So I have a velocity vector b in world space and I have a "pointing" vector a in world space. I need to know how to make the velocity vector to be relative to the "pointing" vector. Better explanation in image below:
So on the left you can see the vectors in world space, but on the right - vectors in a space. I need to get the vector b in the _a_space from the worldspace.
you can just convert the coodinates.
to do this, you have to use the normalized â as an base vector for the new coodinating system then use the 90º rotated norm(a.y,-a.x) as another base.
lets name they as e1 and e2.
now you can express your vector b as an linear combination of e1 and e2 with the linear system of equations:
b = m*e1+n*e2
b.x = m*e1.x+n*e2.x
b.y = m*e1.y+n*e2.y
solve for m and n and the vector (m,n) is your b in a coodinates
m = b / e1; n = b / e2; o = b / e3 right? Yeah, I still need to know the step where you rotate it by 90 degrees in 3D setup.
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Mar 6, 2016 at 5:48
So I solved this in kind of simple way.
As I had the pointing vector(which in my specific case was y, but it doesn't matter that much) and I found out that I can actually get another vector out here. So I made a vector plane of x and y vectors. Than I did x.cross(y) and obtained z vector. Than I made those vectors 4D by setting w to be 0. After that I created a new matrix and plugged each of those vectors into the matrix - x into column 0, y into comun 1 and z into comlumn 2. In the result I had a transformation matrix which I could use as multiplier for the velocity vector and obtained velocity relative to the vector.
Unfourtunately, I should say that vector to vector-space conversion requires 2 vectors. Fourtunately, I had 2:)
The “velocity in a space” can be broken down in two vectors, one that is the same direction as a, and one that is perpendicular to a.
The first vector has direction u and size su, computed as follows:
u = normalize(a)
su = dot(u, b)
The perpendicular vector has direction v and size sv:
v = normalize(b - su * a)
sv = dot(v, b)
