The shortest distance between the lines bar{r | vedclass

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(C) The given lines are $\ell_{1}: \bar{r} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})$ and $\ell_{2}: \bar{r} = (\hat{i}-\hat{j}+\ha

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$\frac{2}{\sqrt{5}} 	ext{ units}$

Vedclass · Answer

(C) The given lines are $\ell_{1}: \bar{r} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})$ and $\ell_{2}: \bar{r} = (\hat{i}-\hat{j}+\hat{k}) + p(\hat{i}+2\hat{j}+2\hat{k})$.Here,$\bar{a}_{1} = \hat{i}-2\hat{j}+3\hat{k}$,$\bar{b}_{1} = -\hat{i}+\hat{j}-2\hat{k}$,$\bar{a}_{2} = \hat{i}-\hat{j}+\hat{k}$,and $\bar{b}_{2} = \hat{i}+2\hat{j}+2\hat{k}$.First,calculate $\bar{a}_{2}-\bar{a}_{1} = (\hat{i}-\hat{j}+\hat{k}) - (\hat{i}-2\hat{j}+3\hat{k}) = \hat{j}-2\hat{k}$.Next,calculate the cross product $\bar{b}_{1} 	imes \bar{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & -2 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(2+4) - \hat{j}(-2+2) + \hat{k}(-2-1) = 6\hat{i} - 3\hat{k}$.The magnitude is $|\bar{b}_{1} 	imes \bar{b}_{2}| = \sqrt{6^2 + (-3)^2} = \sqrt{36+9} = \sqrt{45} = 3\sqrt{5}$.The shortest distance $d = \left| \frac{(\bar{b}_{1} 	imes \bar{b}_{2}) \cdot (\bar{a}_{2}-\bar{a}_{1})}{|\bar{b}_{1} 	imes \bar{b}_{2}|} ight| = \left| \frac{(6\hat{i}-3\hat{k}) \cdot (\hat{j}-2\hat{k})}{3\sqrt{5}} ight| = \left| \frac{0 + 0 + 6}{3\sqrt{5}} ight| = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$ units.

The shortest distance between the lines $\bar{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\bar{r}=(p+1) \hat{i}+(2 p-1) \hat{j}+(2 p+1) \hat{k}$ is

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