寻找三维向量的方位角

N is the normal of the plane (0,0,1)
P is the line that I am working with. It has coordinates Psx,Psy,Psz and Pex,Pey,Pez. where Ps refers to the starting point and Pe refers to the end point.
V is a vector in the direction of P. Vx = Pex-Psx;Vy = Pey-Psy;Vz = Pez-Psz
Now, project V onto the xy plane:
U = V - (V.N)*N; where V.N is the dot product
Now we have a vector in the xy plane, we can find the angle of the this vector and the vector M (0,1,0) which defines North.
To determine the angle:
theta = arccos(U.M/(|U|*|M|))

N = 0,0,1; this is the direction of the normal describing the xy plane
M = 0,1,0; This vector defines north
U = Ux,Uy,Uz; this is the vector that I want to find the angle of relative to M
For an example, lets say U = -1,1,0; this would put the vector in the 2nd quadrant and it should have an angle of 315 degrees. If I use the procedure in my last post I get an angle of 45 degrees.
Now if I find the determinant of:
|-1 1 0|
| 0 1 0|  = -1
| 0 0 1|
I believe, according to the sign that the angle has been calculated the wrong way. So If I take 360 and subtract 45 I arrive at 315 degrees. Which is what I want.
Say I used U=2,2,0 which is 45 degrees, I would get
| 2 2 0|
| 0 1 0|  = 2
| 0 0 1|
The positive result would indicate that I do not need a correction.
If the result is 0, it means that the vectors are parallel.

Option Explicit
Private Const x As Integer = 0
Private Const y As Integer = 1
Private Const z As Integer = 2
Private Const PI As Double = 3.14159
Public Sub findAzimuth()
Dim pickedEntity As AcadEntity
Dim pickedPoint As Variant
Dim myLine As AcadLine
ThisDrawing.Utility.GetEntity pickedEntity, pickedPoint, "Please select a line: " & vbCr
If pickedEntity Is Nothing Then Exit Sub
If TypeOf pickedEntity Is AcadLine Then
    Set myLine = pickedEntity
Else
    ThisDrawing.Utility.Prompt vbCrLf & "Be sure to select a line next time!" & vbCr
    Exit Sub
End If
Dim N(2) As Double 'the normal of the plane
Dim M(2) As Double 'the direction of the azimuth line
Dim v(2) As Double 'the vector representing the line
Dim U(2) As Double 'the projection of V onto the plane
N(x) = 0
N(y) = 0
N(z) = 1
M(x) = 0
M(y) = 1
M(z) = 0
v(x) = myLine.EndPoint(x) - myLine.StartPoint(x)
v(y) = myLine.EndPoint(y) - myLine.StartPoint(y)
v(z) = myLine.EndPoint(z) - myLine.StartPoint(z)
Dim dpResult As Double 'dot product result
dpResult = dotProduct(v, N)
U(x) = v(x) - dpResult * N(x)
U(y) = v(y) - dpResult * N(y)
U(z) = v(z) - dpResult * N(z)
Dim theta As Double
Dim dpVectorNormal As Double
Dim vectMagnitude As Double
vectMagnitude = (vectorMagnitude(U) * vectorMagnitude(M)) 'pre-calculate this value - if it is zero, are angle calculation is greatly simplified
If vectMagnitude > 0 Then
    dpVectorNormal = dotProduct(U, M) 'calculate the dot product of the projected vector and the azimuth vector
    
    theta = aCos(dpVectorNormal / vectMagnitude)
    
    'now we need to figure out which way the angle was measured from
    Dim D As Double
    D = determinant(U, M, N)
    
    If D < 0 Then
        theta = 2 * PI - theta
    End If
Else
    'the hole looks vertical in that plane
    theta = 0
End If
'convert the angle to degrees
theta = theta * 180 / PI
ThisDrawing.Utility.Prompt vbCr & "The Azimuth is: " & theta
Set myLine = Nothing
Set pickedEntity = Nothing
End Sub
Private Function determinant(v1 As Variant, v2 As Variant, v3 As Variant) As Double
'this will calculate the 3x3 determinant of the passed vectors
'v1(x),v1(y),v1(z)
'v2(x),v2(y),v2(z)
'v3(x),v3(y),v3(z)
Dim a As Double
Dim b As Double
Dim c As Double
a = v1(x) * (v2(y) * v3(z) - v2(z) * v3(y))
b = v1(y) * (v2(x) * v3(z) - v2(z) * v3(x))
c = v1(z) * (v2(x) * v3(y) - v2(y) * v3(x))
determinant = a - b + c
End Function
Private Function aCos(v As Double) As Double
aCos = Atn(-v / Sqr(-v * v + 1)) + 2 * Atn(1)
End Function
Private Function vectorMagnitude(v As Variant) As Double
    vectorMagnitude = v(x) * v(x) + v(y) * v(y) + v(z) * v(z)
    vectorMagnitude = Sqr(vectorMagnitude)
End Function
Private Function dotProduct(v1 As Variant, v2 As Variant) As Double
dotProduct = v1(0) * v2(0) + v1(1) * v2(1) + v1(2) * v2(2)
End Function