Vector After Reflection

[Sidelines]

Alright, so I have a vector with velocities x y and z reflecting off of a flat surface. It is a perfect reflection (no velocity is lost). I have 3 points on the flat surface (each with x,y,z.. a triangle) which i took the Normal angle of using the vector of the 3 points.
There's really two parts to this problem, solving for the new Vxy and solving for the new Vz.

Found Variables:
fx, fy, fz (Normal vector from the surface of the triangle)
d = pythagorean using fx and fy
Vt (Velocity of the particle)
Vx, Vy, Vz (Initial velocities of the particle)
AngleXY = Atan(fy/fx) (XY Angle of the normal vector = XY Normal Angle)
AngleZ = Atan(fz/d) (Z Angle of the normal vector = Z Normal Angle)
VelAngleXY (XY Angle of the intial Velocities, found through atan)
VelAngleZ (Z Angle of the initial Velocities, found through atan)

My Attempt:

Solving for Vx, Vy
FinalAngleXY = (-XYAngle)-(VelAngleXY+AngleXY)
Vd = Pythag using Vt and Vz to find adjacent component
FinalVx = Vd*Cos(FinalAngleXY)
FinalVy = Vd*Sin(FinalAngleXY)

Solving for Vz
FinalAngleZ = -AngleZ-(VelAngleZ+AngleZ)
FinalVz = Vt*Sin(FinalAngleZ)

What I found:
When the particle/vector hits a slope, no matter the slopes direction it will reverse the particle's direction. Example: You throw a ball down a hill, it bounces back up instead of down (mainly due to the new XY velocities, but the new Z velocity is off too)

Any help is much appreciated! Been stuck on this for awhile now
I put this in the trig section mainly because it involves mostly trig, and I believe my attempt won't work because of the trig i used.

 

[DrMike]

In terms of vectors, you could say :

if v reflects off a plane with normal n, the component of v in the n direction is reversed, the remaining components stays the same.

Lets assume n is a unit vector, so |n|=1.

The component of v in the n direction is (v.n)n

The component of v parallel to the plane is therefore v-(v.n)n

We want to reverse the first component, and retain the second component. This gives us Rv = (v - (v.n)n) - (v.n)n = v - 2(v.n)n.

Note this is a linear function of v, so R can be expressed as a matrix. You might like to find a formula for that matrix in terms of n1,n2,n3. As a test, if n=(2,3,6)/7, then

      [-12  31 -36 ]
49R = [ 41 -12 -24 ]
      [-28 -42 -35 ]

unless I made a silly arithmetic mistake