If vec{i}+vec{j}-vec{k}, -vec{i}+2vec{j}+vec{ | vedclass

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(C) Let the position vectors be $\vec{a} = \vec{i}+\vec{j}-\vec{k}$,$\vec{b} = -\vec{i}+2\vec{j}+\vec{k}$,$\vec{c} = \vec{j}+2\vec{k}$,and $\vec{d} =

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$\frac{1}{3}$

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(C) Let the position vectors be $\vec{a} = \vec{i}+\vec{j}-\vec{k}$,$\vec{b} = -\vec{i}+2\vec{j}+\vec{k}$,$\vec{c} = \vec{j}+2\vec{k}$,and $\vec{d} = 2\vec{i}-\vec{j}+2\vec{k}$.Line $AB$ passes through $A$ with direction vector $\vec{p} = \vec{b}-\vec{a} = (-1-1)\vec{i} + (2-1)\vec{j} + (1-(-1))\vec{k} = -2\vec{i} + \vec{j} + 2\vec{k}$.Line $CD$ passes through $C$ with direction vector $\vec{q} = \vec{d}-\vec{c} = (2-0)\vec{i} + (-1-1)\vec{j} + (2-2)\vec{k} = 2\vec{i} - 2\vec{j} + 0\vec{k}$.The shortest distance $d$ between two lines is given by $d = \frac{|(\vec{c}-\vec{a}) \cdot (\vec{p} 	imes \vec{q})|}{|\vec{p} 	imes \vec{q}|}$.First,$\vec{c}-\vec{a} = (0-1)\vec{i} + (1-1)\vec{j} + (2-(-1))\vec{k} = -\vec{i} + 0\vec{j} + 3\vec{k}$.Next,$\vec{p} 	imes \vec{q} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ -2 & 1 & 2 \ 2 & -2 & 0 \end{vmatrix} = \vec{i}(0 - (-4)) - \vec{j}(0 - 4) + \vec{k}(4 - 2) = 4\vec{i} + 4\vec{j} + 2\vec{k}$.$|\vec{p} 	imes \vec{q}| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6$.$(\vec{c}-\vec{a}) \cdot (\vec{p} 	imes \vec{q}) = (-1)(4) + (0)(4) + (3)(2) = -4 + 0 + 6 = 2$.Thus,$d = \frac{|2|}{6} = \frac{1}{3}$.

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

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