Question

Let ${{L}_{1}}\And {{L}_{2}}$ denote the lines $\overrightarrow{r}=\overset{\wedge }{\mathop{i}}\,+\lambda \left( -\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right),\lambda \in R$ and $\overrightarrow{r}=\mu \left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right),\mu \in R$ respectively. If ${{L}_{3}}$ is a line which is perpendicular to both ${{L}_{1}}\And {{L}_{2}}$ and cut both of them, then which of the following describe(s) ${{L}_{3}}$?
(a) $\overrightarrow{r}=\dfrac{1}{3}\left( 2\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
(b) $\overrightarrow{r}=\dfrac{2}{9}\left( 4\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
(c) $\overrightarrow{r}=\dfrac{2}{9}\left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
(d) $\overrightarrow{r}=t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$

Answer 1

Hint: We have given the two lines ${{L}_{1}}\And {{L}_{2}}$ and it is also given that there is a line ${{L}_{3}}$ which is perpendicular to both of these two lines which we are going to find the cross product of the direction ratios of line ${{L}_{1}}\And {{L}_{2}}$. After the cross product calculation, reduce the answer of cross product to the simplest integers then those reduced coefficients of $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,,\overset{\wedge }{\mathop{k}}\,$ are the direction ratios of the lines ${{L}_{3}}$. As you can see that lines ${{L}_{1}}\And {{L}_{2}}$ are skew lines so take one point on the line ${{L}_{1}}$ and other point on line ${{L}_{2}}$ and then find the direction ratios of the two points which is proportional to the one that we solved from the cross product. Then, find the values of the points that you have taken on ${{L}_{1}}\And {{L}_{2}}$ and also the midpoint of the two points. This will give you three points. And we can write a line as $\overrightarrow{a}+\lambda \overrightarrow{b}$ so $\overrightarrow{a}$ is one of the three points and $\overrightarrow{b}$ is the direction ratios of the cross product of the two lines.

Complete step-by-step answer:
We have given the two lines ${{L}_{1}}\And {{L}_{2}}$ as follows:
Line ${{L}_{1}}$ is given as:
$\overrightarrow{{{r}_{1}}}=\overset{\wedge }{\mathop{i}}\,+\lambda \left( -\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right),\lambda \in R$
Line ${{L}_{2}}$ is given as:
$\overrightarrow{{{r}_{2}}}=\mu \left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right),\mu \in R$
Now, we have to find the line ${{L}_{3}}$ such that it is perpendicular to both ${{L}_{1}}\And {{L}_{2}}$ so the direction ratio of the line ${{L}_{3}}$ is the cross product of the direction ratios of ${{L}_{1}}\And {{L}_{2}}$.
In the following, we have written the direction ratios of lines ${{L}_{1}}\And {{L}_{2}}$ as:
$\begin{align}
  & {{L}_{1}}=\left( -1,2,2 \right) \\
 & {{L}_{2}}=\left( 2,-1,2 \right) \\
\end{align}$
 Now, taking the cross product of ${{L}_{1}}\And {{L}_{2}}$ we get,
${{L}_{1}}\times {{L}_{2}}=\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   -1 & 2 & 2 \\
   2 & -1 & 2 \\
\end{matrix} \right|$
Expanding the determinant along the first row we get,
$\begin{align}
  & {{L}_{1}}\times {{L}_{2}}=\overset{\wedge }{\mathop{i}}\,\left( 2\left( 2 \right)-2\left( -1 \right) \right)-\overset{\wedge }{\mathop{j}}\,\left( -1\left( 2 \right)-2\left( 2 \right) \right)+\overset{\wedge }{\mathop{k}}\,\left( \left( -1 \right)\left( -1 \right)-2\left( 2 \right) \right) \\
 & \Rightarrow {{L}_{1}}\times {{L}_{2}}=\overset{\wedge }{\mathop{i}}\,\left( 6 \right)-\overset{\wedge }{\mathop{j}}\,\left( -6 \right)+\overset{\wedge }{\mathop{k}}\,\left( -3 \right) \\
 & \Rightarrow {{L}_{1}}\times {{L}_{2}}=6\overset{\wedge }{\mathop{i}}\,+6\overset{\wedge }{\mathop{j}}\,-3\overset{\wedge }{\mathop{k}}\, \\
\end{align}$
Now, taking 3 as common from the above we get,
${{L}_{1}}\times {{L}_{2}}=3\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right)$
Hence, the direction ratios of the line ${{L}_{3}}$ are equal to the coefficients of $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,,\overset{\wedge }{\mathop{k}}\,$ which is:
$\left( 2,2,-1 \right)$
As you can see, the direction ratios of the two lines ${{L}_{1}}\And {{L}_{2}}$ are not the same so these two lines are not parallel to each other.
Now, if the two lines will meet each other then:
$\overrightarrow{{{r}_{1}}}=\overrightarrow{{{r}_{2}}}$ for some values of $\lambda \And \mu $
$\begin{align}
  & \overset{\wedge }{\mathop{i}}\,+\lambda \left( -\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right)=\mu \left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overset{\wedge }{\mathop{i}}\,\left( 1-\lambda \right)+2\lambda \overset{\wedge }{\mathop{j}}\,+2\lambda \overset{\wedge }{\mathop{k}}\,=2\mu \overset{\wedge }{\mathop{i}}\,-\mu \overset{\wedge }{\mathop{j}}\,+2\mu \overset{\wedge }{\mathop{k}}\, \\
\end{align}$
Now, equating $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,,\overset{\wedge }{\mathop{k}}\,$ on both the sides we get,
$\begin{align}
  & 1-\lambda =2\mu , \\
 & 2\lambda =-\mu , \\
 & 2\lambda =2\mu \\
\end{align}$
Solving the above equations, we get,
$\begin{align}
  & \lambda =\mu , \\
 & \lambda =-\dfrac{1}{2}\mu \\
\end{align}$
As you can see that we have two different types of $\lambda $ are getting so $\overrightarrow{{{r}_{1}}}\ne \overrightarrow{{{r}_{2}}}$ so these two lines won’t meet at some point.
Hence, the two lines ${{L}_{1}}\And {{L}_{2}}$ are not parallel and not intersecting with respect to each other so the two lines are skew.
In the below diagram, we have drawn two lines ${{L}_{1}}\And {{L}_{2}}$ and the line ${{L}_{3}}$ which is perpendicular to both the lines

Let us assume two points each on lines ${{L}_{1}}\And {{L}_{2}}$ as point E and F with coordinates:
$\begin{align}
  & E\left( -\lambda +1,2\lambda ,2\lambda \right), \\
 & F\left( 2\mu ,-\mu ,2\mu \right) \\
\end{align}$
Now, EF is perpendicular to both lines ${{L}_{1}}\And {{L}_{2}}$ so EF lies on the line ${{L}_{3}}$ so the direction ratios of EF are equal to:
$\left( 2\mu +\lambda -1,-\mu -2\lambda ,2\mu -2\lambda \right)$
As we have already calculated the direction ratios for line ${{L}_{3}}$ as $\left( 2,2,-1 \right)$ so these direction ratios and the one that we have calculated above are proportional to each other.
$\dfrac{2\mu +\lambda -1}{2}=\dfrac{-\mu -2\lambda }{2}=\dfrac{2\mu -2\lambda }{-1}$
Solving the above equations, by taking first and seconds fraction and first and third fraction we get,
$\begin{align}
  & \dfrac{2\mu +\lambda -1}{2}=\dfrac{-\mu -2\lambda }{2} \\
 & \Rightarrow 2\mu +\lambda -1=-\mu -2\lambda \\
 & \Rightarrow 3\mu +3\lambda =1...........Eq.(1) \\
\end{align}$
$\begin{align}
  & \dfrac{2\mu +\lambda -1}{2}=\dfrac{2\mu -2\lambda }{-1} \\
 & \Rightarrow -2\mu -\lambda +1=4\mu -4\lambda \\
 & \Rightarrow 6\mu -3\lambda =1.........Eq.(2) \\
\end{align}$
Adding eq. (1) and eq. (2) we get,
$\begin{align}
  & 9\mu =2 \\
 & \Rightarrow \mu =\dfrac{2}{9} \\
\end{align}$
Substituting the above value of $\mu $ in eq. (1) we get,
\[\begin{align}
  & 3\left( \dfrac{2}{9} \right)+3\lambda =1 \\
 & \Rightarrow 3\lambda =1-\dfrac{6}{9} \\
 & \Rightarrow 3\lambda =1-\dfrac{2}{3} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 3\lambda =\dfrac{1}{3} \\
 & \Rightarrow \lambda =\dfrac{1}{9} \\
\end{align}\]
Hence, we got the value of $\lambda \And \mu $ as $\left( \dfrac{1}{9},\dfrac{2}{9} \right)$.
Substituting the above values in E and F we get,
$\begin{align}
  & E\left( -\dfrac{1}{9}+1,2\left( \dfrac{1}{9} \right),2\left( \dfrac{1}{9} \right) \right) \\
 & =E\left( \dfrac{8}{9},\left( \dfrac{2}{9} \right),\left( \dfrac{2}{9} \right) \right) \\
 & F\left( 2\left( \dfrac{2}{9} \right),-\left( \dfrac{2}{9} \right),2\left( \dfrac{2}{9} \right) \right) \\
 & =F\left( \left( \dfrac{4}{9} \right),-\left( \dfrac{2}{9} \right),\left( \dfrac{4}{9} \right) \right) \\
\end{align}$
Now, take the midpoint of E and F by adding the coordinates of E and F and then dividing them by 2.
$\begin{align}
  & \left( \dfrac{8}{9}+\dfrac{4}{9},\dfrac{2}{9}-\dfrac{2}{9},\dfrac{2}{9}+\dfrac{4}{9} \right) \\
 & =\left( \dfrac{12}{9},0,\dfrac{6}{9} \right) \\
 & =\left( \dfrac{4}{3},0,\dfrac{2}{3} \right) \\
\end{align}$
Now, the equation of line ${{L}_{3}}$ is equal to:
$\overrightarrow{r}=\overrightarrow{a}+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right)$
Now, $\overrightarrow{a}$ can be any of E, F and the midpoint of E and F.
 By substituting the point E in the above line ${{L}_{3}}$ equation we get,
$\overrightarrow{r}=\dfrac{2}{9}\left( 4\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
By substituting the point F in the above line ${{L}_{3}}$ equation we get,
$\overrightarrow{r}=\dfrac{2}{9}\left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
By substituting the midpoint we get,
$\overrightarrow{r}=\dfrac{1}{3}\left( 2\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{k}}\, \right)+t\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \right),t\in R$
Hence, the correct options are (a), (b) and (c).

Note: You can extract the concepts that you used in this problem and can use in other questions like if we have two lines and third line is perpendicular to both of these lines then how can we find the direction ratios of the third line and also how to know whether the two lines are parallel, intersecting or skew.