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) $Q$ on $L_2$ 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. $\hat{k}$
• B. $\hat{k}+\hat{j}$
• C. $\hat{k}+\frac{1}{2}\hat{j}$
• D. $\hat{k}-\frac{1}{2}\hat{j}$

Answer 1

Answer: (C), (D)

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

Equation of lines $L_1:\overrightarrow{r}=\lambda \hat{i}\Rightarrow \frac{x-0}{\lambda }=\frac{y-0}{0}=\frac{z-0}{0}$
$L_2:\overrightarrow{r}=\hat{k}+\mu \hat{j}\Rightarrow \frac{x-0}{0}=\frac{y-0}{\mu }=\frac{z-1}{0}$
$L_3:\overrightarrow{r}=\hat{i}+\hat{j}+\nu \hat{k}\Rightarrow \frac{x-1}{0}=\frac{y-1}{0}=\frac{z-0}{\nu }$
Let $P(\lambda ,0,0),Q(0,\mu ,1),R(1,1,\nu )$ be points on $L_1,L_2,L_3$ respectively.
$\because P,Q,R$ are collinear, $\overrightarrow{PQ}$ is collinear with $\overrightarrow{QR}$
$\therefore \overrightarrow{PQ}\parallel \overrightarrow{QR}$
$\Rightarrow \overrightarrow{PQ}=k\overrightarrow{QR}$
$\text{Hence,}\frac{-\lambda }{1}=\frac{\mu }{1-\mu }=\frac{1}{\nu -1}=k$
$\Rightarrow \lambda =-k,\mu =\frac{k}{k+1}\text{and}\nu =\frac{k+1}{k}$
$\therefore \mu =\frac{-\lambda }{1-\lambda }=\frac{1}{\nu }$
$\Rightarrow \mu \ne 0,1$
For every $\mu \in R–\left\{0,1\right\}$ there exist unique $\lambda ,\nu \in R$,
$\text{For}\mu =0\text{and}\mu =1,$ we get the points $\hat{k}\text{and}\hat{k}+\hat{j}$ respectively.
Hence, $Q\ne \hat{k}\text{and}Q\ne \hat{k}+\hat{j}$