Orientation of a 3D plane using three points

Question

I have a 3D plane and three points on this plane with known coordinates of these three points. How can I find the orientation of this plane i.e. angles of this plane with X, y and Z-axis.

Answer 1

Find the unit vector $n$ that is normal to the plane defined by the three points $p_1, p_2, p_3$. Then use projection to calculate the angles, i.e. the projection of $u$ on the plane is $v=u-n(n^Tu)$. So for $u=[1\,\,0\,\,0]^T$, the angle between plane and the $x$-axis is $\cos(\theta_x)=v_x$.

To find $n$, note that the dot product of $n$ with all the three points should be equal. Assuming that the points are column vectors, form the matrix $A=[p_1 \,\, p_2\,\, p_3]^T$. Now solve $An=[1\,\,1\,\,1]^T$ for $n$ and normalize $n$.

This only works if the plane does not pass through the origin. If it does the rank of $A$ will be two. In this case, $n$ is the normalized cross product of any two (non-zero, non-parallel) points.

Answer 2

Let the three points be $A, B, C$, then the normal vector is

$N = (B - A) \times (C - A) $

Now, normalize $N$ as follows

$U = \dfrac{N}{\| N \| } $

$U$ is the unit normal vector to the plane.

The angles between the plane and the three axes $X, Y, Z$ are computed as follows

Angle with $X$-axis $= \dfrac{\pi}{2} - \cos^{-1} U_x = \sin^{-1} U_x $

Angle with $Y$-axis $= \dfrac{\pi}{2} - \cos^{-1} U_y = \sin^{-1} U_y$

Angle with $Z$-axis $= \dfrac{\pi}{2} - \cos^{-1} U_z = \sin^{-1} U_z$