1. Introduction

In this tutorial, we’ll explain how to project a three-dimensional point onto a (two-dimensional) plane with its coordinate system.

2. Setup

Let’s say we have a point with coordinates Z=[z_1, z_2, z_3] and a plane whose equation is:

(1)   \begin{equation*}  a x_1 + b x_2 + c x_3 = d \qquad (a, b, c, d \in R, \quad  a^2 + b^2 + c^2 \neq 0, \quad x_1, x_2, x_3 \in R) \end{equation*}

where [x_1, x_2, x_3] is any point in the plane, and a, b, c, d are its parameters. Geometrically, [a, b, c] is the vector normal to the plane. It doesn’t have to be unitary but needs to be non-zero. Parameter d is the x_3-coordinate of the plane’s intersection with the x_3-axis. Visually:

A point and a plane

The plane has a two-dimensional coordinate system of its own, defined by two base unit vectors e_1 and e_2.

So, our goal is to find the coordinates of the \boldsymbol{Z}‘s projection \boldsymbol{Z'} onto the plane in the coordinate system defined by \boldsymbol{e_1} and \boldsymbol{e_2}.

3. How to Project a Point

Let’s start by projecting Z onto the plane and finding the coordinates in the original 3D system. Since the projection \boldsymbol{Z'} belongs to the plane, its coordinates fit Equation (1):

(2)   \begin{equation*}  a z_1' + b z_2' + c z_3' = d \end{equation*}

Further, the line connecting \boldsymbol{Z} and \boldsymbol{Z'} is perpendicular to the plane, so it’s parallel to the plane’s normal vector. The line’s vector is [z_1' - z_1, z_2' - z_2, z_3' - z_3]. So, the following should hold:

(3)   \begin{equation*}  \begin{aligned} z_1' - z_1 &= k a \\ z_2' - z_2 &= k b \\ z_3 '- z_3 &= k c \end{aligned} \end{equation*}

for some real number k.

By solving (3) for z_1', z_2', and z_3', and plugging the obtained expressions into Equation (2), we get:

    \[\begin{aligned} a(z_1 + k a) + b(z_2 + k b) + c( z_3 + k c) &= d \\ a z_1 + b z_2 + c z_3 + k(a^2 + b^2 + c^2) & = d \\ k = \frac{d - a z_1 - b z_2 - c z_3}{a^2 + b^2 + c^2} \end{aligned}\]

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 k in Equations (2), we can easily calculate the coordinates:

    \[\begin{aligned} z_1' &= z_1 + k a \\ z_2' &= z_2 + k b \\ z_3' &= z_3 + k c \end{aligned}\]

3.1. Example

Let Z=[1, 0, 0] and let the plane’s equation be:

    \[x_1 + 3 x_2 + 3 x_3 = 20\]

For k, we get:

    \[k = \frac{20 - 1}{1 + 9 + 9} = \frac{19}{19} = 1\]

So, the projection Z' is [1 - 1, 0 - 3, 0 - 3] = [0, -3, -3].

3.2. An Alternative Parameterization of the Plane

The plane can be specified by the normal vector [a, b, c] and a point X_0=[p, q, r] in it. For an arbitrary point X=[x_1, x_2, x_3] in the plane, it must hold that the vector X - X_0=[x_1 - p, x_2 - q. x_3 - r] is perpendicular to [a, b, c]:


    \[\begin{aligned} a(x_1 - p) + b(x_2 - q) + c(x_3 - r) &= 0 \\ a x_1 + b x_2 + c x_3  - ap - bq - cr & = 0 \\ a x_1 + b x_2 + c x_3 &= ap + bq + cr \end{aligned}\]

This is the same as Equation (??) if we set d=ap +bq +cr. So, after computing d, we can proceed as in the sections above.

4. Finding New Coordinates

We get the new coordinates of \boldsymbol{Z'} by projecting it onto the unit vectors \boldsymbol{e_1} and \boldsymbol{e_2}:

Project a point onto unit vectors


To do that, it’s sufficient to find the dot products \langle Z', e_1 \rangle and \langle Z', e_2 \rangle.

Now, let’s suppose that e_1 = [e_{11}, e_{12}, e_{13}] and e_2=[e_{21}, e_{22}, e_{23}]. If we treat all the vectors (including points) as column vectors, we can get the new coordinates of Z' by matrix multiplication:

(4)   \begin{equation*} \begin{bmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ \end{bmatrix} \times \begin{bmatrix} z_1' \\ z_2' \\ z_3' \end{bmatrix} = \begin{bmatrix} e_{11} z_1' + e_{12} z_2' + e_{13} z_3' \\ e_{21} z_1' + e_{22} z_2' + e_{23} z_3' \\ \end{bmatrix} \end{equation*}

4.1. Example

Let’s say that Z' is [0, 1, 0], e_1 = [0, \sqrt{2}/2, \sqrt{2}/2], and e_2=[0, -\sqrt{2}/2, \sqrt{2}{2}]:

The coordinates of Z' in the new coordinate system are [\sqrt{2}/2, -\sqrt{2}/2].

5. Conclusion

In this article, we showed how to project a point onto a plane and find its coordinates in the plane’s coordinate system.

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