QJEEAdv19Ma7738: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma7738 Solution: Three lines , R, , R and , v R are given. For which point(s) Q…

Given lines,

${L_1}:r = \lambda \widehat i$ , $\lambda$ $\in$ R .......(i)

${L_2}:r = \widehat k + \mu \widehat j$ , $\mu$ $\in$ R .......(ii)

and ${L_3}:r = \widehat i + \widehat j + v\widehat k$ , v $\in$ R .......(iii)

Now, let the point P on L₁ = ( $\lambda$ , 0, 0)

the point Q on L₂ = (0, $\mu$ , 1), and

the point R on L₃ = (1, 1, v)

For collinearity of points P, Q and R, there should be a non-zero scalar 'm', such that PQ = m PR

$\Rightarrow ( - \lambda \widehat i + \mu \widehat j + \widehat k) = m[(1 - \lambda )\widehat i + \widehat j + v\widehat k]$

$\Rightarrow {\lambda \over {\lambda - 1}} = {\mu \over 1} = {1 \over v}$

$\Rightarrow v = {1 \over \mu }$ and $\lambda = {\mu \over {\mu - 1}}$

where, $\mu \ne 0$ and $\mu \ne 1$

$\Rightarrow$ $Q \ne \widehat k$ and $Q \ne \widehat k + \widehat j$

Hence, Q can not have coordinator (0, 0, 1 ) and (0, 1, 1)

Hence, options (c) and (d) are correct.