# rotation angles

I have two lines:
1) First one is drawn somewhere line in 3d space.  I know coordinates of this line.
2) Second line is a vertical line starts in center of the coordinate system (0,0,0) .

I need to move the second line from 0,0,0 to the exact position of the first line in the 3D space, I have to use gltranslate and glrotate functions.

I'm using OpenGL + VB

Can you please help me to find what is wrong with my code? It doesn't draw the line in the right position

Public Sub CalculateRotAngles(x1 As Single, y1 As Single, z1 As Single, _
x2 As Single, y2 As Single, z2 As Single, _
rX As Single, rZ As Single, rY As Single, length As Single)

Dim P1 As tPoint, P2 As tPoint, P3 As tPoint
Dim L1 As Single
Dim L2 As Single
Dim LineSlope As Single
Dim dx As Single, dZ As Single, dy As Single
Dim P2P3 As Single, LineLen As Single

P1.X = x1
P1.Y = y1
P1.z = z1

P2.X = x2
P2.Y = y2
P2.z = z2

'L1 = Sqr((P1.X - P2.X) ^ 2 + (P1.z - P2.z) ^ 2)
'length = L1

dx = Abs(P1.X - P2.X)
dy = Abs(P1.Y - P2.Y)
dZ = Abs(P1.z - P2.z)

L1 = Sqr(dx * dx + dy * dy + dZ * dZ)
length = L1

End Sub

'Draw

glPushMatrix

glTranslatef begX, begY, begZ
glRotatef rX, 1, 0, 0
'glRotatef rY, 0, 1, 0
glRotatef rZ, 0, 0, 1

glColor3f 1, 1, 0
DrawLineCoordinate 0, 0, 0, _
0, length, 0
glPopMatrix

Thank you.
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Commented:
2) Second line is a vertical line starts in center of the coordinate system (0,0,0) .
does this mean from (0,0,0) to (0,0,1)?

dx = P1.X - P2.X
dy = P1.Y - P2.Y
dZ = P1.z - P2.z

vx = 0-0
vy = 0-0
vz = 1-0

cx = vy*dz-dy-vz
cy = vz*dx-dz*vx
cz = vx*dy-dy*vx

r = atan2(sqrt(cx*cx+cy*cy+cz*cz)/(dx*cx+dy*cy+dz*cz)
End Sub

glRotatef r, cx, cy, cz
0
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