Question

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

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

Answer 1

Answer: (B)

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

Let

$a=\hat{1}+2\hat{j}+3\hat{k}$

$b=2\hat{i}+3\hat{j}+4\hat{k}$

$c=2\hat{i}+4\hat{j}+5\hat{k}$

$d=3\hat{i}+4\hat{j}+5\hat{k}$

We get

$\overrightarrow{r}=a+s\overrightarrow{b}\overline{r}=\overline{c}+t\overrightarrow{d}$

So shortest distance $=\frac{\left|\right(\overrightarrow{a}-\overrightarrow{c}).\overrightarrow{b}\times \overrightarrow{d}|}{|\overrightarrow{b}\times \overrightarrow{d}|}$
$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\hat{i}+2\hat{j}-\hat{k}$

$\overrightarrow{a}-\overrightarrow{c}=-\hat{i}-2\hat{j}-\alpha \hat{k}$

Shortest Distance$=\frac{|1-4+2|}{\sqrt{6}}=\frac{1}{\sqrt{6}}$