从给定点垂直于线段 [英] Perpendicular on a line segment from a given point

问题描述

我想计算与给定点垂直的给定线上的一个点.

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

我有一条线段 AB，线段外有一个点 C.我想计算 AB 上的点 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.

我必须找到 D 点.

它非常类似于这个，但我想要还要考虑 Z 坐标，因为它在 3D 空间中无法正确显示.

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:D点在垂直于AB的一条CD线上，当然D属于AB.写出两个向量CD.AB = 0的点积，将D属于AB的事实表示为D=A+t(B-A).

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).

我们最终得到 3 个等式:

We end up with 3 equations:

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

在第三个方程中代入前两个方程给出:

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

分布求解 t 给出:

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

给出:

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

去掉负号:

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

一旦你有了 t，你就可以从前两个方程中找出 D 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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