Question

The shortest distance between the lines $\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k}),\lambda ϵR$ and $\overrightarrow{r}=(-\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k}),\mu ϵR$ is

• A. $\frac{6}{5}$
• B. $\frac{1}{\sqrt{5}}$
• C. $\frac{10}{\sqrt{5}}$
• D. none of these

Answer 1

The correct option is C $\frac{10}{\sqrt{5}}$

Let $\overrightarrow{a_1}=4\hat{i}-\hat{j},a_2=-\hat{i}-\hat{j}+2\hat{k}$
$\overrightarrow{b_1}=\hat{i}+2\hat{j}-3\hat{k},\overrightarrow{b_2}=2\hat{i}+4\hat{j}-5\hat{k}$
Now, $\overrightarrow{a_2}-\overrightarrow{a_1}=-5\hat{i}+0\hat{j}+2\hat{k}$
and $\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|=2\hat{i}-\hat{j}+0\hat{k}$
Again now, $(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})$
$=(-5\hat{i}+0\hat{j}+2\hat{k}).(2\hat{i}-\hat{j}+0\hat{k})=-10$
and $|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{4+1+0}=\sqrt{5}$
$\therefore$ shortest distance $=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1}).(\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$
$=|-\frac{10}{\sqrt{5}}|=\frac{10}{\sqrt{5}}$