The distance between parallel lines vec{r}=(2 | vedclass

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(D) The distance $d$ between two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is given by $d = \frac{|

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$\frac{\sqrt{2}}{3}$ units

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(D) The distance $d$ between two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) 	imes \vec{b}|}{|\vec{b}|}$.Here,$\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.First,calculate $\vec{a}_2 - \vec{a}_1 = (1-2)\hat{i} + (-1 - (-1))\hat{j} + (2-1)\hat{k} = -\hat{i} + \hat{k}$.Next,calculate the cross product $(\vec{a}_2 - \vec{a}_1) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 0 & 1 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(2-2) + \hat{k}(-1-0) = -\hat{i} - \hat{k}$.The magnitude is $|(\vec{a}_2 - \vec{a}_1) 	imes \vec{b}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.Therefore,$d = \frac{\sqrt{2}}{3}$ units.

The distance between parallel lines $\vec{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda(2\hat{i}+\hat{j}-2\hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(2\hat{i}+\hat{j}-2\hat{k})$ is

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