The image of the line `(x-1)/(3)=(y-3)/(1)=(z-4)/(-5)` in the plane `2x-y+z+3=0` is the line

Options

A. `(x-3)/3=(y+5)/1=(z-2)/(-5)`
B. `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`
C. `(x+3)/3=(y-5)/1=(z-2)/(-5)`
D. `(x+3)/(-3)=(y-5)/(-3)=(y-5)/(-1)=(z+2)/5`

Answer

C

Explanation

Clearly, given line is parallel to the given plane. So, its image will also be parallel to the plane and so its direction ratios are proportional to 3,1,-5. Given line passesthrough (1,3,4). Let `(x_(1),y_(1),z_(1)` be its image in the given plane.
Then `x_(1),y_(1),z_(1)` are given by
`(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-((2xx1-3+4+3))/(2^(2)+(-1)^(2)+1^(2))`
`implies(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-2`
`impliesx_(1)=-3,y_(1)=5,z_(1)=2`
So image of (1,3,4) in the given planeis (-3,5,2).
Hence the equations of the required line are
`(x+3)/3=(y-5)/1=(z-2)/(-5)`.