Find the shortest distance between the lines | vedclass

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(C) The formula for the shortest distance $(SD)$ between two lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ i

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$\sqrt{2}$

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(C) The formula for the shortest distance $(SD)$ between two lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ is given by $SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.Given $\vec{a}_1 = \hat{i} + \hat{j} - \hat{k}$,$\vec{b}_1 = 2\hat{i} - \hat{j} - \hat{k}$,$\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$,and $\vec{b}_2 = \hat{i}$.First,calculate $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & -1 \ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0 - (-1)) + \hat{k}(0 - (-1)) = -\hat{j} + \hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.Next,calculate $\vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + \hat{j} - \hat{k}) = -2\hat{j}$.Now,calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (-2\hat{j}) \cdot (-\hat{j} + \hat{k}) = 2$.Finally,$SD = \frac{|2|}{\sqrt{2}} = \sqrt{2}$.

Find the shortest distance between the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i})$.

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