JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 4)
Three lines $${L_1}:r = \lambda \widehat i$$, $$\lambda $$ $$ \in $$ R,
$${L_2}:r = \widehat k + \mu \widehat j$$, $$\mu $$ $$ \in $$ R and
$${L_3}:r = \widehat i + \widehat j + v\widehat k$$, v $$ \in $$ R 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?
$${L_2}:r = \widehat k + \mu \widehat j$$, $$\mu $$ $$ \in $$ R and
$${L_3}:r = \widehat i + \widehat j + v\widehat k$$, v $$ \in $$ R 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?
$$\widehat k$$
$$\widehat k$$ + $$\widehat j$$
$$\widehat k$$ + $${1 \over 2}$$$$\widehat j$$
$$\widehat k$$ $$-$$ $${1 \over 2}$$$$\widehat j$$
Explanation
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.
$${L_1}:r = \lambda \widehat i$$, $$\lambda $$ $$ \in $$ 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.
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