Finding Tangent Vectors for Unit Normal (nx=ny=nz)

[lost1]

hi,

given a unit normal (nx,ny,nz) to a surface, the two tangent vectors
are: s = (-ny+nz, nz-nz, -nx+ny) and t = n x s (with s, t normalized).
now, if nx=ny=nz, then s & t become trivial.
how do i find these tangent vectors for this special case?

thanks alot.

 

[AKG]

hi,

given a unit normal (nx,ny,nz) to a surface, the two tangent vectors
are: s = (-ny+nz, nz-nz, -nx+ny) and t = n x s (with s, t normalized).
now, if nx=ny=nz, then s & t become trivial.
how do i find these tangent vectors for this special case?

thanks alot.

I'm not exactly sure of what you're asking, but it seems that if you have a surface with normal (nx,ny,nz) such that nx=ny=nz, let a = nx, and you can express you normal as (a,a,a). The vectors tangent to this surface should also be tangent to the plane with normal (a,a,a). It shouldn't be hard to figure out two vectors that would lie on this plane. You should be able to see easily that there is some b such that the points (b,0,0), (0,b,0), and (0,0,b) lie on the plane, so the vectors (b,-b,0) and (0,b,-b) are acceptable candidates for your tangent vectors.

You can express the plane with a cartesian equation:

ax + ay + az + d = 0.

Knowing (a,a,a) lies on the plane, d = -3a². Now:

ab + a(0) + a(0) - 3a² = 0.
b = 3a

So, you can choose your tangent vectors to be:

s = (3nx,-3nx,0) and t = (0,3nx,-3nx). [or replace x with y or z ... or anything other than zero]

Of course, you can simplify this further and choose instead:

u = (1,-1,0) and v = (0,1,-1)

 

[tacman]

Hello,

In the case where nx=ny=nz, the unit normal vector is pointing in the direction of one of the coordinate axes (x, y, or z). In this case, the surface is flat and there are an infinite number of tangent vectors that can be chosen. However, we can still find two tangent vectors that are perpendicular to each other and to the normal vector.

One way to find these tangent vectors is to choose any two vectors that are perpendicular to the normal vector. For example, if the normal vector is (1,1,1), we can choose s=(1,-1,0) and t=(1,0,-1). These two vectors are perpendicular to each other and to the normal vector, and they can be normalized to become unit tangent vectors.

Another way to find the tangent vectors is to use the cross product. Since the normal vector is pointing in the direction of one of the coordinate axes, the cross product of the normal vector with any other vector will result in a vector that is perpendicular to both. For example, if the normal vector is (1,1,1), we can choose s=(1,0,0) and use the cross product to find t: t = n x s = (1,1,1) x (1,0,0) = (0,1,-1). Again, these two vectors can be normalized to become unit tangent vectors.

I hope this helps. Let me know if you have any further questions.