检查一个点是否在两个点之间

问题描述：

我刚刚认识到我的数学有点生疏..我想检查一下Point C is between Point A and Point B. C可以在A和B的线段上，也可以不在.可能有三种情况，我必须全部识别出来:

I just recognized my math is a bit rusty.. I wanna check if Point C is between Point A and Point B. C can be on the line segment of A and B, or not. There can be three cases and I have to identify all of them:

C在A和B之间

C is between A and B

   C  
  / \  
 A---B

C在A和B前面

C is in front of A and B

C  
 \  \
  A--B

C在A和B的后面

C is in the back of A and B

       C  
   /  /
  A--B

最后两点的草图"应该是三角形.

The "sketch" in the last two points should be a triangle.

我用点积检查C是否在A和B之间.

I used the dotproduct to check if C is between A and B.

if (VectorOf(AB) * VectorOf(BC)) >= 0)

要检查C是否在A和B的后面，请使用以下代码:

To check if C is in the back of A and B i use this:

if (VectorOf(AB) * VectorOf(BC)) < 0)

但是如何识别C是否在A和B前面?

But how to identify if C is in front of A and B?

答

只需使用从点B开始的点积.

Just use the dot product starting from point B.

if (VectorOf(AC) * VectorOf(AB) < 0) {
    // C is on the left of A
}
else {
    if (VectorOf(BC) * VectorOf(BA) < 0) {
        // C is on the right of B
    }
    else {
        // C is between A and B
    }
}

或者，您可以计算相对于矢量AB的投影距离:

Alternatively, you can compute the projected distance, relative to vector AB :

(VectorOf(AC) * VectorOf(AB)) / (VectorOf(AB) * VectorOf(AB))

结果将是< 0，介于0和1之间，或者在三种情况下为> 1，如下所示:

The result would be < 0, between 0 and 1, or > 1 in your three cases, as shows the math below :

      C
     /│
    / │
   /  │
──A── H ─────B─────

点积的定义是

AC · AB = AC×AB×cos(Â)= AH×AB(带符号:如果C留在A处，则为负；如果C留在A处，则为正.右).

AC · AB = AC×AB×cos(Â) = AH×AB (signed : negative if C is left of A, positive if C is to the right).

AB · AB =AB²(正数)

AB · AB = AB² (positive)

除法的结果是符号比AH/AB:

The result of the division is the signed ratio AH/AB :

-   0          1   >1
────A── H ─────B─────

检查一个点是否在两个点之间

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