The shortest distance between lines bar{r}=(h | vedclass

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(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \h

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$\frac{2 \sqrt{2}}{\sqrt{19}}$ units

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(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \hat{k}$,$\bar{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}$,$\bar{a}_2 = 2\hat{i} - \hat{j} + 2\hat{k}$,and $\bar{b}_2 = \hat{i} - \hat{j} + \hat{k}$. First,calculate $\bar{a}_2 - \bar{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (2-(-1))\hat{k} = \hat{i} - 3\hat{j} + 3\hat{k}$. Next,calculate the cross product $\bar{b}_1 	imes \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1-3) - \hat{j}(2+3) + \hat{k}(-2-1) = -2\hat{i} - 5\hat{j} - 3\hat{k}$. The magnitude is $|\bar{b}_1 	imes \bar{b}_2| = \sqrt{(-2)^2 + (-5)^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$. The shortest distance is $d = \left| \frac{(\bar{b}_1 	imes \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)}{|\bar{b}_1 	imes \bar{b}_2|} ight| = \left| \frac{(-2\hat{i} - 5\hat{j} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 3\hat{k})}{\sqrt{38}} ight| = \left| \frac{-2 + 15 - 9}{\sqrt{38}} ight| = \frac{4}{\sqrt{38}} = \frac{4}{\sqrt{2} 	imes \sqrt{19}} = \frac{2 	imes 2}{\sqrt{2} 	imes \sqrt{19}} = \frac{2 \sqrt{2}}{\sqrt{19}}$ units.

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

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