2025年判断一个点是否在三角形内

科技前沿 • 2025-02-26 08:57 • 阅读 40

大家好，我是讯享网，很高兴认识大家。

算法1：通过面积法判断点P是否在三角形ABC内，如果P点在三角形内，则Sabc = Sapc + Sapb + Spbc（S代表面积）

Sabc = 向量AB ^ 向量AC / 2;//面积公式 ^ 代表叉乘除以2 是得到三角形面积，否则得到四边形面积

unity中代码：

 Vector3 a = new Vector3(0, 0); Vector3 b = new Vector3(4, 0); Vector3 c = new Vector3(0, -5); Vector3 p = new Vector3(1, -3.7501f); Vector3 pa = p - a; Vector3 pb = p - b; Vector3 pc = p - c; Vector3 t1 = Vector3.Cross(pa, pb); Vector3 t2 = Vector3.Cross(pb, pc); Vector3 t3 = Vector3.Cross(pc, pa); float sumMianJi = Vector3.Cross(a - b, a - c).magnitude * 0.5f; float mianji1 = t1.magnitude * 0.5f; float mianji2 = t2.magnitude * 0.5f; float mianji3 = t3.magnitude * 0.5f; //这里可设置阈值，在多少范围内算相等 if (Mathf.Approximately(sumMianJi, (mianji1 + mianji2 + mianji3))) { Debug.Log("在三角形内"); } else { Debug.Log("不在三角形内"); }

讯享网

算法2：若P点在ABC内部，则∠apc,∠bpc,∠bpa都是小于180度的，或者同时大于180度，用此原理则可以计算出p点是否在三角形内，

unity内代码：

讯享网 Vector3 a = new Vector3(0, 0); Vector3 b = new Vector3(4, 0); Vector3 c = new Vector3(0, -5); Vector3 p = new Vector3(1, -3.7501f); Vector3 pa = p - a; Vector3 pb = p - b; Vector3 pc = p - c; Vector3 t1 = Vector3.Cross(pa, pb); Vector3 t2 = Vector3.Cross(pb, pc); Vector3 t3 = Vector3.Cross(pc, pa); bool isIn2 = t1.z >= 0 && t2.z >= 0 && t3.z >= 0 || t1.z <= 0 && t2.z <= 0 && t3.z <= 0; Debug.Log("isIn:" + isIn2); Debug.Log("t1:" + t1 + " t2:" + t2 + " t3:" + t3);

参考链接：https://www.cnblogs.com/TenosDoIt/p/4024413.html

2025年判断一个点是否在三角形内

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