垂直于从给定的点的线段 [英] Perpendicular on a line segment from a given point

问题描述

我要计算在给定的行是垂直从给定的点的点。

I want to calculate a point on a given line that is perpendicular from a given point.

我有一个线段AB，并有外线段中的C点。我想计算的AB点D，使得CD垂直于AB。

I have a line segment AB and have a point C outside line segment. I want to calculate a point D on AB such that CD is perpendicular to AB.

我得找点D.

这非常类似于<一href="http://stackoverflow.com/questions/1811549/perpendicular-on-a-line-from-a-given-point">this,但我要考虑到Z坐标也因为它没有正确地在三维空间中显示出来。

It quite similar to this, but I want to consider to Z coordinate also as it does not show up correctly in 3D space.

推荐答案

证明： 点D是CD线垂直于AB在，当然D属于另外一个AB型。 写下两个向量CD.AB = 0的点积和EX preSS的事实D属于另外一个AB为D = A + T（BA）。

Proof: Point D is on a line CD perpendicular to AB, and of course D belongs to AB. Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).

我们结束了3公式：

 Dx=Ax+t(Bx-Ax)
 Dy=Ay+t(By-Ay)
(Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0

Subtitute前两个方程中的第三个给出：

Subtitute the first two equations in the third one gives:

(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0

分发解决在t给出了：

Distributing to solve for t gives:

(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0

这给：

t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]

摆脱了负号：

getting rid of the negative signs:

t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]

一旦你有了T，你可以从第一两个方程计算出的D中的坐标。

Once you have t, you can figure out the coordinates for D from the first two equations.

 Dx=Ax+t(Bx-Ax)
 Dy=Ay+t(By-Ay)

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