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)/(-1)=(z-2)/(5)`

Answer

A

Explanation

Here DR's of line and plane are `(a_(1),b_(1),c_(1))=(3,1,-5) and (a_(2),b_(2),c_(2))=(2,-1,1)`
Here, plane and line are parallel to each other. Therefore equation of normal to plane through the point (1,3,4) is
`(x-1)/(2)=(y-3)/(-1)=(z-4)/(1)=k" "(say)"`
Any point on this normal is
`Q(2k+1,-k+3,4+k)`
Then, mid point of P and Q is R
`((2k+1+1)/(2),(-k+3+3)/(2),(4+k+4)/(2))`, which lie on plane.
`:." "2(k+1)-((6-k)/(2))+((8-k)/(2))+3=0`
`rArr" "2(k+1)-((6-k)/(2))+((8-k)/(2))+3=0`
`rArr" "k=-2`
So, image of a point is
`(2k+1,3-k,4+k)`
i.e. `[2(-2)+1,3+2,4-2=(-3,5,2)]`
Hence, equation of image line is
`(x+3)/(3)=(y-5)/(1)=(z-2)/(-5)`