The shortest distance between the lines x+1=2 | vedclass

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(A) First,express the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $x+1 = 2y = -12z \Rightarrow \

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$2$

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(A) First,express the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $x+1 = 2y = -12z \Rightarrow \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$. Point $A = (-1, 0, 0)$,direction vector $\vec{p} = (1, 1/2, -1/12)$.For the second line: $x = y+2 = 6z-6 \Rightarrow \frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$. Point $B = (0, -2, 1)$,direction vector $\vec{q} = (1, 1, 1/6)$.The vector $\vec{B}-\vec{A} = (0-(-1), -2-0, 1-0) = (1, -2, 1) = \hat{i} - 2\hat{j} + \hat{k}$.The cross product $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1/2 & -1/12 \ 1 & 1 & 1/6 \end{vmatrix} = \hat{i}(\frac{1}{12} + \frac{1}{12}) - \hat{j}(\frac{1}{6} + \frac{1}{12}) + \hat{k}(1 - \frac{1}{2}) = \frac{1}{6}\hat{i} - \frac{1}{4}\hat{j} + \frac{1}{2}\hat{k}$.Scaling the vector $\vec{p} 	imes \vec{q}$ by $12$,we get $2\hat{i} - 3\hat{j} + 6\hat{k}$.The magnitude $|\vec{p} 	imes \vec{q}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.The shortest distance is $\left| \frac{(\vec{B}-\vec{A}) \cdot (\vec{p} 	imes \vec{q})}{|\vec{p} 	imes \vec{q}|} ight| = \left| \frac{(1, -2, 1) \cdot (2, -3, 6)}{7} ight| = \left| \frac{2 + 6 + 6}{7} ight| = \frac{14}{7} = 2$.

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is

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