CODEWITHSUNDEEP
c#math3dgeometry
user1222017
user1222017

Reputation: 121

3D Perpendicular Point on Line From 3D point

This question has been asked before in reference to 2D. This question extends it to 3D. How do I find the perpendicular intersected point on a line from a point in 3D space?. If my line is defined by points (x1,y1,z1) & (x2,y2,z2) and I have a point (x3,y3,z3) in space.

How do I find the perpendicular intersection of point (x4,y4,z4) on the line from (x3,y3,z3)?

Upvotes: 11

Views: 12702

Answers (4)

daniel
daniel

Reputation: 167

I did the calculation for the answer marked as Best Answer:
let alpha = [(x3-x1)(x2-x1) + (y3-y1)(y2-y1) + (z3-z1)(z2-z1)] / [(x2-x1)(x2-x1) + (y2-y1)(y2-y1) + (z2-z1)(z2-z1)]
point of intersection = (x1+alpha*(x2-x1), y1+alpha*(y2-y1), z1+alpha*(z2-z1)).

Upvotes: 1

VivekParamasivam
VivekParamasivam

Reputation: 1154

We Can also Use AAS Triangle Method.

A - (x1,y1,z1), B - (x2,y2,z3) , C - (x3,y3,z3)

We need to find the Point D in Line AB which is perpendicular to point C

Now we have Directional Vectors

VectorAC = normalize(A - C),
VectorCA = normalize(C - A),
VectorAB = normalize(A - B),

Lets Consider ADC as a triangle

Dot product gives the angle between 2 vectors 

AngleA = Angle Between VectorAC and vectorAB
AngleD = Angle Between VectorDC and vector DA, Always 90 deg, As VectorDC and DA are perpendicular to each other.
AngleC =  180 - (AngleA + AngleD), Angle between VectorCD and VectorCA.

So now we have 3 angle of the Triangle

Use AAS triangle method and find the distance between A and D.

http://www.mathsisfun.com/algebra/trig-solving-aas-triangles.html

 D = (A + (VectorAB * Distance between A and D)).

Upvotes: 2

Tigran
Tigran

Reputation: 62246

You're asking, in practice, the calculation of a distance between the point and line, so the Length of the vector from the (x3,y3,z3) point to the line which is orthogonal to the vector. If we immagine the line like a vector, that means that dot-product of both vectors is equal to 0. (this is just to give you a hint) .

Upvotes: 2

Robert Rouhani
Robert Rouhani

Reputation: 14678

For starters, you pretty much need some implementation of a Vector3 class, whether you write your own, find a standalone implementation on the internet somewhere, or use a library that contains one like XNA or Sharp3D.Math.

Typically lines in 3d space are not represented by two points, but by parametric equations and operated on by vectors and not scalars. Your parametric equation would be of the form:

x = x1 + t(x2-x1), y = y1 + t(y2-y1), z = z1 + t(z2-z1)

The vector u is defined by the coefficients of t. <x2-x1, y2-y1, z2-z1>.

The vector PQ is defined by your chosen point Q minus a point P on the line. Any point on the line can be chosen, so it would be simplest to just use the line t = 0, which simplifies to x1, y1, and z1. <x3-x1, y3-y1, z3-z1>

The definition of the shortest distance between a point and a line in 3-space is as follows:

D = ||PQ x u|| / ||u||

Where x is the cross product operator, and || ... || gets the magnitude of the contained vector. Depending on which library you choose, your code may vary, but it should be very similar:

Vector3 u = new Vector3(x2 - x1, y2 - y1, z2 - z1);
Vector3 pq = new Vector3(x3 - x1, y3 - y1, z3 - z1);

float distance = Vector3.Cross(pq, u).Length / u.Length;

Edit: I just realized you wanted the actual point of intersection, and not the distance. The formula to find the actual point is a bit different. You need to use inner product space to get the component of u perpendicular to PQ. To do that, you need to find the component of u in the direction of PQ:

((PQ · u) / ||u||^2) * u

This gets us the w1 component, but we want w2, which is the component between Q and the line:

PQ = w1 + w2

w2 = PQ - w1

From there, we take w2 and add it to the point Q to get the point on the line nearest Q. In code this would be:

Vector3 p1 = new Vector3(x1, y1, z1);
Vector3 p2 = new Vector3(x2, y2, z2);
Vector3 q = new Vector3(x3, y3, z3);

Vector3 u = p2 - p1;
Vector3 pq = q - p1;
Vector3 w2 = pq - Vector3.Multiply(u, Vector3.Dot(pq, u) / u.LengthSquared);

Vector3 point = q - w2;

Where point.X is x4, point.Y is y4, and point.Z is z4.

Upvotes: 11

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