Simple Collision Response for Plane

Lets say we have a plane (B) with a slope of (delta Y / delta X), the slope of the normal (n) must be (-delta X / delta Y). In vector form, thats the same as B = [delta X, delta Y] and n = [delta Y, -delta X].

Let's assume the incoming velocity as U, the result velocity as R.
Now, it's time to find the projection (P) of -U onto the normalized normal (N).
By using dot product of -U and N, we will get the length of the projection on N.
Thus, P = (-U . N) * N --> Eq. 1

To find the reflection, we add in V where V = U + P --> Eq. 2.
Result velocity (R) = P + V --> Eq.3

Substitute Eq.2 into Eq.3,
R = P + U + P
= 2 * P + U

Substitute Eq.1 into it,
R = U + 2 * (-U . N)N or R = U - 2 * (U . N)N

Dot product and Cross Product

Dot Product
A.B = |A||B|cos(angle)

A.B = a1b1 + a2b2 + a3b3
where A = [a1 a2 a3], B = [b1 b2 b3]

If A.B = 0, then |A|B|cos(angle) = 0,
acos(0) = 90degree.
Thus, A is perpendicular to B,

If A.B <> 90 degree
If A.B > 0, angle <>

** Dot product provides the length of the projection of one vector to another.
If A is going to project on to B, B must be normalised.

Cross Product
A x B = [(a2b3 - a3b2) (a3b1 - a1b3) (a1b2 - a2b1)]
where A = [a1 a2 a3], B = [b1 b2 b3]

A x B is perpendicular to both A and B.
A x B = -(B x A)

|A x B| = |A||B| sin(angle)