检查3D点是否位于给定的3D线上（两个3D点之间）

简单的点积可以很容易地做到这一点...所以让我们考虑一下 line 定义了两个点 p0，p1 任意点 | dot（p1-p0，p-p0）| /（| p1-p0 | * | p-p0 |）= 1.0

使它更像浮点数比较稳健：

其中 1e-10 足够小epsilon ...重写为代码：

正如你所看到的，没有必要使用sqrt，我们可以比较它的功耗......

然而，如果 p == p0 然后使用 p1 或返回真正的马上。

I have two 3D Points viz. (x1,y1,z1) and (x2,y2,z2) and a 3D Point (x,y,z). I would like to know if (x,y,z) lies on the line connecting (x1,y1,z1) and (x2,y2,z2).

I tried the following algorithm:

But,what if my x1 = x2 (or) y1 = y2 (or) z1=z2? Then I would be getting an error saying "Division by zero" is not possible.

I would be glad,if someone can propose some alternative method.

Thanks in Advance.

simple dot product can do this easily ... so let consider we got line defined by two points p0,p1. Any point p on that line will have the same or negative slope to any of the endpoints so

to make it more robust with floating point compare like this:

Where 1e-10 is small enough epsilon ... rewriten to code:

As you can see no need for the sqrt we can compare the power instead ...

However you should handle edge case when p==p0 then either use p1 or return true right away.

In case you want points only inside the line segment (not outside the edge points) then you need a slight change in code

So:

btw the result of the dot product gives you ratio of one vector projected to another perpendicularly or cos of the angle between them (if they are normalized) so for parallel vectors the result is 100% of length or 1.0. If you tweak the 1e-10 value using goniometry and p-p0 you can convert this to detect points up to some perpendicular distance to line (which might get handy for thick lines and or mouse selecting).