Question

我正在尝试在C＃中为vector rejection实现一个函数。那就是：

enter image description here

我一直在尝试用C＃编写这个公式，但由于某种原因，它总是返回零。这就是我现在所拥有的：

private Vector3 Projection(Vector3 vectorA, Vector3 vectorB) {
    Vector3 a2 = Vector3.Scale ( Divide (Vector3.Scale (vectorA, vectorB), Vector3.Scale (vectorB, vectorB)), vectorB);
    return a2;
}

private Vector3 Rejection(Vector3 vectorA, Vector3 vectorB) {
    Vector3 a2 = vectorA - Projection(vectorA, vectorB);
    return a2;
}

private Vector3 Divide(Vector3 a, Vector3 b) {
    Vector3 c = new Vector3 ();
    c.x = a.x / b.x;
    c.y = a.y / b.y;
    c.z = a.z / b.z;
    return c;
}

private void Example() {
    Vector3 a = new Vector3 (5, 5, 0);
    Vector3 b = new Vector3 (0, 10, 0);
    Vector3 c = Rejection(a, b); // Returns (NaN, 0, NaN)
}

// The Vector3 class represents a 3D vector and it's x, y and z components are of float type. It's [scale method][3] multiplies two vectors component wise.

示例：

想象一下，矢量A是应用于沿地面移动的物体的力，矢量B是重力（垂直于地面）。当向量A应用于对象时，它移动的轨迹应该是垂直于重力的第三个向量，可以被认为是垂直于B的行进。这是从B中拒绝A。

enter image description here

Answer 1

符号a·b代表vector dot product，结果是标量值。

所以你需要这样的代码：

public struct Vector3
{
    public readonly double x, y, z;
    public Vector3(double x, double y, double z)
    {
        this.x=x;
        this.y=y;
        this.z=z;
    }
    public double Dot(Vector3 other)
    {
        return x*other.x+y*other.y+z*other.z;
    }
    public Vector3 Scale(double factor)
    {
        return new Vector3(factor*x, factor*y, factor*z);
    }
    public Vector3 Add(Vector3 other)
    {
        return new Vector3(x+other.x, y+other.y, z+other.z);
    }
    public static Vector3 operator+(Vector3 a, Vector3 b) { return a.Add(b); }
    public static Vector3 operator-(Vector3 a) { return a.Scale(-1); }
    public static Vector3 operator-(Vector3 a, Vector3 b) { return a.Add(-b); }
    public static Vector3 operator*(double f, Vector3 a) { return a.Scale(f); }
    public static Vector3 operator/(Vector3 a, double d) { return a.Scale(1/d); }
    public static double operator*(Vector3 a, Vector3 b) { return a.Dot(b); }

    public Vector3 Projection(Vector3 other)
    {
        // (scalar/scalar)*(vector) = (vector)
        return (other*this)/(other*other)*other;
    }
    public Vector3 Rejection(Vector3 other)
    {
        // (vector)-(vector) = (vector)
        return this-Projection(other);
    }
}

class Program
{
    static void Main(string[] args)
    {
        var A=new Vector3(5, 5, 0);
        var B=new Vector3(0, 10, 0);
        var C=A.Rejection(B);
        // C = { 5,0,0}, expected answer from math {5,5,0}-0.5*{0,10,0}
    }
}

修改

如果您无法控制代码，可以将代码移到Vector3类之外

// Vector3 defined elsewhere with .x, .y and .z fields class VectorAlgebra { public static Vector3 Subtract(Vector3 a, Vector3 b) { return new Vector3(a.x-b.x, a.y-b.y, a.z-b.z); } public static Vector3 Scale(float f, Vector3 a) { return new Vector3(f*a.x, f*a.y, f*a.z); } public static float Dot(Vector3 a, Vector3 b) { return (a.x*b.x)+(a.y*b.y)+(a.z*b.z); } public static Vector3 Projection(Vector3 a, Vector3 b) { return Scale(Dot(a, b)/Dot(b, b), b); } public static Vector3 Rejection(Vector3 a, Vector3 b) { return Subtract(a, Projection(a, b)); } static void Main(string[] args) { var A=new Vector3(5, 5, 0); var B=new Vector3(0, 10, 0); var C=Rejection(A, B); // C = { 5,0,0}, expected answer from math {5,5,0}-0.5*{0,10,0} } }

当您对矢量代数规则进行编码时，投影和拒绝的编码与数学书籍中的公式相同。您可以使用Wolfram Alpha检查答案。

Answer 2

好的，现在我们有一些数字让我们分解步骤

Vector3.Scale ( Divide (Vector3.Scale (vectorA, vectorB), Vector3.Scale (vectorB, vectorB)), vectorB);

是

var numerator = Vector3.Scale (vectorA, vectorB);
var denominator = Vector3.Scale (vectorB, vectorB);
var divided = Divide(numerator, denominator);
var result = Vector3.Scale(divided, vectorB);

通过你的号码来处理

vectorA = (5,5,0)
vectorB = (0,10,0)
numerator = (0,50,0)
denominator = (0,100,0)
divided = (NaN,.5,NaN) <- since you can't divide by zero
result = (NaN,5,NaN)

不确定为什么你为第二个维度获得0而不是5（一个拼写错误？某种类型的转换问题？）但希望这可以解释为什么你会弹出NaN。

C＃中的矢量投影/拒绝？

2 个答案: