Question

Consider two straight lines $L_1:-2x+4=2y-4=z-4$ and $L_2:2x=y+1=-z+3$. Let the points $P$ and $Q$ be on $L_1$ and $L_2$ respectively and point $M$ divides $PQ$ 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

• A. $2x+z=6$
• B. $y+2z=8$
• C. $x+y=2$
• D. $3x+y+z=8$

Answer 1

Answer: (A)

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