Question
Find the distance between parallel lines $\bar{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}-2 \hat{k})$ and$\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-2 \hat{k})$
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Answer
The distance between parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\lambda \bar{b}$ given by $d=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|$
Here $\bar{a}_1=2 \hat{i}-\hat{j}+\hat{k}, \quad \bar{a}_2=\hat{i}-\hat{j}+2 \hat{k}, \quad \bar{b}=2 \hat{i}+\hat{j}-2 \hat{k}$
$
\begin{aligned}
& \therefore \bar{a}_2-\bar{a}_1=(\hat{i}-\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}+\hat{k} \text { and } \hat{b}=\frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3} \\
& \therefore\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 0 & 1 \\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}
\end{array}\right|=\frac{1}{3}\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 0 & 1 \\
2 & 1 & -2
\end{array}\right|=\frac{1}{3}\{-\hat{i}-\hat{k}\} \\
& d=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|=\frac{\sqrt{2}}{3} \text { unit }
\end{aligned}
$
Here $\bar{a}_1=2 \hat{i}-\hat{j}+\hat{k}, \quad \bar{a}_2=\hat{i}-\hat{j}+2 \hat{k}, \quad \bar{b}=2 \hat{i}+\hat{j}-2 \hat{k}$
$
\begin{aligned}
& \therefore \bar{a}_2-\bar{a}_1=(\hat{i}-\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}+\hat{k} \text { and } \hat{b}=\frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3} \\
& \therefore\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 0 & 1 \\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}
\end{array}\right|=\frac{1}{3}\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 0 & 1 \\
2 & 1 & -2
\end{array}\right|=\frac{1}{3}\{-\hat{i}-\hat{k}\} \\
& d=\left|\left(\bar{a}_2-\bar{a}_1\right) \times \hat{b}\right|=\frac{\sqrt{2}}{3} \text { unit }
\end{aligned}
$
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