Find the image of the point `(1,3,4)` in the plane `2x-y+z+3=0.`

Options

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

Answer

B

Explanation

We know that the image of the point `(x_(1),y_(1),z_(1))` in the plane `ax+by+cz+d=0` is given by
`(x-x_(1))/a=(y-y_(1))/b=(z-z_(1))/c=(-2(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))`
So the image of the point `P(1,3,4)` in the plane `2x-y+z+3=0` is given by
`(x-2)/2=(y-3)/(-1)=(z-4)/1=(-2(2-3+4+3))/(4+1+1)`
`implies(x-1)/2=(y-3)/(-1)=(z-4)/1=-2impliesx=-3,y=5,z=2`
Hence the required pointis `(-3,5,2)`.