确定3D空间中直线的角度 [英] Determine angle of a straight line in 3D space

问题描述

我在空间中有一条直线，有一个起点和终点（x，y，z），我试图获得角度，但我无法弄清楚这个公式。我正在使用VB.NET



这是我的3D环境中的线条图片（以红色圈出的线条）



如果有人可以提供帮助，我们将不胜感激。再次，我有起点和终点的值（x，y，z）



现在设置为70度的角度。

I have a straight line in space with an start and end point (x,y,z) and I am attempting to get the angle but I cannot figure out the formula for this. I am using VB.NET

Here is a picture of the line in my 3d environment (the one that is circled in red)

If someone can help, it would be appreciated. Again, i have the values of the start and end points (x,y,z)

It is set to an angle of 70 degrees right now.

推荐答案

Dim x, y, z As Double

        x = EndPoint.X - StartPoint.X
        y = EndPoint.Y - StartPoint.Y
        z = EndPoint.Z - StartPoint.Z

        'a = arcsin(v.z / sqrt(v.x²+v.y²+v.z²))
        MsgBox((Math.Asin(z / Math.Sqrt((x ^ 2) + (y ^ 2) + (z ^ 2)))) * (180 / Math.PI))

是解决方案。有人在另一个网站上帮助我了！

Is the solution. Someone was kind enough to assist me on another site!

这是一个数学问题，与VB.NET或任何其他语言无关。据我所知，它与两侧的长度，正弦，余弦等有关。

That is a mathematical question, nothing to do with VB.NET, or any other language. As far as I recall it has something to do with the length of the sides, and sines, cosines etc.

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