (-7), 2 - 6, 6 - 0) = (14, -4, 6)$.The cross product $\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -6 & 7 & 1 \ -2 & 1 & 1 \end{vmatrix} = \hat{i}(7-1) - \hat{j}(-6+2) + \hat{k}(-6+14) = 6\hat{i} + 4\hat{j} + 8\hat{k} = (6, 4, 8)$.The magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{6^2 + 4^2 + 8^2} = \sqrt{36 + 16 + 64} = \sqrt{116} = 2\sqrt{29}$.The shortest distance $d = \left| \frac{(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} 	imes \vec{b}_{2})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight| = \left| \frac{(14, -4, 6) \cdot (6, 4, 8)}{2\sqrt{29}} ight| = \left| \frac{84 - 16 + 48}{2\sqrt{29}} ight| = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.

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

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