c++ - 找到与给定 3D 线正交的远离 d 距离的点

Question

I want to find the coordinate of a point (say P) which is away from a known distance (say d). the case is: I want the point that lie on a line (3d) which is perpendicular to the another given line segment (3d) and passing through the one end of that given line segment (say A). So, I know 2 end points (also vector along the line) of the given line segment and distance d and vector of the perpendicular line. Also point C where the perpendicular line pass through is also known. I am having vector3 class and line3 class.

解决这个问题的方法很难弄清楚，所以请在这方面帮助我指点迷津。

是的，由于该线上有 2 个相反方向的点，我正在寻找更接近点 C(而不是 Q)的点 (P)。

提前谢谢

Answer 1

找到方向 vector ，然后将其乘以d，然后添加到起点:

Vector A, C;
float d = 100;

Vector dir = C - A;
dir.normalize();
dir *= d;
Vector P = A + dir;