Convert 2 3D Points to Directional Vectors to Euler Angles

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 |