# Determing the distance from a line segment to a point in 3-space

Imagine I have a line segment defined by endpoints $p_1$ and $p_2$, and some 3-space coordinate $q$.

Is there a robust (in the sense of never giving divide-by-zero errors) way to quickly determine the distance between the point and line segment?

Update - The neat answer provided by julien seems to provide the distance to a line, not a line segment as specified in the problem description.

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Looks like the sphere plays no role whatsoever in the question. Why mention it? –  Dan Shved Mar 6 '13 at 18:40
@DanShved No argument there - I forgot to change it. –  AffiDavid Mar 6 '13 at 18:57
Have you considered trying Google? The first hit for "distance point line segment" is the StackOverflow question Shortest distance between a point and a line segment. Also pretty high in the results list is David Eberly's article Distance Between Point and Line, Ray, or Line Segment. –  Rahul Mar 6 '13 at 22:28
@RahulNarain I did not google either, I did not even read your comment...Unfortunately. –  1015 Mar 6 '13 at 23:38
@RahulNarain I looked through StackOverflow and also the other link you mention. However, Julien's answer is clearer and more helpful than anything at these locations IMO. –  AffiDavid Mar 7 '13 at 5:14

1) I will first show how to compute the distance between $p$ and the line $(p_1,p_2)$.

Let $u$ be the vector $\vec{p_1p_2}$ and let $v$ be the vector $\vec{p_1q}$.

You want to find the orthogonal projection $p$ of $q$ on the line.

This is given by the formula $$p=p_1+\frac{(u,v)}{\|u\|^2}u.$$

Once you have $p$, you distance is simply the distance between $q$ and $p$, namely $$d(q,p)=\|\vec{qp}\|.$$

Note: $(u,v)$ denotes the Euclidean inner-product and $\|u\|=\sqrt{(u,u)}$ the Euclidean norm.

2) Now let us consider the distance to the segment $[p_1,p_2]$. Recall that $p$ is the orthogonal projection of $q$ on the line. There are three cases:

a) The projection $p$ belongs to $[p_1,p_2]$, then your distance is $d(q,p)=\|\vec{qp}\|$.

b) The projection $p$ belongs to $(-\infty,p_1)$, the infinite portion of the line which starts at $p_1$ excluded and does not contain $p_2$. Then your distance is $d(q,p_1)=\|\vec{qp_1}\|$.

c) The projection $p$ belongs to $(p_2,+\infty)$, the infinite portion of the line which starts at $p_2$ excluded and does not contain $p_1$. In this case, it is $d(q,p_2)=\|\vec{qp_2}\|$.

3) How to make this an algorithm.

3.1) Compute $$\frac{(u,v)}{\|u\|^2}.$$

3.2) If this is in $[0,1]$, you are in case a), so compute $p$ and return $d(q,p)=\|\vec{qp}\|$.

3.3)If this is negative, you are in case b), so return $d(q,p_1)=\|\vec{qp_1}\|$.

3.4) If this is greater than $1$, you are in case c), so the answer is $d(q,p_2)=\|\vec{qp_2}\|$.

I believe this is robust, since this never leads to a division by $0$.

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This is for a line, not a line segment, right? –  AffiDavid Mar 6 '13 at 20:27
In Mathematica, computing: EuclideanDistance[PointCoordinate, (LineEndPointOne + Dot[LineVector, LinePointVector]/Norm[LineVector]^2*LineVector)], seems to give the distance to a line, not the line segment. –  AffiDavid Mar 6 '13 at 20:29
@AffiDavid Correct, this is for the distance to the line. I forgot the word segment. I'll edit. –  1015 Mar 6 '13 at 23:03
@AffiDavid Here you go. I think this should answer your question now. –  1015 Mar 6 '13 at 23:19
Thanks, I appreciate the answer for the line segment as well as your previous one! Sorry for the lack of etiquette, I've just been having a difficult time chasing down an actual solution for the line segment question. –  AffiDavid Mar 7 '13 at 5:13

Well for a segment you have to consider three situations. If the solution p is in the segment then it is the projection. Otherwise the closest point is either p_1 or p_2 depending on which is closer to p.

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