c# - 找到两条线之间的交点

Question

所以我一直在研究这个相对简单的算法。我不确定我的代码有什么问题，但我没有得到它们实际相交的交点。

我正在使用 Unity3D，我试图在 x,z 平面中找到两条线相交的点，但不在 x,y 平面中。我假设适用于 x,y 的算法应该适用于 x,z；

我的代码：

Vector3 thisPoint1 = thisCar.position + (2 * thisCar.forward);
Vector3 thisPoint2 = thisCar.position + (20 * thisCar.forward);

Debug.DrawLine(thisPoint1, thisPoint2, Color.white, 2);

Vector3 otherPoint1 = threateningCar.position + (2 * threateningCar.forward);
Vector3 otherPoint2 = threateningCar.position + (20 * threateningCar.forward);

Debug.DrawLine(otherPoint1, otherPoint2, Color.white, 2);

float A1 = thisPoint2.z - thisPoint1.z;
float B1 = thisPoint1.x - thisPoint2.x;
float C1 = A1 * thisPoint1.x + B1 * thisPoint1.z;

float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x;
float C2 = A2 * otherPoint1.z + B2 * otherPoint1.z;

float det = A1 * B2 - A2 * B1;

float x = (B2 * C1 - B1 * C2) / det;
float z = (A1 * C2 - A2 * C1) / det;

return new Vector3(x, this.transform.position.y, z);

谁能帮忙指出我做错了什么？

thisCar.forward并且threateningCar.forward通常是[0,0,1], [0,0,-1]或者[1,0,0], [-1,0,0]

Answer 1

找到了！！！

float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x;
float C2 = A2 * otherPoint1.z + B2 * otherPoint1.z;

应该：

float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x; 

float C2 = A2 * otherPoint1.x + B2 * otherPoint1.z ;

浪费了很多时间：/。

无论如何，这将帮助任何想要进行线交叉的人。

Answer 2

您可以将一个空的 GameObject 作为汽车的父对象，并将其稍微放在它的前面（在 IDE 中或在启动时）。这样您就可以安全地获得绘制一条线所需的两个点。

Answer 3

如果你知道t这辆车和另一辆车需要行驶的距离u，那么交叉点就很简单了。

Vector3 intersection = thisCar.position + (t * thisCar.forward);
Vector3 intersection = threateningCar.position + (u * threateningCar.forward);

但是您可以求解t并u使用一些代数和 2D 矢量叉积，[x_1,z_1] × [x_2,z_1] = (x_2 z_1 - z_2 x_1)它是y3D 叉积的标量值。

t = Vector3.Cross(threateningCar.forward, thisCar.position - threateningCar.position).y 
    / Vector3.Cross(thisCar.forward, threateningCar.forward).y;

u = Vector3.Cross(thisCar.forward, thisCar.position - threateningCar.position).y 
    / Vector3.Cross(thisCar.forward, threateningCar.forward).y;