Convert 2 3D Points to Directional Vectors to Euler Angles - algorithm

Question

Here's essentially my problem. Also maybe I am not familiar enough with Euler angles and what I'm attempting to do is not possible.
I have 2 points in 3d space.
p1 (1,2,3)
p2 (4,5,6)
In order to get the unit vectors for these two points I'm doing this basically.
var productX = (position.X2 - position.X1);
var productY = (position.Y2 - position.Y1);
var productZ = (position.Z2 - position.Z1);
var normalizedTotal = Math.sqrt(productX * productX + productY * productY + productZ * productZ);
var unitVectorX, unitVectorY, unitVectorZ;
if(normalizedTotal == 0)
{
unitVectorX = productX;
unitVectorY = productY;
unitVectorZ = productZ;
}
else
{
unitVectorX = productX / normalizedTotal;
unitVectorY = productY / normalizedTotal;
unitVectorZ = productZ / normalizedTotal;
}
So now I have a unit vector x y z for these 2 3d points.
I'm attempting now to convert from directional vector to euler angle. Is this possible. What am I missing here as I can't find any good resource on how to do this.
Thanks for the help.
Sometimes a picture helps.
maybe this will give a better example of what i'm trying to solve for.
Given 2 points, I have determined a midpoint, length, and now i'm trying to figure out hte angles to set so that the cylinder is correctly oriented around the x,y,z axis. I think I need to figure out all 3 angles not just 1 and 2 is that correct? I think the euler angles from a directional vector bit through you off.

Answer 1

What you want is a transformation from Cartesian coordinates of the vector
v = (v_x, v_y, v_z)
to the spherical coordinates r, ψ and θ where
v = ( r*COS(ψ)*COS(θ), r*SIN(θ), r*SIN(ψ)*COS(θ) )
This is done with the following equations
r = SQRT(v_x^2+v_y^2+v_z^2)
TAN(ψ) = (v_z)/(v_x)
TAN(θ) = (v_y)/(v_x^2+v_z^2)
To get the angles ψ and θ, use the ATAN2(dy,dx) function as in
ψ = ATAN2(v_z, v_x)
θ = ATAN2(v_y, SQRT(v_x^2+v_z^2))
Now that you have the along direction vector
j = ( COS(ψ)*COS(θ), SIN(θ), SIN(ψ)*COS(θ) )
you can get the two perpendicular vectors from
i = ( SIN(ψ), 0, -COS(ψ) )
k = ( COS(ψ)*SIN(θ), -COS(θ), SIN(ψ)*SIN(θ) )
These three vectors make up the columns of the 3×3 rotation matrix
| SIN(ψ) COS(ψ)*COS(θ) COS(ψ)*SIN(θ) |
E =[i j k] = | 0 SIN(θ) -COS(θ) |
| -COS(ψ) SIN(ψ)*COS(θ) SIN(ψ)*SIN(θ) |
In terms of Euler angles the above is equivalent to
E = RY(π/2-ψ)*RX(π/2-θ)
Example
Two points p_1=(3,2,3) and p_2=(5,6,4) define the vector
v = (5,6,4) - (3,2,3) = (2,4,1)
NOTE: I am using the notation of v[i] for the i-th element of the vector, as in v[1]=2 above. This is neither like C, Python which is zero based, nor like VB, FORTRAN or MATLAB which uses parens () for the index.
Using the expressions above you get
r = √(2^2+4^2+1^2) = √21
TAN(ψ) = 1/2
TAN(θ) = 4/√(2^2+1^2) = 4/√5
ψ = ATAN2(1,2) = 0.463647
θ = ATAN2(4,√5) = 1.061057
Now to find the direction vectors
j = ( COS(ψ)*COS(θ), SIN(θ), SIN(ψ)*COS(θ) ) = (0.4364, 0.87287, 0.21822 )
i = ( SIN(ψ), 0, -COS(ψ) ) = (0.44721, 0, -0.89443 )
k = ( COS(ψ)*SIN(θ), -COS(θ), SIN(ψ)*SIN(θ) ) = (0.78072, -0.48795, 0.39036)
Put the direction vectors as columns of the local to world coordinate transformation (rotation)
E[1,1] = i[1] E[1,2] = j[1] E[1,3] = k[1]
E[2,1] = i[2] E[2,2] = j[2] E[2,3] = k[2]
E[3,1] = i[3] E[3,2] = j[3] E[3,3] = k[3]
| 0.447213595499957 0.436435780471984 0.780720058358826 |
| |
E = | 0 0.872871560943969 -0.487950036474266 |
| |
| -0.894427190999915 0.218217890235992 0.390360029179413 |