Question

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

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{6}}$

The shortest distance between $\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{c}+t\overrightarrow{d}$ is
$|(\overrightarrow{c}-\overrightarrow{a}).\hat{n}|$ where $\hat{n}=\frac{\overrightarrow{b}\times \overrightarrow{d}}{|\overrightarrow{b}\times \overrightarrow{d}|}$ and $\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$
$=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)$
$=-\hat{i}+2\hat{j}-\hat{k}$
$\therefore \hat{n}=\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{1+4+1}}=\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}$
$\Rightarrow$ shortest distance $=|(\overrightarrow{c}-\overrightarrow{a}).\hat{n}|$
We have $\overrightarrow{c}-\overrightarrow{a}=-\hat{i}-\hat{i}-\hat{j}-\hat{k}=-(\hat{j}+\hat{k})$
$=\frac{\left|(\hat{j}+\hat{k})(-\hat{i}+2\hat{j}-\hat{k})\right|}{\sqrt{6}}$
$=\frac{2-1}{\sqrt{6}}=\frac{1}{\sqrt{6}}$