1. Tardigrade
  2. Question
  3. Mathematics
  4. Consider two straight lines L1:-2 x+4=2 y-4=z-4 and L2: 2 x=y+1=-z+3. Let the points P and Q be on L1 and L2 respectively and point M divides P Q in the ratio 2: 3 (internally). Also R and S be points on lines L 1 and L 2 respectively which are nearest to each other. Given the line L 2 intersects xy , yz , zx planes at A , B , C respectively. The locus of M is

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Q. Consider two straight lines $L_1:-2 x+4=2 y-4=z-4$ and $L_2: 2 x=y+1=-z+3$. Let the points $P$ and $Q$ be on $L_1$ and $L_2$ respectively and point $M$ divides $P Q$ in the ratio $2: 3$ (internally). Also $R$ and $S$ be points on lines $L _1$ and $L _2$ respectively which are nearest to each other. Given the line $L _2$ intersects $xy , yz , zx$ planes at $A , B , C$ respectively.
The locus of $M$ is

Vector Algebra

Solution:

image
Now,
$5 u =6-3 \lambda+2 \mu $
$5 v =3 \lambda+6-4 \mu-2 $
$5 w =6 \lambda+12+6-4 \mu $
$\therefore 5 u =2 \mu-3 \lambda+6 $ .....(1)
$ 5 v =4 \mu+3 \lambda+4 $.....(2)
$5 w =-4 \mu+6 \lambda+18$.....(3)
$(2)+(3)$
$5( v + w )=9 \lambda+22$.....(4)
Also, $ 10 u -5 v =-9 \lambda+8$.....(5)
(4) $+(5)$
$5 w +10 u =30 \Rightarrow w +2 u =6$
Hence locus is $2 x + z =6$