The shortest distance between the lines vec{r | vedclass

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(C) The shortest distance $d$ 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 $d = \le

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$\frac{1}{\sqrt{3}}$

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(C) The shortest distance $d$ 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 $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$.Here, $\vec{a_1} = 4\hat{i} - \hat{j}$, $\vec{a_2} = \hat{i} - \hat{j} + 2\hat{k}$, $\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$, and $\vec{b_2} = \hat{i} + 4\hat{j} - 5\hat{k}$.$\vec{a_2} - \vec{a_1} = (1-4)\hat{i} + (-1 - (-1))\hat{j} + (2-0)\hat{k} = -3\hat{i} + 2\hat{k}$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 1 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2) = 2\hat{i} + 2\hat{j} + 2\hat{k}$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$.$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (-3\hat{i} + 0\hat{j} + 2\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 2\hat{k}) = -6 + 0 + 4 = -2$.$d = \left| \frac{-2}{2\sqrt{3}} ight| = \frac{1}{\sqrt{3}}$.

The shortest distance between the lines $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(\hat{i} + 4\hat{j} - 5\hat{k})$ is:

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