need newbie math help

[ChaiRuiPeng]

okay so i have a system, call it system SlideFactor

SlideFactor has two inputs, both are 3d vectors. one of those is a normalized vector, call it Normal. call the other vector force. vector force may or may not come in normalized.

SlideFactor outputs a single 3d vector, vector Output, it is always normalized before being outputed.

Now, here is what I need to do:

vector Output must satisfy these reqs.

1. Be _a_ vector perpendicular to vector Normal. <-- infinite number of solutions

then

2. Be coplanar to vectors Normal and Force <-- two solutions at this point

3. Be inside the smaller angle formed by Normal and Force. <-- one solution at this point.

If the angle between Normal and Force is less than or equal to 90 degrees it returns vector Force

If the angle is equal to 180 degrees it returns vector Output = (0.0,0.0,0.0)

I'm just barely starting to write the code for this... will probably have it and several other features for my game that were waiting on this ready by the end of the day... but just making math discussion, and incase anyone else needed something like this. you can use it to vector forces and velocities along a surface of an object for correct sliding.

UPDATE: okay i will just take vector Normal, and rotate within the plane created by vector Normal and Vector Force by plus/minus 90 degrees, depending on the handed system.

[Luben]

Hi, i ment to reply faster but stuff got in the way.
What you'r looking for is essentially vector projection, which is the act of figuring out how much of a vector is parallel to another. I'll explain more:

The vector N is the Normal, F is the Force, P is the part of F that is parallel to the Normal and O is the part that is Orthogonal(perpendicular) to the normal. d the length of P.
We begin by using the Dot-product (Denoted as (A|B) ) which is equal to |A|*|B|*cos(alpha) where alpha is the smallest angle between the vectors. In this case, we take the dot-product between F and N, (F|N) = |F|*|N|*cos(a). If the length of N is 1, it can be rewritten as |F|*cos(a). (Otherwise we divide with the length of N to obtain d as below).
As we can see in the picture, this is equal to the value d. And by using d, we can create P as d*N/|N| or d*Ñ (Where Ñ denotes normalized N).
Now that we have P, we can remove this part from F. Vector addition/subtraction thus gives us the vector O, which is F without parts that point in the direction N.
The whole formula for P is thus (F|N)*N/(|N|²) and when |N| = 1 it can be simplified as (F|N)*N.

To check wether the force points along the normal or not, we can make use of the dot-product once again; Since the length of a vector is always positive, the only thing which affects the sign of the result from a dot-product is the cosine-term.
If the dot-product is >0 then the angle is less than 90 degrees or pi/2 radians. If it is 0, the angle is exactly 90 degrees or pi/2. And if the result is <0 then the angle is more than 90 degrees or pi/2.
Using this knowledge of sign and angles, we can determine that if the dot-product is >0 then the vectors point in roughly the same direction, and if it is <0 then they point on opposite directions.