给定 3 点，我如何计算法线向量?

问题描述

给定三个 3D 点(A、B 和 C)，我如何计算法线向量?这三个点定义了一个平面，我想要垂直于这个平面的向量.

Given three 3D points (A,B, & C) how do I calculate the normal vector? The three points define a plane and I want the vector perpendicular to this plane.

我可以获得演示此功能的示例 C# 代码吗?

Can I get sample C# code that demonstrates this?

推荐答案

这取决于点的顺序.如果从与法线的方向看，这些点是按逆时针顺序指定的，那么计算起来很简单:

It depends on the order of the points. If the points are specified in a counter-clockwise order as seen from a direction opposing the normal, then it's simple to calculate:

Dir = (B - A) x (C - A)
Norm = Dir / len(Dir)

其中 x 是叉积.

如果您使用的是 OpenTK 或 XNA(可以访问 Vector3 类)，那么只需:

If you're using OpenTK or XNA (have access to the Vector3 class), then it's simply a matter of:

class Triangle {
    Vector3 a, b, c;
    public Vector3 Normal {
        get {
            var dir = Vector3.Cross(b - a, c - a);
            var norm = Vector3.Normalize(dir);
            return norm;
        }
    }
}

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