Find the shortest distance between the lines | vedclass

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(A) Comparing $(1)$ and $(2)$ with $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$ respectively,we get $\vec{a}_{1

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$\frac{10}{\sqrt{59}}$

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(A) Comparing $(1)$ and $(2)$ with $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$ respectively,we get $\vec{a}_{1}=\hat{i}+\hat{j}, \quad \vec{b}_{1}=2 \hat{i}-\hat{j}+\hat{k}$$\vec{a}_{2}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}_{2}=3 \hat{i}-5 \hat{j}+2 \hat{k}$Therefore,$\vec{a}_{2}-\vec{a}_{1}=(2-1)\hat{i}+(1-1)\hat{j}+(-1-0)\hat{k} = \hat{i}-\hat{k}$Now,$\vec{b}_{1} 	imes \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2+5) - \hat{j}(4-3) + \hat{k}(-10+3) = 3\hat{i} - \hat{j} - 7\hat{k}$Magnitude $|\vec{b}_{1} 	imes \vec{b}_{2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9+1+49} = \sqrt{59}$The shortest distance $d$ is given by $d = \left| \frac{(\vec{b}_{1} 	imes \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} 	imes \vec{b}_{2}|} ight|$$d = \left| \frac{(3\hat{i} - \hat{j} - 7\hat{k}) \cdot (\hat{i} - \hat{k})}{\sqrt{59}} ight| = \left| \frac{3(1) + (-1)(0) + (-7)(-1)}{\sqrt{59}} ight| = \left| \frac{3+0+7}{\sqrt{59}} ight| = \frac{10}{\sqrt{59}}$

Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are
$\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})$ $(1)$
and $\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$ $(2)$

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Find the shortest distance between the lines $l_{1}$ and $l_{2}$ whose vector equations are$\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k})$ $(1)$and $\vec{r}=2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$ $(2)$