How can I project a 3D point onto a 3D line?

Question

Let's say I have a line defined by two points, A and B, both in the form (x, y, z). These points represent a line in 3D space.

I also have a point P, defined in the same format, that isn't on the line.

How would I calculate the projection of that point on to the line? I'm aware of how to do this in 2D but 3D seems to have bugger all resources on it.

Answer 1

You simply need to project vector AP onto vector AB, then add the resulting vector to point A.

Here is one way to compute it:

A + dot(AP,AB) / dot(AB,AB) * AB

This formula will work in 2D and in 3D. In fact it works in all dimensions.

Answer 2

Here's a fast and easy way to do it in python:

from numpy import *
def ClosestPointOnLine(a, b, p):
    ap = p-a
    ab = b-a
    result = a + dot(ap,ab)/dot(ab,ab) * ab
    return result

Use floats; If your vectors contain integers the division will be an integer division, and the results will be incorrect.

Answer 3

Just for an explanation about the formula from Sam Hocevar:

1. If A,B are on the line, then the vector

$$\vec{u} = \vec{AB} / \left\|\vec{AB}\right\| $$

is the unit vector for this line (don't forget that \$\left\|\vec{AB}\right\| = \sqrt {\vec{AB} \cdot \vec{AB}} \$ or respectively \$\vec{AB} \cdot \vec{AB} = \left\|\vec{AB}\right\|^2\$).

The line equation follows:

$$k \cdot{\vec{u}} + A $$ for any \$k\$ scalar.

Then when projecting P on this line, the projected point \$I\$ is defined as perpendicular to the line's unit vector, that is:

$$ \vec{IP} \cdot \vec{u} = 0 $$ and $$ I = \vec{u} \cdot \left\|\vec{AI}\right\| + A $$

Since projecting P on the line is the dot product of \$\vec{AP}\$ and \$\vec{AB}\$, we can compute:

$$ \left\|\vec{AI}\right\| = \vec{AP} \cdot \vec{AB} / \left\|\vec{AB}\right\| $$

since \$\vec{AP} = \vec{AI} + \vec{IP}\$ and \$\vec{IP} \cdot \vec{u} = \vec{IP} \cdot \vec{AB} = 0\$, then:

$$ \vec{AP} \cdot \vec{AB} = \vec{AI} \cdot \vec{AB}=\vec{AI}\cdot\vec{u} *\left\|\vec{AB}\right\| = \left\|\vec{AI}\right\| * \left\|\vec{AB}\right\|$$

That means that only the colinear component of \$\vec{AP}\$ to \$\vec{AB}\$ is participating to the scalar result

Which finally gives:

$$ \begin{matrix} I_{x} \\ I_{y} \\ I_{z} \end{matrix} = \begin{matrix} A_{x} \\ A_{y} \\ A_{z} \end{matrix} + \begin{pmatrix} P_{x} - A_{x} \\ P_{y} - A_{y} \\ P_{z} - A_{z} \end{pmatrix} \cdot \begin{pmatrix} B_{x} - A_{x} \\ B_{y} - A_{y} \\ B_{z} - A_{z} \end{pmatrix} \cdot \begin{pmatrix} u_{x} \\ u_{y} \\ u_{z} \end{pmatrix} / \left\|\vec{AB}\right\|$$

This is the formula from Sam above.

Answer 4

on this one (i cannot cooment)

    ap = p-a
    ab = b-a
    result = a + dot(ap,ab)/dot(ab,ab) * ab
    return result

If you first make the "versor" of ab you do not need dot(ab,ab) right? or you just do the normal somehow you can rewrite as:

    ap = p-a
    ab = normal(b-a)
    result = a + dot(ap,ab) * ab
    return result

(I learned it with versors (normalized vectors) so was kind of weird.