Three lines $L_{1}: \vec{r}=\lambda \hat{i}, \lambda \in \mathbb{R}$, $L_{2}: \vec{r}=\vec{k}+\mu \hat{j}, \mu \in \mathbb{R}$ and $L_{3}: \vec{r}=\hat{i}+\hat{j}+v \hat{k}, v \in \mathbb{R}$ 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 ?

(C) $\hat{\mathrm{k}}+\frac{1}{2} \hat{\mathrm{j}}$; (D) $\hat{\mathrm{k}}-\frac{1}{2} \hat{\mathrm{j}}$

(A) $\hat{\mathrm{k}}+\hat{\mathrm{j}}$

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Three Dimensional Geometry

Three lines

L_{1} : r = λ \hat{i}, λ \in R

L_{2} : r = k + μ \hat{j}, μ \in R

and

L_{3} : r = \hat{i} + \hat{j} + v \hat{k}, v \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 ?

Answer: (C), (D)

Let

P (λ, 0, 0)

Q (0, μ, 1)

R (1, 1, v)

be points on

L_{1}

L_{2}

, and

L_{3}

respectively. For P, Q, R to be collinear,

P Q

must be collinear with

QR

. This gives

\frac{- λ}{1} = \frac{μ}{1 - μ} = \frac{1}{v - 1}

. For every

μ \in R - {0, 1}

, there exist unique

λ, v \in R

. Hence Q cannot have coordinates

(0, 1, 1)

and

(0, 0, 1)

, which correspond to options (A) and (B). Options (C) and (D) are valid.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options