1
$\begingroup$

I have 100 points on $z=0$ plane. I want to rotate those points, such that they lie on any plane $P(a,b,c,d)$, preserving distances.

Hence, I need a rotation matrix.

For instance, if my points are $(2,5,0)$, $(4,4,0)$, $(1,3,0)$ and $(3,5,0)$, how do I rotate them and put them onto $x + 3y - z + 1 = 0$ plane?

This is what I have tried:
First, I have calculated the angle between the normal vector of $P$ and all the normal vectors of $x=0$, $y=0$ and $z=0$ plane $(1,0,0)$, $(0,1,0)$, $(0,0,1)$
Let us refer to those angles as $\alpha$, $\beta$ and $\theta$ respectively.
Now, I rotate my points $\alpha$ around $x-axis$, $\beta$ around $y-axis$ and $\theta$ around $z-axis$.

I think my approach is wrong. Distances are preserved but the new plane is not the plane I want.

$\endgroup$
  • $\begingroup$ See this and this for a potential approach. You will need an axis of rotation and an angle of rotation. $\endgroup$ – Biswajit Banerjee May 2 '14 at 0:00
  • $\begingroup$ @BiswajitBanerjee Thanks for the links! However, what I try to do is vice-versa. $\endgroup$ – padawan May 2 '14 at 0:13
1
$\begingroup$

To expand on my earlier comment, the approach I suggested does work and is relatively straightforward. See the plots produced by the Matlab script below. enter image description here enter image description here

function rotatePoints

  origin = [0 0 0];

  % Points on z = 0 plane
  pts = [[2 5 0];[4 4 0];[1 3 0];[3 5 0]];

  % Normal to z = 0 plane
  zaxis = [0 0 1];

  % Normal to x + 3y - z + 1 = 0 plane
  normal0 = [1 3 -1];
  normal = normal0/norm(normal0);

  % Angle and axis of rotation
  rot_angle = acos(dot(normal,zaxis))
  rot_axis = cross(normal, zaxis)

  % Rotation matrix
  R = rotation_matrix(rot_angle, rot_axis);

  % Rotate points
  [npts, m] = size(pts);
  pts_rotated = zeros(npts,3);
  for ii=1:npts
   pts_rotated(ii,:) = (R*pts(ii,:)')';
  end

  % Plot the two normals
  plot3([origin(1) zaxis(1)],[origin(2) zaxis(2)], [origin(3) zaxis(3)],'k'); hold on;
  plot3([origin(1) normal0(1)],[origin(2) normal0(2)], [origin(3) normal0(3)],'r');

  % Plot reference points
  for ii=1:npts
    plot3(pts(ii,1), pts(ii,2), pts(ii,3), 'ko'); hold on;
  end
  fill3(pts(:,1),pts(:,2),pts(:,3),'k');

  % Plot rotated points
  for ii=1:npts
    plot3(pts_rotated(ii,1), pts_rotated(ii,2), pts_rotated(ii,3), 'ro');
  end
  fill3(pts_rotated(:,1),pts_rotated(:,2),pts_rotated(:,3),'r','EdgeColor','r');
  grid on;
  axis equal;

  % Check distances
  for ii=1:npts-1
    xdiff = pts(ii+1,1)-pts(ii,1);
    ydiff = pts(ii+1,2)-pts(ii,2);
    zdiff = pts(ii+1,3)-pts(ii,3);
    dist(ii) = sqrt(xdiff^2+ydiff^2+zdiff^2);
  end
  for ii=1:npts-1
    xdiff = pts_rotated(ii+1,1)-pts_rotated(ii,1);
    ydiff = pts_rotated(ii+1,2)-pts_rotated(ii,2);
    zdiff = pts_rotated(ii+1,3)-pts_rotated(ii,3);
    dist_rotated(ii) = sqrt(xdiff^2+ydiff^2+zdiff^2);
  end
  dist_err = dist_rotated - dist

function [R] = rotation_matrix(angle, axis)

   axis = axis/norm(axis);
   ca = cos(angle);
   sa = sin(angle);
   I = [[1 0 0];[0 1 0];[0 0 1]];
   aa = axis'*axis
   A = [[0 axis(3) -axis(2)];[-axis(3) 0 axis(1)];[axis(2) -axis(1) 0]]
   R = (I - aa)*ca + aa + A*sa;
$\endgroup$
  • $\begingroup$ This is perfect! But why are the planes so far from each other? $\endgroup$ – padawan May 2 '14 at 1:35
  • $\begingroup$ Because you haven't specified what point the plane passes through. In this case the plane passes through the origin and the rotation is around the origin. To put on the points on another parallel plane you just have to translate them by the appropriate vector. $\endgroup$ – Biswajit Banerjee May 2 '14 at 3:15
  • $\begingroup$ Please do me just one more favor and tell how do I find the appropriate vector? $\endgroup$ – padawan May 21 '14 at 11:31
  • $\begingroup$ First find a point on the plane. For example, if $x=0$ and $y=0$ then $z=1$. Therefore the position vector of the point on the plane is $\mathbf{p}=(0,0,1)$. Let $\mathbf{p}_1$ be one of the points on the plane that passes through $(0,0,0)$. Then the translated point is $\mathbf{q} = \mathbf{p}_1 + \mathbf{p}$. $\endgroup$ – Biswajit Banerjee May 27 '14 at 1:54
1
$\begingroup$

No need to compute angles anywhere.

The columns of the transformation matrix are the images of the unit vectors under said transformation. Since you didn't specify placement of your points in that plane, you can use any pair of orthogonal uint vectors in that plane. A nice trick to obtain orthogonal vectors in 3d is using the cross product.

So start with the normal vector $(1,3,-1)$ of the plane, and an arbitrarily chosen second vector $(0,0,1)$ Compute the cross products

$$v_1=\begin{pmatrix}1\\3\\-1\end{pmatrix}\times\begin{pmatrix}0\\0\\1\end{pmatrix} =\begin{pmatrix}3\\-1\\0\end{pmatrix} \\ v_2=\begin{pmatrix}1\\3\\-1\end{pmatrix}\times\begin{pmatrix}3\\-1\\0\end{pmatrix} =\begin{pmatrix}-1\\-3\\-10\end{pmatrix}$$

To preserve distances, you'll have to scale them to unit length. Then you can plug them into the columns of a transformation matrix, and choose an offset such that the image of the origin lies within your plane. For example like this:

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix} \frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{110}} & \frac{1}{\sqrt{11}} \\ -\frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{110}} & \frac{3}{\sqrt{11}} \\ 0 & -\frac{10}{\sqrt{110}} & -\frac{1}{\sqrt{11}} \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} +\begin{pmatrix}0\\0\\1\end{pmatrix}$$

If you really want to rotate around the axis where both your planes intersect, so that points on that axis remain fixed, you'll need a bit more work. First find the direction of the common line. It is perpendicular to both the normals of the two planes. Since I used $(0,0,1)$ as the arbitrary vector in the first step above, and that's the normal of the $z=0$ plane, you already have $v_1$ as the direction of that axis. You can get the transformation matrix like this:

$$\begin{pmatrix} \frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{110}} & \frac{1}{\sqrt{11}} \\ -\frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{110}} & \frac{3}{\sqrt{11}} \\ 0 & -\frac{10}{\sqrt{110}} & -\frac{1}{\sqrt{11}} \end{pmatrix}\begin{pmatrix} \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} & 0 \\ -\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}$$

The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially.

To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed.

In the end you obtain

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \frac1{110}\begin{pmatrix} -\sqrt{11} + 99 & -3 \, \sqrt{11} - 33 & 10 \, \sqrt{11} \\ -3 \, \sqrt{11} - 33 & -9 \, \sqrt{11} + 11 & 30 \, \sqrt{11} \\ -10 \, \sqrt{11} & -30 \, \sqrt{11} & -10 \, \sqrt{11} \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}- \frac1{110}\begin{pmatrix} \sqrt{11} + 11 \\ 3 \, \sqrt{11} + 33 \\ 10 \, \sqrt{11} \end{pmatrix}$$

$\endgroup$
  • $\begingroup$ I'm now trying to implement those steps. Just a little question: The right matrix (prior to being inverted) you wrote, but I could not understand where does this matrix come from? $[3, -1, 0]$ is $v1$ but there is nothing like $[1, 3, 0]$. $\endgroup$ – padawan May 12 '14 at 19:46
  • $\begingroup$ @cagirici: The $[1,3,0]$ is characterized by 1. being in the $z=0$ plane and 2. being orthogonal to $[3,-1,0]$. So it's a local orthogonal coordinate system in the $z=0$ plane. This still leaves two choiced for the sign of that second vector; $[-1,-3,0]$ would satisfy both conditions as well. Using that instead, as I did at first, will lead to a reflection not a rotation, though, since it reverses signs. You can tell by checking out the determinant of the final matrix; it will be $-1$ for the reflection but $1$ for the rotation. $\endgroup$ – MvG May 13 '14 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.