Can anyone provide me with a computer program or algorithm that calculates the distance between a point and a line in a three-dimensional space? The program can be written in JavaScript or any other programming language. In this scenario, P represents the point, while A and B represent two points on the line.
P=(4,2,1);
A=(1,0,1);
B=(1,2,0);
When looking for the point X that is closest to the point P on the line (A, V), it is important to find the point where the line (X, P) is normal to the line (A, B).
For a line defined by two points A and B, a Unit vector D that gives the direction of the line can be calculated as follows (Note that the length of a unit vector is 1):
D = normalize(B-A);
To use the following formula, a point O on the line is required, such as O = A.
Now, the closest point X to the point P on the line (O, D) needs to be determined.
First, calculate a vector V from O to P:
V = P - O;
The distance d from O to the intersection point (closest point) X can be calculated using the Dot product.
In general, the dot product of two vectors is equal to the cosine of the angle between the two vectors multiplied by the magnitude (length) of both vectors.
dot( A, B ) == | A | * | B | * cos( angle_A_B )
https://i.sstatic.net/IoXGA.png
Since D is a unit vector, the dot product of V and D is equal to the cosine of the angle between the line (O, D) and the vector V, multiplied by the magnitude (length) of V:
d = dot(V, D);
The intersection point X can be calculated by moving the point O along the line (D) by the distance d:
X = O + D * d;
Therefore, the formula for the intersection point is:
O ... any point on the line
D ... unit vector pointing in the direction of the line
P ... the "Point"
X = O + D * dot(P-O, D);
https://i.sstatic.net/3s0V5.png
The calculation involving points on the line A, B, and the point P is as follows:
D = normalize(B-A);
X = A + D * dot(P-A, D);
The dot Product for 3-dimensional cartesian coordinates can be expressed as:
dot(A, B) = Ax*Bx + Ay*By + Az*Bz
A normalized vector (unit vector) can be calculated by:
len(A) = sqrt(Ax*Ax + Ay*Ay + Az*Az)
normalize(A) = A / len(A)
In pure Javascript, this can be calculated as follows:
var P=[4,2,1];
var A=[1,0,1];
var B=[1,2,0];
var AB = [B[0]-A[0], B[1]-A[1], B[2]-A[2]];
var lenAB = Math.sqrt(AB[0]*AB[0] + AB[1]*AB[1] + AB[2]*AB[2]);
var D = [AB[0]/lenAB, AB[1]/lenAB, AB[2]/lenAB];
var AP = [P[0]-A[0], P[1]-A[1], P[2]-A[2]];
var d = D[0]*AP[0] + D[1]*AP[1] + D[2]*AP[2];
var X = [A[0] + d * D[0], A[1] + d * D[1], A[2] + d * D[2]];
This can also be achieved using Three.js - Vector3, where vector arithmetic operations like add, sub, dot, and normalize are provided:
var P = new THREE.Vector3(4, 2, 1);
var A = new THREE.Vector3(1, 0, 1);
var B = new THREE.Vector3(1, 2, 0);
var D = B.clone().sub(A).normalize();
var d = P.clone().sub(A).dot(D);
var X = A.clone().add(D.clone().multiplyScalar(d));
If you need to find the distance between a point and a line in a 3D space, you can rely on the Line3 class provided by three.js to handle the calculation for you.
var line = new THREE.Line3(); // initialize the Line3 object for reuse
var P = new THREE.Vector3(); // create Vector3 objects for points
var A = new THREE.Vector3();
var B = new THREE.Vector3();
var C = new THREE.Vector3();
P.set( 4, 2, 1 );
A.set( 1, 0, 1 );
B.set( 1, 2, 0 );
// find the closest point C on the infinite line AB to point P
line.set( A, B ).closestPointToPoint( P, false, C );
var distance = P.distanceTo( C ); // calculate the distance between P and C
console.log( distance ); // output the calculated distance
Using version r.97 of three.js
Point=Pnew THREE.Vector3(4,2,1);
Start=new THREE.Vector3(1,0,1);
End=new THREE.Vector3(1,2,0);
Normal=End.clone().sub(Start).normalize()
Distance = Normal.multiplyScalar(Point.sub(Start).dot(Normal)).add(Start).sub(Point).length()
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