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(D) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ is given by the

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$\frac{6}{\sqrt{5}}$

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(D) The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ is given by the formula:$d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$Comparing the given equations with the standard form,we have:$\vec{a_1} = 4\hat{i} - \hat{j}$,$\vec{a_2} = \hat{i} - \hat{j} + 2\hat{k}$$\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$,$\vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$First,calculate $\vec{a_2} - \vec{a_1} = (1-4)\hat{i} + (-1 - (-1))\hat{j} + (2-0)\hat{k} = -3\hat{i} + 0\hat{j} + 2\hat{k}$.Next,calculate the cross product $\vec{b_1} 	imes \vec{b_2}$:$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 - (-12)) - \hat{j}(-5 - (-6)) + \hat{k}(4 - 4) = 2\hat{i} - 1\hat{j} + 0\hat{k}$.Now,calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})$:$(-3\hat{i} + 0\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 1\hat{j} + 0\hat{k}) = (-3)(2) + (0)(-1) + (2)(0) = -6 + 0 + 0 = -6$.Calculate the magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5}$.Finally,the shortest distance $d = \left| \frac{-6}{\sqrt{5}} ight| = \frac{6}{\sqrt{5}}$.

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

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