Question

If the shortest distance between the lines $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( 2\hat{i}+3\hat{j}+4\hat{k} \right)$ and $\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu \left( 3\hat{i}+4\hat{j}+5\hat{k} \right)$ is k, then the value of ${{\tan }^{-1}}\tan \left( 2\sqrt{6}k \right)$ should be given by
(A). 1
(B). 2
(C). $2-\pi $
(D). $\dfrac{\pi }{2}-2$

Answer 1

Hint: In the question, we are already given the equations of the straight lines and we have to find the shortest distance between them which will give us the value of k. Using that value of k, we can find ${{\tan }^{-1}}\tan \left( 2\sqrt{6}k \right)$, which will be our required answer.

Complete step-by-step solution -
Let
\[\begin{align}
  & {{{\vec{a}}}_{1}}=\hat{i}+2\hat{j}+3\hat{k} \\
 & {{{\vec{a}}}_{2}}=2\hat{i}+3\hat{j}+4\hat{k} \\
 & {{{\vec{a}}}_{3}}=2\hat{i}+4\hat{j}+5\hat{k} \\
 & {{{\vec{a}}}_{4}}=3\hat{i}+4\hat{j}+5\hat{k}..............(1.1) \\
\end{align}\]
Then the straight lines given in the question can be written as
$\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{a}}_{2}}$
and
$\vec{r}={{\vec{a}}_{3}}+\lambda {{\vec{a}}_{4}}$
respectively.
We know that the shortest distance between two lines represented by $\vec{r}=\vec{p}+\lambda \vec{q}$ and
$\vec{r}=\vec{t}+\lambda \vec{s}$ is given by
$\text{dist=}\left| \dfrac{\left( \vec{t}-\vec{p} \right).\left( \vec{q}\times \vec{s} \right)}{\left| \vec{q}\times \vec{s} \right|} \right|..............(1.2)$
Therefore, in this case, the shortest between the lines should be
$\text{dist=k=}\left| \dfrac{\left( {{{\vec{a}}}_{3}}-{{{\vec{a}}}_{1}} \right).\left( {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right)}{\left| {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right|} \right|..............(1.3)$
From (1.1), we get
$\begin{align}
  & {{{\vec{a}}}_{3}}-{{{\vec{a}}}_{1}}=\left( 2-1 \right)\hat{i}+\left( 4-2 \right)\hat{j}+\left( 5-3 \right)\hat{k} \\
 & =\hat{i}+2\hat{j}+2\hat{k}..........................(1.4) \\
\end{align}$
And the cross product should be given by
$\begin{align}
  & {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}}=\left| \begin{matrix}
   {\overset{\scriptscriptstyle\frown}{i}} & {\overset{\scriptscriptstyle\frown}{j}} & {\overset{\scriptscriptstyle\frown}{k}} \\
   2 & 3 & 4 \\
   3 & 4 & 5 \\
\end{matrix} \right|=\left( 3\times 5-4\times 4 \right)\hat{i}+\left( 3\times 4-2\times 5 \right)\hat{j}+\left( 2\times 4-3\times 3 \right)\hat{k} \\
 & =-\hat{i}+2\hat{j}-\hat{k}..................(1.5) \\
\end{align}$
Also, we know that the dot product of two vectors is given by
\[\left( a\hat{i}+b\hat{j}+c\hat{k} \right).\left( c\hat{i}+d\hat{j}+e\hat{k} \right)=ac+bd+ce...........(1.6)\]
Therefore, using equation (1.6) in equation (1.4) and (1.5), we obtain
\[\begin{align}
  & \left( {{{\vec{a}}}_{3}}-{{{\vec{a}}}_{1}} \right).\left( {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right)=\left( \hat{i}+2\hat{j}+2\hat{k} \right).\left( -\hat{i}+2\hat{j}+-\hat{k} \right) \\
 & =1\times -1+2\times 2+2\times -1=1................(1.7) \\
\end{align}\]
And also we know that the magnitude of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is given by
$\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$. Therefore, from equation (1.5), we have
$\left| {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right|=\left| -\overset{\scriptscriptstyle\frown}{i}+2\overset{\scriptscriptstyle\frown}{j}-\overset{\scriptscriptstyle\frown}{k} \right|=\sqrt{-{{1}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{6}..................(1.8)$
Therefore, using equations (1.7) and (1.8) in equation (1.3), we have
$\text{k=}\left| \dfrac{\left( {{{\vec{a}}}_{3}}-{{{\vec{a}}}_{1}} \right).\left( {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right)}{\left| {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right|} \right|=\left| \dfrac{1}{\sqrt{6}} \right|=\dfrac{1}{\sqrt{6}}..............(1.10)$
Therefore, we have $2\sqrt{6}k=2\sqrt{6}\times \dfrac{1}{\sqrt{6}}=2................(1.11)$
As for any angle, ${{\tan }^{-1}}\tan \left( \theta \right)=\theta $, therefore taking $\theta =2\sqrt{6}k$ and using equation (1.11), we have
${{\tan }^{-1}}\tan \left( 2\sqrt{6}k \right)=2\sqrt{6}k=2$
Which matches option (b) of the question. Hence (b) is the correct answer to this question.

Note: We should note that in equation (1.3), we could have taken $\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{3}} \right)$ instead of $\left( {{{\vec{a}}}_{3}}-{{{\vec{a}}}_{1}} \right)$ or $\left( {{{\vec{a}}}_{4}}\times {{{\vec{a}}}_{2}} \right)$ instead of $\left( {{{\vec{a}}}_{2}}\times {{{\vec{a}}}_{4}} \right)$ as there is an overall mod and thus the sign of obtained value gets cancelled out. This is done because the shortest distance should be a positive number.