Rotate XYZ frame in 3D space

Question

Given a XYZ frame in 3D space at origin O(0,0,0). And given a plane equation:

L = 0.5774x + 0.1155y + 0.8083z = 0.

I need to derive the matrix which rotate the XYZ frame until Z-axis is vertical to the plane L.

Answer 1

Your plane is given by the equation $$ L = 0.5774 x + 0.1155 y + 0.8083 z = 0 $$

from this one can derive the normal vector

$$ n = \frac{(0.5774, 0.1155, 0.8083)^t}{|(0.5774, 0.1155, 0.8083)^t|} = (0.5774, 0.1155, 0.8083)^t $$

Thus $L$ was already given as

$$ L = n \cdot r = 0 $$

for a vector $r = (x,y,z)^t$.

Looking for a vector $u = (1, y, 0)^t \in L$:

$$ u \in L \iff n \cdot u = n_x + n_y y = 0 \iff y = -\frac{n_x}{n_y} = -5 \iff u = (1, -5, 0)^t \\ n_u = \frac{u}{|u|} = (0.19612,-0.98058, 0)^t $$

So we have the normal unit vector $n$ of the plane and a unit vector $n_u$ within the plane. Looking for another unit vector $n_v \perp n_u \wedge n_v \perp n$:

$$ n_v = n_u \times n = \epsilon_{i,j,k} \, e_i \, n_{u j} \, n_k = (-0.79260, -0.15852, 0.58884)^t $$

That is a nice orthonormal system, with two vectors in the plane, and one perpendicular to it.

The matrix $$ R = (n_u, n_v, n) $$

fullfills

$$ R e_x = n_u \quad R e_y = n_v \quad R e_z = n $$

where $e_x = (1, 0, 0)^t$, $e_y = (0,1,0)^t$ and $e_z = (0, 0, 1)^t$.

However $\det(R) = -1$, which means this is not a proper rotation, therefore we switch to $$ R = (n_v, n_u, n) = \left( \begin{matrix} -0.79260 & 0.19612 & 0.57740 \\ -0.15852 & -0.98058 & 0.11550 \\ 0.58884 & 0.00000 & 0.80830 \end{matrix} \right) $$

Answer 2

You have single condition to have the Z axis oriented the same direction as the normal of the plane. Which means, that you'll have infinitely many possibilities.

You have the vector of $Z$ axis $\vec{z}$ and vector of the normal of the plane $\vec{n}$. If you rotate the coordinate system around the vector $\vec{z} + \vec{n}$ by $\pi$, you'll have the Z axis to the original direction of normal vector.

I used that in OpenGL, where is a nice function glRotatef.

If you want some specific orientation of XY plane, you'll have to rotate again around the Z axis.