## Shortest distance between a line segement and a point in three dimensions (3D)

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I've been building a program in MATLAB off of an excellent thread describing how to find the shortest distance between a point and a line segment in 2D (Shortest distance between a point and a line segment). I need a function that does essentially the same thing as this previously answered question but in 3D instead of 2D and in MATLAB.

None of the top comments for answers to this previous post are in MATLAB so I'm having some trouble understanding what's going on behind the scenes in this code. Maybe some of you smarter or more skilled out there can help me convert this to a 3D MATLAB code?

The line segment would be defined as two points S1 (x1,y1,z1) and S2 (x2,y2,z2) and the point is simply a single coordinate Pnt (x3,y3,z3).

EDIT: There seems to be a little confusion here. I really do mean line segments not an infinite line. I've attached the code that I'm working with. I'd like to add that this code which I've modified was originally written as part of a comment in the above linked thread and the original author Peter Karasev deserves credit for it. As is, the code works in 2D, and I've commented in 3 lines which are a start to make it 3D (vz, uz, and at lenSqr). My specific question is that I really don't understand what's going on mathematically with detP and how I can make detP and the subsequent if statements work in 3D.

The inputs are as defined above in the original question text.

function r = PointToLineSegment3D( S1, S2, Pnt ) % r = PointToLineSegment3D( S1, S2, Pnt ) vx = S1(1)-Pnt(1); vy = S1(2)-Pnt(2); % vz = S1(3)-Pnt(3); ux = S2(1)-S1(1); uy = S2(2)-S1(2); % uz = S2(3)-S1(3); lenSqr= (ux*ux+uy*uy); % +uz*uz detP= -vx*ux + -vy*uy; if( detP < 0 ) r = norm(S1-Pnt,2); elseif( detP > lenSqr ) r = norm(S2-Pnt,2); else r = abs(ux*vy-uy*vx)/sqrt(lenSqr); end end

Just define `A`

and `B`

and `P`

as column vectors. Then any point `X`

on the line `AB`

has the form

X = A+t*(B-A)

for some value of `t`

.

And certainly the line `XP`

must be perpendicular to the line `AB`

, that means the corresponding scalar product must be zero:

0 == (A+t*(B-A) - P)' * (B-A) == (A-P)'*(B-A)+t*norm(B-A)^2

This implies

t = (A-P)'*(B-A) / norm(B-A)^2

Then it is just a matter of calculating the distance `XP`

which is

d = norm(X-P)

so

d = norm(A+t*(B-A)-P)

So you just need to use the third and the fifth line of code that I've posted here and if I made no mistake you're good to go.

**3D Coordinate Geometry,** In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition. In three-dimensional space, points are represented by their positions along the Distance from a point to a line in space formula. If M 0 (x 0, y 0, z 0) point coordinates, s = {m; n; p} directing vector of line l, M 1 (x 1, y 1, z 1) - coordinates of point on line l, then distance between point M 0 (x 0, y 0, z 0) and line l can be found using the following formula:

C++ im not sure that it is right answer but some times it work) test data: Pnt=[1 1 1]; S1=[0 0 0]; S2=[0 3 3]; ans=1.0

#include<iostream> #include<math.h> #include<stdio.h> #include<vector> #include<iterator> #include <iomanip> using namespace std; int main() { double vx,vy,vz,ux,uy,uz,r=0,lenSqr,detP, c,tmp; int i; vector<double>copy; vector<double>Pnt; vector<double>S1; vector<double>S2; for(i=0; i<9; i++) { cin>>c; copy.push_back(c); } for(i=0; i<3; i++) { Pnt.insert(Pnt.begin(), copy[i]); // cout<<copy[i]<<endl; } copy.erase(copy.begin(),copy.begin()+3); copy.shrink_to_fit(); for(i=0; i<3; i++) { S1.insert(S1.begin(), copy[i]); } copy.erase(copy.begin(),copy.begin()+3); copy.shrink_to_fit(); for(i=0; i<3; i++) { S2.insert(S2.begin(), copy[i]); copy.erase(copy.begin()); } copy.shrink_to_fit(); /* vector<float>Pnt(3,1.0); //for(i=0; i<3; i++) //cout<<Pnt[i]; vector<float>S1(3,0.0); //for(i=0; i<3; i++) //cout<<S1[i]; vector<float>S2; S2.insert(S2.begin(), 3.0); S2.insert(S2.begin(), 3.0); S2.insert(S2.begin(), 0.0); //for(int i=0; i<3; i++) //cout<<S2[i]; //cout<<endl; */ vx = S1[0]-Pnt[0]; vy = S1[1]-Pnt[1]; vz = S1[2]-Pnt[2]; //cout<<"V: "<<vx<<vy<<vz<<endl; ux = S2[0]-S1[0]; uy = S2[1]-S1[1]; uz = S2[2]-S1[2]; //cout<<"U: "<<ux<<uy<<uz<<endl; lenSqr= (ux*ux+uy*uy+uz*uz); //cout<<"lenSqr "<<lenSqr<<endl; detP= (-vx*ux ) + (-vy*uy) + (-vz*uz); //cout<<"detP "<<detP<<endl; if( detP < 0 ) { // r = norm(S1-Pnt,2) for(i=0; i<3; i++) { tmp=pow((S1[i]-Pnt[i]),2); r += tmp; // cout<<"r: "<<r; } r = sqrt(r); cout<<fixed<<r; } else if( detP > lenSqr ) { // r = norm(S2-Pnt,2); for(i=0; i<3; i++) { tmp=pow((S2[i]-Pnt[i]),2); r += tmp; // cout<<"r: "<<r; } r = sqrt(r); cout<<fixed<<r; } //if(detP <= lenSqr || detP>=0) else { // r =norm( abs(cross((S2-S1),(S1-Pnt)))/sqrt(lenSqr)); float i1,j1,k1; i1 = uz*vy-uy*vz; j1 = ux*vz-uz*vx; k1 = uy*vx-ux*vy; //cout<<"I J k: "<<i1<<j1<<k1<<endl; r=sqrt(pow(i1,2)+pow(j1,2)+pow(k1,2))/sqrt(lenSqr); cout<<fixed<<r; } return 0; }

**Shortest distance between a Line and a Point in a 3-D plane ,** Shortest distance between a Line and a Point in a 3-D plane when D does not fall on line segment AB there will be a point D' such that PD' is perpendicular to AB and D' lies on line segment AB with CD = PD'. 3D Coordinates of the Vector . I've been building a program in MATLAB off of an excellent thread describing how to find the shortest distance between a point and a line segment in 2D (Shortest distance between a point and a line segment). I need a function that does essentially the same thing as this previously answered question but in 3D instead of 2D and in MATLAB.

For future users who find this question, this is the code that I've made to work in 3D in MATLAB. This does not work for an infinite line only a line segment.

function r = PointToLineSegment3D( S1, S2, Pnt ) % r = PointToLineSegment3D( S1, S2, Pnt ) vx = S1(1)-Pnt(1); vy = S1(2)-Pnt(2); vz = S1(3)-Pnt(3); ux = S2(1)-S1(1); uy = S2(2)-S1(2); uz = S2(3)-S1(3); lenSqr= (ux*ux+uy*uy+uz*uz) detP= -vx*ux + -vy*uy + -vz*uz; if( detP < 0 ) r = norm(S1-Pnt,2); elseif( detP > lenSqr ) r = norm(S2-Pnt,2); else r =norm( abs(cross((S2-S1),(S1-Pnt)))/sqrt(lenSqr)); end end

**Point-Line Distance--3-Dimensional -- from Wolfram MathWorld,** Let a line in three dimensions be specified by two points x_1=(x_1,y_1,z_1) and x_2=(x_2 The squared distance between a point on the line with parameter t and a point The minimum distance can then be found by plugging t back into (2 )� Demonstration of 3 methods of finding the shortest distance from a point to a line in 3D space. Thereby also revising a good amount of 3D vector content and application. 1. Using a perpendicular

**Find shortest distance between lines in 3D,** So you have two lines defined by the points r1=(2,6,−9) and r2=(−1,−2,3) and the (non unit) direction vectors e1=(3,4,−4) and e2=(2,−6,1). The coordinates of all� Determining the distance between a point and a plane follows a similar strategy to determining the distance between a point and a line. Consider a plane defined by the equation. a x + b y + c z + d = 0 ax + by + cz + d = 0 a x + b y + c z + d = 0. and a point (x 0, y 0, z 0) (x_0, y_0, z_0) (x 0 , y 0 , z 0 ) in space. Then the normal vector to

**Distance of a point to a line in 3D using 3 different techniques ,** Demonstration of 3 methods of finding the shortest distance from a point to a line in 3D Duration: 17:38
Posted: Apr 5, 2013 In three-dimensional Cartesian space, points have three coordinates each. To find the distance between A ( x 1 , y 1 , z 1 ) and B ( x 2 , y 2 , z 2 ) , use the formula:

**Line segments in 3D: Finding the midpoint! (KristaKingMath ,** The endpoints of the line segment will be given as coordinate points in three- dimensional Duration: 2:27
Posted: Feb 24, 2016 This command can help you design for a minimum distance between an alignment centerline and the right-of-way, for example. The blue lines in the following illustration show the minimum distance found. This command calculates the 2D distance between entities. Elevations are not considered in the calculations. If the selected entities cross or are collinear, the distance is displayed as zero

##### Comments

- What have you tried so far in 3-D? This site is not a code writing service. Please show us your code and describe what specific problem you have. Your problem (assuming you mean a line that is defined by two points) is fully described on MathWorld here, including very simple formulae.
- stackoverflow.com/questions/43207514 works for 2d and 3d
- Thank you @horchler for your reply, I've added the information you requested.