几何向量：空间三角形内心

          之前计算过外心了，这次我们计算一下空间三角形内心（内切圆的圆心）。

          问：已知世界空间中任意三角形和顶点坐标，求其内心坐标。如下：

          

          一般平面三角形内心求法：任意两角的平分线相交点即为内心，如图：

          

          这也好理解：△ABC任意两角角度平分后交于P点，做垂线PD、PE、PF，可以组成四个直角三角形，则垂线PD、PE、PF的长度相同，所以交点P即为内心。

          那么求空间三角形内心的关键：求空间角平分线，例如求∠BAC平分线AP，如下：

          

           我们得到单位平分线AG的表达式。

           接下来就是求△ABC平分线的交点P(x,y,z)，也就是内心：

          

           主要是通过直角三角形性质，求得r2或者r1的长度，即可得到P点的坐标表达式。其中∠ABC和∠BAC的求法如下：

          

          主要就是通过点积求得角度。

          接下来就是通过代码来具体实现一下，看下我们推导是否正确：

using UnityEngine;

public class TriangleInsideCenter : MonoBehaviour
{
    public Transform TA;
    public Transform TB;
    public Transform TC;
    public Transform TP;

    void Start()
    {
        float deg = 30f;
        float sin = Mathf.Sin(deg * Mathf.Deg2Rad);
        float cos = Mathf.Cos(deg * Mathf.Deg2Rad);

        float arcsin = Mathf.Asin(sin) * Mathf.Rad2Deg;
        float arccos = Mathf.Acos(cos) * Mathf.Rad2Deg;
    }

    void Update()
    {
        Vector3 p = CalculateInsideCenter(TA.position, TB.position, TC.position);
        TP.position = p;
#if UNITY_EDITOR
        Debug.DrawLine(TA.position, TB.position, Color.black);
        Debug.DrawLine(TB.position, TC.position, Color.black);
        Debug.DrawLine(TC.position, TA.position, Color.black);

        Debug.DrawLine(TA.position, p, Color.white);
        Debug.DrawLine(TB.position, p, Color.white);
        Debug.DrawLine(TC.position, p, Color.white);
#endif
    }

    private Vector3 CalculateInsideCenter(Vector3 A, Vector3 B, Vector3 C)
    {
        Vector3 AB = B - A;
        Vector3 AC = C - A;
        Vector3 BA = A - B;
        Vector3 BC = C - B;

        Vector3 nBA = BA.normalized;
        Vector3 nBC = BC.normalized;

        float radBAC = Mathf.Acos(Vector3.Dot(AB, AC) / (AB.magnitude * AC.magnitude));
        float radABC = Mathf.Acos(Vector3.Dot(BA, BC) / (BA.magnitude * BC.magnitude));

        float halfRadBAC = radBAC / 2f;
        float halfRadABC = radABC / 2f;

        float r2 = AB.magnitude / (Mathf.Cos(halfRadBAC) * Mathf.Sin(halfRadABC) / Mathf.Sin(halfRadBAC) + Mathf.Cos(halfRadABC));

        Vector3 P = ((nBA + nBC) / 2f).normalized * r2 + B;
        return P;
    }
}

           效果如图：

          

           估计这种看不出来结果是否正确，我们得将内心P到△ABC每条边的垂点和模长计算出来才行，如下：

          

          可以根据向量表达式和点积以及一元二次方程求根得到垂点D的表达式，接下来通过代码实现如下：

using UnityEngine;

public class TriangleInsideCenter : MonoBehaviour
{
    public Transform TA;
    public Transform TB;
    public Transform TC;
    public Transform TP;

    void Start()
    {
        float deg = 30f;
        float sin = Mathf.Sin(deg * Mathf.Deg2Rad);
        float cos = Mathf.Cos(deg * Mathf.Deg2Rad);

        float arcsin = Mathf.Asin(sin) * Mathf.Rad2Deg;
        float arccos = Mathf.Acos(cos) * Mathf.Rad2Deg;
    }

    void Update()
    {
        Vector3 A = TA.position;
        Vector3 B = TB.position;
        Vector3 C = TC.position;
        Vector3 P = CalculateInsideCenter(TA.position, TB.position, TC.position);
        TP.position = P;
#if UNITY_EDITOR
        Debug.DrawLine(A, B, Color.black);
        Debug.DrawLine(B, C, Color.black);
        Debug.DrawLine(C, A, Color.black);

        Vector3 D = CalculateVerticalVertex(A, B, P);
        Vector3 E = CalculateVerticalVertex(B, C, P);
        Vector3 F = CalculateVerticalVertex(C, A, P);

        Debug.DrawLine(D, P, Color.white);
        Debug.DrawLine(E, P, Color.white);
        Debug.DrawLine(F, P, Color.white);

        Debug.LogFormat("DP = {0} EP = {1} FP = {2}", Vector3.Distance(D, P), Vector3.Distance(E, P), Vector3.Distance(F, P));
#endif
    }

    private Vector3 CalculateInsideCenter(Vector3 A, Vector3 B, Vector3 C)
    {
        Vector3 AB = B - A;
        Vector3 AC = C - A;
        Vector3 BA = A - B;
        Vector3 BC = C - B;

        Vector3 nBA = BA.normalized;
        Vector3 nBC = BC.normalized;

        float radBAC = Mathf.Acos(Vector3.Dot(AB, AC) / (AB.magnitude * AC.magnitude));
        float radABC = Mathf.Acos(Vector3.Dot(BA, BC) / (BA.magnitude * BC.magnitude));

        float halfRadBAC = radBAC / 2f;
        float halfRadABC = radABC / 2f;

        float r2 = AB.magnitude / (Mathf.Cos(halfRadBAC) * Mathf.Sin(halfRadABC) / Mathf.Sin(halfRadBAC) + Mathf.Cos(halfRadABC));

        Vector3 P = ((nBA + nBC) / 2f).normalized * r2 + B;
        return P;
    }

    private Vector3 CalculateVerticalVertex(Vector3 A, Vector3 B, Vector3 P)
    {
        Vector3 AB = B - A;
        Vector3 nAB = AB.normalized;

        float Xn = nAB.x;
        float Yn = nAB.y;
        float Zn = nAB.z;

        float Xa = A.x;
        float Ya = A.y;
        float Za = A.z;

        float Xp = P.x;
        float Yp = P.y;
        float Zp = P.z;

        float a = Xn * Xn + Yn * Yn + Zn * Zn;
        float b = Xn * Xa - Xn * Xp + Yn * Ya - Yn * Yp + Zn * Za - Zn * Zp;
        float c = 0;

        float[] rs = CalculateQuadraticEquation(a, b, c);
        Vector3 D = Vector3.zero;
        //剔除模长r为0的情况
        if (rs[0] != 0)
        {
            D = nAB * rs[0] + A;
        }
        else if (rs[1] != 0)
        {
            D = nAB * rs[1] + A;
        }
        return D;
    }

    private float[] CalculateQuadraticEquation(float a, float b, float c)
    {
        float x1 = (-b + Mathf.Sqrt(b * b - 4 * a * c)) / (2 * a);
        float x2 = (-b - Mathf.Sqrt(b * b - 4 * a * c)) / (2 * a);
        return new float[]
        {
            x1,
            x2
        };
    }
}

          最后效果如下：

         

         可以通过图形和打印看得出来结果正确。