The image of the point (1, 3, 4) in the plane `2x-y+z+3=0` is `(3,\ 5,\ 2)` b. `(-3,5,\ 2)` c.`(3,\ 5,\ -2)` d. `(3,\ -5,\ 2)`

Options

A. (3,-5,2)
B. (3,5,-2)
C. (3,5,2)
D. (-3,5,2)

Answer

d

Explanation

The equation of a line through the point P(1,3,4) and perpendicular to the plane `2x-y+z+3=0` is
`(x-1)/2=(y-3)/-1=(z-4)/1=k` (say)
A general point on this line is `(2k+1, -k+3, k+4)`.
For some value of k let `N(2k+1, -k+3, k+4)` be a point of the line lying on the plane. Then,
`2(2k+1)-(-k+3)+(k+4)+3=0 rArr 6k =-6 rArr k=-1`.
`therefore` coordinates of N are `(-2+1,1+4,-1+4,-1+4)`, i.e., `N(-1,4,3)`.
Let `Q(alpha,beta,gamma)` be the image of P in the given plane. Then,
`(1+alpha)/2=-1, (3+beta)/2=4` and `(4+gamma)/2`. So, `alpha=-3, beta=5` and `gamma=2k`.
Hence, the required image of `P(1,3,4)` in the given plane is `Q(-3,5,2)`.