Question
Find the shortest distance between the following pairs of parallel lines whose equations are:
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(4\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$

Answer

$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(4\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ or $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+2\mu\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
These two lines pass through the points having position vectors
$\vec{\text{a}}_1=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{a}}_2=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and are parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=\hat{\text{i}}-\hat{\text{k}}$
and
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}-\hat{\text{k}}\big)\times\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&0&-1\\2&-1&1 \end{vmatrix}$
$=-\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$\Rightarrow\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|=\sqrt{(-1)^2+(-3)^2+(-1)^2}$
$=\sqrt{1+9+1}$
$=\sqrt{11}$
The shortest distance between the two lines is given by
$\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}}\big|=\frac{\sqrt{11}}{\sqrt{6}}$

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