Given a plane S: x+y+z=8 and two points A(6,5,3) and B(10,5,-1), find the point P on plane S such that a line from point A bounces off of the plane S at point P and passes through point B.
The Attempt at a Solution
If I know that the angle of incidence = the angle of reflectance, then I can also say that a normal vector from plane S bisects will these two points at point P. So if I can proove that the angle between vectors AP and n is the same as the angle between vectors BP and n then this should solve my problem. Well, this is what I tried to show at least, and did not come up with the angle between the two vectors to be the same.
I found my normal vector n to be <1,1,1> and I took my point P to be (8,0,0). So, from there, I found vectors AP and BP to be <2,-5,-3> and <-2,-5,1> respectively.
Any help on which direction to go towards next would be a great help! Thank you!