The vector equations of two lines are:r = i +2j +3k + (i-3j +2k) … - Vidyadip

Question

The vector equations of two lines are:$\overrightarrow{r} = \hat{i} +2\hat{j} +3\hat{k} +\lambda (\hat{i}-3\hat{j} +2\hat{k}) \text{and} \overrightarrow{r} = 4\hat{i} +5\hat{j} +6\hat{k} + \mu (2\hat{i}-3\hat{j} +\hat{k})$
Find the shortest distance between the above lines.

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Answer

$\text{Here} \overrightarrow{a}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}, \overrightarrow{b}_{1} = \hat{i} - 3\hat{j} + 2\hat{k}$$\overrightarrow{a}_{2} = 4\hat{i} + 5\hat{j} + 6\hat{k}, \overrightarrow{b}_{2} = 2\hat{i} + 3\hat{j} + \hat{k}$
$\therefore \overrightarrow{a}_{2} - \overrightarrow{a}_{1} = 3\hat{i} + 3\hat{j} +3\hat{k}$
$\overrightarrow{b}_{1} \times\overrightarrow{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = -9\hat{i} + 3\hat{j} + 9\hat{k} $
$\therefore \begin{vmatrix} \overrightarrow{b}_{1}\times&\overrightarrow{b}_{2}\end{vmatrix} = {\sqrt{9^{2} + 3^{2} +9^{2}}} = \sqrt{171} $
$\therefore \text{Shorted distance d is given by}$
$d=\Bigg| \frac{\big(\overrightarrow{b_{1}}\times\overrightarrow{b_{1}}\big).{\big(\overrightarrow{a_{2}} - \overrightarrow{a_{1}}\big)}}{\big|\overrightarrow{b_{1}}\times\overrightarrow{b_{1}}\big|}\Bigg|$
$= \bigg|\frac{-27 + 9 + 27}{\sqrt{171}}\bigg|=\frac{9}{\sqrt{171}} =\frac{3}{\sqrt{19}}$

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