找到一条线的交点

Question

所以我一直在用这个相对简单的算法来敲打我的头。我不知道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，我试图找到两条线在x，z平面相交的点，但不在x，y平面。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); 

谁能帮成指出我在做什么错？

thisCar.forward和threateningCar.forward通常是要么[0,0,1], [0,0,-1]或[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; 

应该是：

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

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

很多的浪费时间没有：/。

反正这将帮助任何想要做线路交叉口的人。

Answer 2

你可以母公司的空游戏物体的车稍微放置在它的（无论是在IDE或启动）前面。这样你就可以安全地得到绘制线条所需的两点。