In this tutorial, we’ll explain how to project a three-dimensional point onto a (two-dimensional) plane with its coordinate system.
Let’s say we have a point with coordinates and a plane whose equation is:
where is any point in the plane, and are its parameters. Geometrically, is the vector normal to the plane. It doesn’t have to be unitary but needs to be non-zero. Parameter is the -coordinate of the plane’s intersection with the -axis. Visually:
The plane has a two-dimensional coordinate system of its own, defined by two base unit vectors and .
So, our goal is to find the coordinates of the ‘s projection onto the plane in the coordinate system defined by and .
3. How to Project a Point
Further, the line connecting and is perpendicular to the plane, so it’s parallel to the plane’s normal vector. The line’s vector is . So, the following should hold:
for some real number .
If the normal vector is also a unit vector (i.e., its length is 1), the denominator is also one since it denotes the vector’s squared length.
By substituting in Equations (2), we can easily calculate the coordinates:
Let and let the plane’s equation be:
For , we get:
So, the projection is .
3.2. An Alternative Parameterization of the Plane
The plane can be specified by the normal vector and a point in it. For an arbitrary point in the plane, it must hold that the vector is perpendicular to :
This is the same as Equation (??) if we set . So, after computing , we can proceed as in the sections above.
4. Finding New Coordinates
We get the new coordinates of by projecting it onto the unit vectors and :
To do that, it’s sufficient to find the dot products and .
Now, let’s suppose that and . If we treat all the vectors (including points) as column vectors, we can get the new coordinates of by matrix multiplication:
Let’s say that is , , and :
The coordinates of in the new coordinate system are .
In this article, we showed how to project a point onto a plane and find its coordinates in the plane’s coordinate system.