Question 389306: If P(1, -2, 4) is reflected in the plane with equation 2x - 3y - 4z + 66 = 0, determine the coordinates of its image point, P'. Note that the plane 2x - 3y - 4z + 66 = 0 is the right bisector of the line joining P(1, -2, 4) with its image.
Any help would be appreciated ^.^
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! First determine the equation of the line perpendicular to the given plane and passing through the point (1,-2,4). Then get the intersection of the line with the given plane. That intersection point is the midpoint of the segment connecting (1,-2,4) to the unknown point (a,b,c).
The normal vector to the given plane is <2,-3,-4>. The symmetric form of the line perpendicular to the given point is then  . Solving for y and z in terms of x, then we get:
 , and z = -2x+6.
Putting these two equations into 2x - 3y - 4z + 66 = 0, we get
 , or  after simplification.
Hence x = -3, y = 4, and z = 12, after substituting back into and z = -2x+6. Thus the intersection point of the normal line and the plane is (-3,4,12).
To find the reflected point (a,b,c), we use the midpoint formula: (Recall the given point is (1,-2,4))
 ----> a = -7
 ----> b = 10
 ------> c = 20.
Therefore the reflected point is (-7,10,20).
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