Question

If the shortest distance between the lines $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left(\begin{array}{c}2\hat{i}+3\hat{j}+4\hat{k}\end{array}\right)$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu \left(\begin{array}{c}3\hat{i}+4\hat{j}+5\hat{k}\end{array}\right)$ is $k$, then the value of ${\tan }^{-1}\tan \left(\begin{array}{c}2\sqrt{6}k\end{array}\right)$ is

• A. $1$
• B. $2$
• C. $2-\pi$
• D. $\frac{\pi }{2}-2$

Answer 1

The correct option is C $2-\pi$

Given equation of lines
$\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left(\begin{array}{c}2\hat{i}+3\hat{j}+4\hat{k}\end{array}\right)$
Here, the line passes through $A(1,2,3)$ and $n_1=2\hat{i}+3\hat{j}+4\hat{k}$
$\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu \left(\begin{array}{c}3\hat{i}+4\hat{j}+5\hat{k}\end{array}\right)$
Here, the line passes through $B(2,4,5)$ and $n_2=3\hat{i}+4\hat{j}+5\hat{k}$
Now, $\overrightarrow{AB}=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})$
$\Rightarrow \overrightarrow{AB}=\hat{i}+2\hat{j}+2\hat{k}$
Now, $n_1\times n_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$
$\Rightarrow n_1\times n_2=-\hat{i}+2\hat{j}-\hat{k}$
Now, $k=\left|\begin{array}{c}\frac{\overrightarrow{AB}.(\overrightarrow{n_1}\times \overrightarrow{n_2})}{\left|\begin{array}{c}\overrightarrow{n_1}\times \overrightarrow{n_2}\end{array}\right|}\end{array}\right|$
$=\left|\begin{array}{c}\frac{\left(\begin{array}{c}\hat{i}+2\hat{j}+2\hat{k}\end{array}\right).\left(\begin{array}{c}-\hat{i}+2\hat{j}-\hat{k}\end{array}\right)}{\sqrt{6}}\end{array}\right|=\frac{1}{\sqrt{6}}$
$\Rightarrow k=\frac{1}{\sqrt{6}}$
$\therefore {\tan }^{-1}\tan \left(\begin{array}{c}2\sqrt{6}k\end{array}\right)={\tan }^{-1}\tan 2$
Since, $2\notin (-\frac{\pi }{2},\frac{\pi }{2})$
So, $={\tan }^{-1}(-\tan (\pi -2\left)\right)$
$=2-\pi$ ($\because {\tan }^{-1}\left(\tan x\right)=x;x\in (-\frac{\pi }{2},\frac{\pi }{2})$)