Finding Point P on Plane S: A Reflective Approach

[prace]

Homework Statement

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.

Homework Equations

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!

 

[pizzasky]

Points A and B are in fact the same distance from plane S. (Can you show this?)

Draw a diagram reflecting this, and add point P to the diagram. What do you realize?

 

[prace]

I worked out the distances to be different than each other. To find the distance of points A and B from plane S, i first found a point on plane S to reference from, call it Q. So, to find the distance of point A from plane S, i found:

h1 = |QA - proj (QA on n)|, where n is the normal vector for plane S.

I came up with a value of ~32.4.

to find the distance of point B from plane S, i found:

h2 = |QB - proj (QB on n)|, where again, n is the normal vector for plane S.

I came up with a value of ~134.28.

What happened? If I don't have the same height, how can I relate the two angles to each other?

 

[pizzasky]

Shouldn't it be $$h_{1} =\ \mid proj\ (\ QA\ on\ n\ ) \mid$$?

Try drawing a diagram, it should help.