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)
这篇关于垂直于从给定的点的线段的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！