Question

Three lines, $L_1:\overrightarrow{r}=\lambda \hat{i},\lambda \in R$, $L_2:\overrightarrow{r}=\hat{k}+\mu \hat{j},\mu \in R$ and $L_3:\overrightarrow{r}=\hat{i}+\hat{j}+\nu \hat{k},\nu \in R$ are given, for which point(s) $Q$ and $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$, so that $P,Q$ and $R$ are collinear?

• A. $\hat{k}+\hat{j}$
• B. $\hat{k}$
• C. $\hat{k}+\frac{1}{2}\hat{j}$
• D. $\hat{k}-\frac{1}{2}\hat{j}$

Answer 1

Answer: (C), (D)

$\hat{k}-\frac{1}{2}\hat{j}$

Explanation for correct option:

Find the points from the given vector lines that satisfy the given condition

For the given line, $L_1:\overrightarrow{r}=\lambda \hat{i},\lambda \in R$, let point $P$ be $(\lambda ,0,0)$

Similarly, for $L_2:\overrightarrow{r}=\hat{k}+\mu \hat{j},\mu \in R$, let point $Q$ be $(0,\mu ,1)$

and, for $L_3:\overrightarrow{r}=\hat{i}+\hat{j}+\nu \hat{k},\nu \in R$, let point $R$ be $(1,1,\nu )$

The given condition is that the three points are collinear, so, $\overrightarrow{PQ}=k\overrightarrow{PR}$, where $k$ is a constant value.

So, $\frac{\lambda -0}{\lambda -1}=\frac{0-\mu }{0-1}=\frac{0-1}{0-\nu }$

From $\frac{\lambda -0}{\lambda -1}=\frac{0-\mu }{0-1}$, we get,

$$\begin{gathered} \Rightarrow \lambda =\lambda ·\mu -\mu \\ \Rightarrow \lambda =\mu (\lambda -1) \end{gathered}$$

Here, $\mu$ cannot be $1$, otherwise the value of $\lambda$ will be not defined which is not possible.

So, $\overrightarrow{Q}\ne \hat{k}$

Now, from $\frac{0-\mu }{0-1}=\frac{0-1}{0-\nu }$, we get

$$\begin{gathered} \Rightarrow \mu =\frac{1}{\nu } \\ \Rightarrow \nu =\frac{1}{\mu } \end{gathered}$$

Here, $\mu$ cannot be $0$, otherwise $\nu$ will be undefined.

So, $\overrightarrow{Q}\ne \hat{k}+\hat{j}$

Thus, other possible values are $\hat{k}+\frac{1}{2}\hat{j}$ and $\hat{k}-\frac{1}{2}\hat{j}$.

Hence, option (C) and (D) are correct answers.