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The given equations are: $\vec{r}=(1-\lambda)\hat{i}+(\lambda-2)\hat{j}+(3-2\lambda)\hat{k}$ $\vec{r}=(\mu+1)\hat{i}+(2\mu-1)\hat{j}-(2\mu+1)\hat{k}$ We can rewrite these as: $\vec{r}=(\hat{i}-2\hat{j}+3\hat{k}) + \lambda(-\hat{i}+\hat{j}-2\hat{k})$ $\vec{r}=(\hat{i}-\hat{j}-\hat{k}) + \mu(\hat{i}+2\hat{j}-2\hat{k})$ Comparing with the standard form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$, we have: $\vec{a_1} = \hat{i}-2\hat{j}+3\hat{k}$ $\vec{b_1} = -\hat{i}+\hat{j}-2\hat{k}$ $\vec{a_2} = \hat{i}-\hat{j}-\hat{k}$ $\vec{b_2} = \hat{i}+2\hat{j}-2\hat{k}$
For the lines to be parallel, $\vec{b_1}$ and $\vec{b_2}$ must be proportional. $\frac{-1}{1} \neq \frac{1}{2} \neq \frac{-2}{-2}$ Since the ratios are not equal, the lines are not parallel.
For the lines to intersect, the shortest distance between them must be zero. The shortest distance is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ If $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$, the lines intersect. $\vec{a_2} - \vec{a_1} = (\hat{i}-\hat{j}-\hat{k}) - (\hat{i}-2\hat{j}+3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$ $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2+4) - \hat{j}(2+2) + \hat{k}(-2-1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$ $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k}) = 0 - 4 + 12 = 8$ Since $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \neq 0$, the lines do not intersect.
Since the lines are neither parallel nor intersecting, they are skew lines.
The shortest distance between the skew lines is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ We already have $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 8$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$ Therefore, the shortest distance is: $d = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{29}$: $d = \frac{8}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{8\sqrt{29}}{29}$
Final Answer: $\frac{8\sqrt{29}}{29}$
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