在三維旋轉(zhuǎn)的時(shí)候  我們很難找到旋轉(zhuǎn)軸 也很難確定 我們要旋轉(zhuǎn)的空間 角度。我通過(guò)旋轉(zhuǎn)對(duì)象的軸線和 旋轉(zhuǎn)方向上的一條直線，找到旋轉(zhuǎn)平面，然后計(jì)算這個(gè)平面的法線方程。再計(jì)算兩條直線的夾角，在平面內(nèi)旋轉(zhuǎn)這個(gè)角度就到了。三維旋轉(zhuǎn)軸就是平面的法線，旋轉(zhuǎn)的角度就是兩條直線的夾角。

Sub Rotate3D_1()
    On Error Resume Next
    Dim Obj As AcadEntity
    Dim PickPnt
    Dim A As Double, B As Double, C As Double, D As Double
    Dim P1 As Variant, P2 As Variant, P3 As Variant
    Dim T(0 To 2) As Double
    Dim L1 As Double, L2 As Double, L3 As Double
    Dim J As Double
    ThisDrawing.Utility.GetEntity Obj, PickPnt, "選擇旋轉(zhuǎn)對(duì)象："
    ThisDrawing.Utility.InitializeUserInput 1, ""
    P1 = ThisDrawing.Utility.GetPoint(, "旋轉(zhuǎn)對(duì)象軸線上一點(diǎn):")
    ThisDrawing.Utility.InitializeUserInput 1, ""
    P2 = ThisDrawing.Utility.GetPoint(, "旋轉(zhuǎn)對(duì)象軸線與旋轉(zhuǎn)方向直線的交點(diǎn):")
    ThisDrawing.Utility.InitializeUserInput 1, ""
    P3 = ThisDrawing.Utility.GetPoint(, "旋轉(zhuǎn)方向直線上的另一點(diǎn):")
   
    KJPMFC P1, P2, P3, A, B, C, D
    MsgBox A
    '過(guò)平面法線的一點(diǎn)
    T(0) = A * 10 + P2(0)
    T(1) = B * 10 + P2(1)
    T(2) = C * 10 + P2(2)
    '計(jì)算空間兩條直線的夾角
    L1 = P2PDistance(P1, P2)
    L2 = P2PDistance(P2, P3)
    L3 = P2PDistance(P3, P1)
           
    '利用余弦定理 a^2=b^2+c^2-2*b*c*cos(A)
           
    J = Arccos((L1 * L1 + L2 * L2 - L3 * L3) / 2 / L1 / L2)
   
    Obj.Rotate3d P2, T, -J

End Sub