Find the shortest distance between the lines | vedclass

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(A) The equations of the given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.Comparing with the given

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$\frac{3 \sqrt{2}}{2}$ units

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(A) The equations of the given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.Comparing with the given lines:$\vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k}$,$\vec{b}_{1}=\hat{i}-\hat{j}+\hat{k}$$\vec{a}_{2}=2 \hat{i}-\hat{j}-\hat{k}$,$\vec{b}_{2}=2 \hat{i}+\hat{j}+2 \hat{k}$The shortest distance $d$ is given by $d=\left| \frac{(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight|$.First,calculate $\vec{a}_{2}-\vec{a}_{1} = (2\hat{i}-\hat{j}-\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = \hat{i}-3\hat{j}-2\hat{k}$.Next,calculate the cross product $\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2-1) - \hat{j}(2-2) + \hat{k}(1+2) = -3\hat{i}+3\hat{k}$.The magnitude is $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.Now,the dot product $(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1}) = (-3\hat{i}+3\hat{k}) \cdot (\hat{i}-3\hat{j}-2\hat{k}) = (-3)(1) + (0)(-3) + (3)(-2) = -3 - 6 = -9$.Finally,$d = \left| \frac{-9}{3\sqrt{2}} ight| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ units.

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

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