C点垂直-线段AB的Ｂ点

兰州人

图示，求Ｃ点到ＡＢ线段的垂线，且交点于Ｂ点．

A点:  X = 0.0000     Y = 1.0000     Z = 0.0000

B点:  X = 2.0000     Y = 3.0000     Z = 0.0000

要求通过解析几何算法,求C点坐标, (通过作图得出Ｃ点坐标为 X = 2.7071     Y = 2.2929     Z = 0.0000)

兰州人

找到解法了

已知Kab=1.7 那么Kbc=-1/Kab=-.588235294117647
再用x^2+y^2=R^2公式推导
x = r/sqrt(1+Kbc^2)
y = r/ sqrt ( 1+ (1/Kbc)^2)
调试通过的程序如下：
Sub LSs()
  Dim Aa(2) As Variant
  Aa(0) = Array(10, 33)
  Aa(1) = Array(20, 50)
  Dim pp(0 To 2) As Double, ppp(0 To 2) As Double
  Dim kkk As Double
  Kab = (Aa(1)(1) - Aa(0)(1)) / (Aa(1)(0) - Aa(0)(0))
  kkk = -1 / Kab
  For jj = 0 To 1
    pp(jj) = Aa(1)(jj)
  Next jj
  ppp(0) = pp(0) + 5 / Sqr(1 + kkk ^ 2) 
  ppp(1) = pp(1) - 5 / Sqr(1 + (1 / kkk) ^ 2)
  Set ll = ThisDrawing.ModelSpace.AddLine(pp, ppp)
  ll.color = 1
End Sub