Question

Three lines
$\begin{align}
  & {{L}_{1}}=\vec{r}=\lambda \hat{i},\lambda \in R \\
 & {{L}_{2}}=\vec{r}=\hat{k}+\mu \hat{j},\mu \in R \\
 & {{L}_{3}}=\vec{r}=\hat{i}+\hat{j}+\nu \hat{k},\mu \in R \\
\end{align}$
are given. For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P, Q and R are collinear
(a) $\hat{k}+\dfrac{1}{2}\hat{j}$
(b) $\hat{k}+\hat{j}$
(c) $\hat{k}$
(d) $\hat{k}-\dfrac{1}{2}\hat{j}$
(This question has multiple correct options)

Answer 1

Hint: To solve this question, we will draw a rough figure, so that we can get the idea of what we have to do. We will find the points P, Q and R on the three given line vectors. Then, for the points to be collinear, the direction ratios of the line derived from 2 points at a line must be proportional. Thus, we will try to find conditions for point Q with the aforementioned relation. Then we will see, which of the given options satisfy those conditions or which of the given options do not satisfy the conditions.

Complete step-by-step answer:
The lines given to us are as follows:
$\begin{align}
  & {{L}_{1}}=\vec{r}=\lambda \hat{i},\lambda \in R \\
 & {{L}_{2}}=\vec{r}=\hat{k}+\mu \hat{j},\mu \in R \\
 & {{L}_{3}}=\vec{r}=\hat{i}+\hat{j}+\nu \hat{k},\mu \in R \\
\end{align}$
It is also given that P is a point on line ${{L}_{1}}$, Q is a point on line ${{L}_{2}}$ and R is a point on line ${{L}_{3}}$.
It is also given that P, Q and R are collinear.
The rough figure of the given lines will be as follows:

Thus, coordinates of any point P on the line ${{L}_{1}}$ will be ($\lambda $, 0, 0).
Similarly, coordinates of any point Q on the line ${{L}_{2}}$ will be (0, $\mu $, 1) and coordinates of any point R on line ${{L}_{3}}$ will be (1, 1, v).
For the three points to be collinear in 3D space, the direction ratios of the line formed by the It is a common way to solve questions in which we have multiple correct options. We try to prove options which are not correct and rest all the options are correct.
first and the second point must be in proportion to direction ratios of the line formed by the first and the third point.
Therefore, $\dfrac{\lambda -0}{\lambda -1}=\dfrac{0-\mu }{0-1}=\dfrac{0-1}{0-\nu }$
Now, we will take the equalities one by one and find the conditions for $\lambda $, $\mu $ and v.
\[\begin{align}
  & \Rightarrow \mu =\dfrac{1}{v} \\
 & \Rightarrow v=\dfrac{1}{\mu } \\
 & \Rightarrow \lambda =\lambda \mu -\mu \\
 & \Rightarrow \lambda =\dfrac{\mu }{\mu -1} \\
\end{align}\]
For the above conditions to be real, $\mu \ne 0$ and $\mu \ne 1$.
This means, Q cannot be (0, 0, 1) and (0, 1, 1).
This means $Q\ne \hat{k},Q\ne \hat{j}+\hat{k}$
Hence Q can be $\hat{k}+\dfrac{1}{2}\hat{j}$ and $\hat{k}-\dfrac{1}{2}\hat{j}$.
Therefore, option (a) and option (d) are the correct option.

Note: It is a common way to solve questions in which we have multiple correct options. We try to prove options which are not correct and rest all the options are correct.