Question

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

• A. $3$
• B. $\frac{1}{\sqrt{6}}$
• C. $\frac{\sqrt{3}}{14}$
• D. $\frac{2}{\sqrt{13}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{6}}$

Given equation of lines
$\overrightarrow{r}=t(\hat{i}+\hat{j}+\hat{k})$
Comparing with $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$, we have
$\overrightarrow{a}=\overrightarrow{0},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$
and $\overrightarrow{r}=\hat{k}+s(\overrightarrow{i}+\widehat{2j}+\widehat{3k})$
Here, $\overrightarrow{c}=\hat{k},\overrightarrow{d}=\hat{i}+2\hat{j}+3\hat{k}$
Now, shortest distance $=\frac{|(\overrightarrow{c}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{d})|}{|\overrightarrow{b}\times \overrightarrow{d}|}$
$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|$
$\Rightarrow \overrightarrow{b}\times \overrightarrow{d}=\hat{i}-2\hat{j}+\hat{k}$
$|\overrightarrow{b}\times \overrightarrow{d}|=\sqrt{6}$
Hence, $d=\frac{|\hat{k}.(\hat{i}-2\hat{j}+\hat{k})|}{\sqrt{6}}$
$d=\frac{1}{\sqrt{6}}$