I have one triangle in $3D$ space that I am tracking in a simulation. Between time steps I have the previous normal of the triangle and the current normal of the triangle along with both the current and previous $3D$ vertex positions of the triangles.

Using the normals of the triangular plane I would like to determine a rotation matrix that would align the normals of the triangles thereby setting the two triangles parallel to each other. I would then like to use a translation matrix to map the previous onto the current, however this is not my main concern right now.

I have found this website that says I must

  • determine the cross product of these two vectors (to determine a rotation axis)
  • determine the dot product ( to find rotation angle)
  • build quaternion (not sure what this means)
  • the transformation matrix is the quaternion as a $3 \times 3$ (not sure)

Any help on how I can solve this problem would be appreciated.

  • 2
    $\begingroup$ This may be helpful: gamedev.stackexchange.com/questions/20097/… $\endgroup$
    – mboratko
    Aug 8, 2012 at 21:21
  • 15
    $\begingroup$ I know it's not your main concern right now, but I suspect it will become a concern later: There's no reason to expect that after applying an arbitrary rotation aligning the normals the triangles will be related by a translation -- you'd still have to rotate around the normal to align them. Stated differently, it's not clear that aligning the normals is a good first step, since you then have to perform two separate rotations. You can get the axis of the full rotation by taking the cross-product of the changes in the differences between two pairs of vertex positions. $\endgroup$
    – joriki
    Aug 9, 2012 at 5:13
  • 2
    $\begingroup$ That is a very good point I had not even though of that $\endgroup$ Aug 9, 2012 at 14:23
  • 3
    $\begingroup$ @user1084113: No, that would be the cross-product of the changes in two vertex positions; I was talking about the cross-product of the changes in the differences between two pairs of vertex positions, which would be $((A-B)-(A'-B'))\times((B-C)\times(B'-C'))$. This gives you the axis of rotation (except if it lies in the plane of the triangle) because the translation drops out due to the differences, so this is purely the change in two different vectors due to rotation. The translation wouldn't drop out in your version. $\endgroup$
    – joriki
    Aug 10, 2012 at 13:51
  • 5
    $\begingroup$ @user1084113: There's no canonical answer to that question. An orientation-preserving isometry of $\mathbb R^3$ can be written as a composition of a rotation and a translation (in either order), but this decomposition is not unique -- you can shift the rotation axis and compensate by performing a different translation. Thus you can only determine the direction of the axis from the data, not its position. You can choose the position arbitrarily, e.g. through the origin, and choose the translation accordingly. $\endgroup$
    – joriki
    Aug 10, 2012 at 14:33

20 Answers 20


Suppose you want to find a rotation matrix $R$ that rotates unit vector $a$ onto unit vector $b$.

Proceed as follows:

Let $v = a \times b$

Let $s = \|v\|$ (sine of angle)

Let $c = a \cdot b$ (cosine of angle)

Then the rotation matrix R is given by: $$R = I + [v]_{\times} + [v]_{\times}^2\frac{1-c}{s^2},$$

where $[v]_{\times}$ is the skew-symmetric cross-product matrix of $v$, $$[v]_{\times} \stackrel{\rm def}{=} \begin{bmatrix} \,\,0 & \!-v_3 & \,\,\,v_2\\ \,\,\,v_3 & 0 & \!-v_1\\ \!-v_2 & \,\,v_1 &\,\,0 \end{bmatrix}.$$

The last part of the formula can be simplified to $$ \frac{1-c}{s^2} = \frac{1-c}{1-c^2} = \frac{1}{1+c}, $$ revealing that it is not applicable only for $\cos(\angle(a, b)) = -1$, i.e., if $a$ and $b$ point into exactly opposite directions.

  • 6
    $\begingroup$ I confirm that this works and gives answers identical to those in my answer. $\endgroup$ Aug 14, 2014 at 22:46
  • 7
    $\begingroup$ @Julien__ what if a == b? $\endgroup$
    – Jerem
    May 30, 2015 at 12:22
  • 5
    $\begingroup$ Indeed... I forgot that case. In that case v = 0 and the formula gives R = I. So here is a better answer : when the cross product is 0, R = I $\endgroup$
    – Julien__
    May 31, 2015 at 13:08
  • 3
    $\begingroup$ I get the wrong answer when trying this: import numpy as np a = np.array([1, 0, 0], dtype=np.float64) b = np.array([0, 0, 1], dtype=np.float64) v = np.cross(a, b) s = np.linalg.norm(v) c = np.dot(a, b) vx = np.array([[0, -v[2], v[1]], [v[2], 0, -v[0]], [-v[1], v[0], 0]]) r = np.eye(3) + vx + vxvx (1-c)/(s**2) print(r) $\endgroup$
    – user972014
    Sep 12, 2018 at 12:38
  • 10
    $\begingroup$ @user972014 note that you should use np.dot when squaring vx matrix, not a regular multiplication. $\endgroup$
    – taketwo
    Feb 7, 2019 at 10:49

Using Kjetil's answer answer, with process91's comment, we arrive at the following procedure.


We are given two unit column vectors, $A$ and $B$ ($\|A\|=1$ and $\|B\|=1$). The $\|\circ\|$ denotes the L-2 norm of $\circ$.

First, note that the rotation from $A$ to $B$ is just a 2D rotation on a plane with the normal $A \times B$. A 2D rotation by an angle $\theta$ is given by the following augmented matrix: $$G=\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$

Of course we don't want to actually compute any trig functions. Given our unit vectors, we note that $\cos\theta=A\cdot B$, and $\sin\theta=||A\times B||$. Thus $$G=\begin{pmatrix} A\cdot B & -\|A\times B\| & 0 \\ \|A\times B\| & A\cdot B & 0 \\ 0 & 0 & 1\end{pmatrix}.$$

This matrix represents the rotation from $A$ to $B$ in the base consisting of the following column vectors:

  1. normalized vector projection of $B$ onto $A$: $$u={(A\cdot B)A \over \|(A\cdot B)A\|}=A$$

  2. normalized vector rejection of $B$ onto $A$: $$v={B-(A\cdot B)A \over \|B- (A\cdot B)A\|}$$

  3. the cross product of $B$ and $A$: $$w=B \times A$$

Those vectors are all orthogonal, and form an orthogonal basis. This is the detail that Kjetil had missed in his answer. You could also normalize $w$ and get an orthonormal basis, if you needed one, but it doesn't seem necessary.

The basis change matrix for this basis is: $$F=\begin{pmatrix}u & v & w \end{pmatrix}^{-1}=\begin{pmatrix} A & {B-(A\cdot B)A \over \|B- (A\cdot B)A\|} & B \times A\end{pmatrix}^{-1}$$

Thus, in the original base, the rotation from $A$ to $B$ can be expressed as right-multiplication of a vector by the following matrix: $$U=F^{-1}G F.$$

One can easily show that $U A = B$, and that $\|U\|_2=1$. Also, $U$ is the same as the $R$ matrix from Rik's answer.

2D Case

For the 2D case, given $A=\left(x_1,y_1,0\right)$ and $B=\left(x_2,y_2,0\right)$, the matrix $G$ is the forward transformation matrix itself, and we can simplify it further. We note $$\begin{aligned} \cos\theta &= A\cdot B = x_1x_2+y_1y_2 \\ \sin\theta &= \| A\times B\| = x_1y_2-x_2y_1 \end{aligned}$$

Finally, $$U\equiv G=\begin{pmatrix} x_1x_2+y_1y_2 & -(x_1y_2-x_2y_1) \\ x_1y_2-x_2y_1 & x_1x_2+y_1y_2 \end{pmatrix}$$ and $$U^{-1}\equiv G^{-1}=\begin{pmatrix} x_1x_2+y_1y_2 & x_1y_2-x_2y_1 \\ -(x_1y_2-x_2y_1) & x_1x_2+y_1y_2 \end{pmatrix}$$

Octave/Matlab Implementation

The basic implementation is very simple. You could improve it by factoring out the common expressions of dot(A,B) and cross(B,A). Also note that $||A\times B||=||B\times A||$.

GG = @(A,B) [ dot(A,B) -norm(cross(A,B)) 0;\
              norm(cross(A,B)) dot(A,B)  0;\
              0              0           1];

FFi = @(A,B) [ A (B-dot(A,B)*A)/norm(B-dot(A,B)*A) cross(B,A) ];

UU = @(Fi,G) Fi*G*inv(Fi);


> a=[1 0 0]'; b=[0 1 0]';
> U = UU(FFi(a,b), GG(a,b));
> norm(U) % is it length-preserving?
ans = 1
> norm(b-U*a) % does it rotate a onto b?
ans = 0
> U
U =

   0  -1   0
   1   0   0
   0   0   1

Now with random vectors:

> vu = @(v) v/norm(v);
> ru = @() vu(rand(3,1));
> a = ru()
a =

> b = ru()
b =

> U = UU(FFi(a,b), GG(a,b));
> norm(U)
ans =  1
> norm(b-U*a)
ans =    2.2888e-16
> U
U =

   0.73680  -0.32931   0.59049
  -0.30976   0.61190   0.72776
  -0.60098  -0.71912   0.34884

Implementation of Rik's Answer

It is computationally a bit more efficient to use Rik's answer. This is also an Octave/MatLab implementation.

ssc = @(v) [0 -v(3) v(2); v(3) 0 -v(1); -v(2) v(1) 0]
RU = @(A,B) eye(3) + ssc(cross(A,B)) + \

The results produced are same as above, with slightly smaller numerical errors since there are less operations being done.

  • 1
    $\begingroup$ Matlab notation for Rik's Implementation: function R=fcn_RotationFromTwoVectors(A, B) v = cross(A,B); ssc = [0 -v(3) v(2); v(3) 0 -v(1); -v(2) v(1) 0]; R = eye(3) + ssc + ssc^2*(1-dot(A,B))/(norm(v))^2; $\endgroup$
    – phyatt
    Aug 20, 2015 at 21:55
  • $\begingroup$ @phyatt Doesn't the lambda syntax work in Matlab, too? $\endgroup$ Aug 20, 2015 at 22:16
  • 1
    $\begingroup$ I've used lamdas in other languages, and I haven't seen it in use in the Matlab projects... and sure enough, its there. mathworks.com/help/matlab/matlab_prog/anonymous-functions.html Thanks @KuberOber. $\endgroup$
    – phyatt
    Aug 20, 2015 at 22:21
  • $\begingroup$ Rik's answer doesn't work if a == b, does your answer work in this case? $\endgroup$
    – Ruts
    Nov 10, 2015 at 8:53
  • 1
    $\begingroup$ @Ruts It can't work, since you can't normalize a zero-length vector a-dot(a,a)*a. You have to check for a==b first :) $\endgroup$ Nov 10, 2015 at 17:06

Rodrigues's rotation formula gives the result of a rotation of a vector $a$ about an axis of rotation $k$ through the angle $\theta$. We can make use of this by realizing that, in order to bring a normalized vector $a$ into coincidence with another normalized vector $b$, we simply need to rotate $a$ about $k=(a+b)/2$ by the angle $\pi$. With this, one gets the beautiful $$ R = 2 \frac{(a+b)(a+b)^T}{(a+b)^T(a+b)} - I. $$ This works in any dimension.

For the case $d=3$, one can derive another matrix from Rodrigues's rotation formula, namely the rotation around the orthogonal $(a\times b) / \|a\times b\|$: $$ R_3 = \frac{1}{\|a\|\|b\|} \left(\langle a, b\rangle I + [a \times b]_\times + \frac{\|a\| \|b\| - \langle a, b\rangle}{\|a \times b\|^2} (a \times b) (a \times b)^T\right). $$ This matrix is the identity for $a = b$.

  • 3
    $\begingroup$ I can confirm that this works. Checked using Wolfram Mathematica that $R\cdot a=b$ with the assumptions that $||a||=||b||=1$. $\endgroup$
    – Ruslan
    Apr 28, 2018 at 19:30
  • 6
    $\begingroup$ While this will map vectors $a$ to $b$, it can add a lot of unnecessary "twist". For example, if $a=b=e_z$ this formula will produce a 180-degree rotation about the $z$-axis rather than the identity one might expect. $\endgroup$ Mar 24, 2019 at 19:54
  • 1
    $\begingroup$ Octave code snippet k = (a / norm(a) + b / norm(b)) / 2; R = 2 * k * k' / (k' * k) - eye(3); det(R) assuming that a and b are arbitrary 3D column vectors. This implements the top formula. $\endgroup$ Mar 23, 2020 at 13:57
  • 2
    $\begingroup$ @Matrefeytontias No, it really does work in any dimension. $\endgroup$ Aug 13, 2020 at 9:38
  • 2
    $\begingroup$ @jspencer this is because there are many possible rotations to bring a to b. 1. The minimal one brings a to b following the shortest path, and it's given by all the methods above. The axis of rotation is perpendicular to both a and b. 2. Then there is this other one in this answer, which brings a to b following a much longer path, which turns pi radians around an axis midway between a and b. This axis of rotation is coplanar with a and b, while the other one was orthogonal to both a and b. The rotation of pi radians is what produces an inversion of axes. $\endgroup$
    – Joan Solà
    Jun 14, 2022 at 22:50

The MATLAB code for any dimension greater than one is

u = a/norm(a);                      % a and b must be column vectors
v = b/norm(b);                      % of equal length
N = length(u);
S = reflection( eye(N), v+u );      % S*u = -v, S*v = -u 
R = reflection( S, v );             % v = R*u


function v = reflection( u, n )     % Reflection of u on hyperplane n.
% u can be a matrix. u and v must have the same number of rows.

v = u - 2 * n * (n'*u) / (n'*n);

See this for background on how this works.


From the top of my head (do the checking yourself)

  1. Let the given vectors in $R^3$ be $A$ and $B$. For simplicity assume they have norm $1$ and are not identical.

  2. Define $C$ as the cross product of $A$ and $B$. We want an orthogonal matrix $U$ such that $UA=B$ and $UC=C$.

  3. First change bases into the new base $(U_1,u_2,u_3)=(A,B,C)$. In this new basis the matrix doing the job is simply $G=\left(\begin{smallmatrix} 0&1&0\\1&0&0\\0&0&1\end{smallmatrix}\right)$.

  4. Then we need the basis shift matrix to the new basis. Write the coordinates of the vectors in the old base as simply $A=(a_1,a_2,a_3), B=(b_1,b_2,b_3), C=(c_1,c_2,c_3)$. Then the basis shift matrix can be seen to be $\left( \begin{smallmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3 \end{smallmatrix}\right)^{-1}$.

  5. The result is now simply $U=F^{-1} G F$, which is an orthogonal matrix rotating $A$ into $B$.

  • 1
    $\begingroup$ Actually, I think there may be further trouble - your matrix $U$ is only guaranteed to be orthogonal if $F$ is, which is only the case if $A$ and $B$ are orthogonal. In other instances, the angles of various other vectors to which $U$ is applied may be stretched or compressed. $\endgroup$
    – mboratko
    Aug 9, 2012 at 19:21
  • $\begingroup$ Your thought process is a good one, though - don't give up on it yet! I think there is a solution this way if you consider different original basis vectors which still have an easily represented form of $A$ and $B$. You could, for instance, choose the projection of $A$ on $B$, the rejection of $A$ on $B$, and $A \times B$. $\endgroup$
    – mboratko
    Aug 9, 2012 at 19:50
  • 5
    $\begingroup$ Just so that people aren't derailed: this answer doesn't work as-is, one has to take process91's comments into account. +1 since it provided valuable inspiration. $\endgroup$ Aug 14, 2014 at 22:26

As you read from the thread: http://forums.cgsociety.org/archive/index.php/t-741227.html

Note that this will not align all $3$ axes of the teapots, only the Z axis. We have no knowledge of the full orientation of the teapots in space, only know where the Z are pointing at. theTM will be the value you are looking for.

There is NO unique Matrix that could rotate one unit vector to another. Simply because the solution to $3$ equations with $9$ arguments does not unique.

Since you have the plane (not only the normal vector), a way to find a unique rotation matrix between two coordinate system would be: do the non-unique rotation twice!

##That is

  1. Find an orthogonal vector in the same plane of interest with $A$ and $B$ respectively. Say $A_o \bot A$ in the original plane and $B_o \bot B$ in the target plane.
  2. use previous answers: This or this to find the rotation matrix within the $A\times B$ plane. Assume matrix is $U_{AB}$
  3. Apply the matrix $U_{AB}$ to $A_o$ result in a new $B'_o=U_{AB}A_o$.
  4. This $B'_o$ can be used as one of the new inputs to the same rotation funtion from the previous answers together with $B_o$. Another rotation matrix $U_{B'B}$ is calculated in $B'\times B$ plane.
  5. Since $B'\times B$ plane is perpendicular to $A$, $U_{B'B}$ would not influence the transform of $A\to B$. i.e. $B=U_{AB}A=U_{B'B}U_{AB}A$. and the final coordinate transform matrix should be $U=U_{B'B}U_{AB}$
  6. consider yourself for corner cases. :)



% Implementation of Rik's Answer:
ssc = @(v) [0 -v(3) v(2); v(3) 0 -v(1); -v(2) v(1) 0];
RU = @(A,B) eye(3) + ssc(cross(A,B)) + ...


> a=[1 0 0]'; b=[0 1 0]';
> ao=[0 1 0]'; bo=[-sqrt(2)/2,0,sqrt(2)/2]';
> Uab = RU(a,b);
> Ucd = RU(Uab*ao,bo);
> U = Ucd*Uab;
> norm(b-U*a) % does Ucd influence a to b?
ans = 0
> norm(bo-U*ao) % does Ucd influence a-orthogonal and b-orthogonal?
ans = 0
> U
U =

     0   -0.7071    0.7071
1.0000         0         0
     0    0.7071    0.7071
  • 1
    $\begingroup$ Please, use Latex. $\endgroup$ Jul 15, 2016 at 9:21
  • 1
    $\begingroup$ @MithleshUpadhyay I don't get it? I feel somewhat difficult when editing answers but I did use some LaTeX for equations. They looks nice on my screen. Is there anything I can do better? If there are tricks and tips I'm willing to hear from you. $\endgroup$ Jul 19, 2016 at 11:47

Use Rodrigues' rotation formula (See the section "Conversion to rotation matrix"). $\cos\theta$ is the dot product of the normalised initial vectors and $\sin\theta$ can be determined from $\sin^2\theta + \cos^2\theta =1$

  • $\begingroup$ I don't think $\sin \theta$ can be determined in that way. There is sign ambiguity. (If it weren't so, then we'd get the same rotation matrix for going from $a$ to $b$ and viceversa.) $\endgroup$
    – leonbloy
    Aug 8, 2017 at 23:59
  • $\begingroup$ Could it be that the formula on Wikipedia at en.wikipedia.org/wiki/… lacks a cos𝜃 before the identity matrix? I do not catch the explanation why this has been dropped. $\endgroup$ Mar 23, 2020 at 10:30

Kuba Ober and Leyu Wang's answer works great. Here, is a python implementation of the same algorithm.

import numpy as np
import math

def rotation_matrix(A,B):
# a and b are in the form of numpy array

   ax = A[0]
   ay = A[1]
   az = A[2]

   bx = B[0]
   by = B[1]
   bz = B[2]

   au = A/(np.sqrt(ax*ax + ay*ay + az*az))
   bu = B/(np.sqrt(bx*bx + by*by + bz*bz))

   R=np.array([[bu[0]*au[0], bu[0]*au[1], bu[0]*au[2]], [bu[1]*au[0], bu[1]*au[1], bu[1]*au[2]], [bu[2]*au[0], bu[2]*au[1], bu[2]*au[2]] ])


The quaternion is a $4$-dimensional complex number: http://en.wikipedia.org/wiki/Quaternion used to describe rotations in space. A quaternion (like a complex number) has a polar representation involving the exponential of the arguments (rotations), and a magnetude multiplier. Building the quaternion comes from the cross product (the product of the complex components), which will give you the argument in those $3$ dimensions, you'll then get a number from that in the form $A+Bi+Cj+Dk$, and write it out in the matrix form described in the article there.

An easier way would be to simply fingure out what your original vectors are in the $4$-space, and take the appropriate inverse operations to get your resultant quaternion (without going through the dot/cross product steps) but that requires a good foundation in hypercomplex algebra.

  • 1
    $\begingroup$ Just to be clear: from the article, the dot product will be your scalar part, and the cross product the vector part, so with dot product: x and scalar product iy + jz + ka the quaternion q would be q = x +iy +jz +ka. $\endgroup$
    – Vilid
    Aug 8, 2012 at 21:51
  • $\begingroup$ Not exactly. If the dot product is a, and the cross product is (x, y, z), it would be cos(a/2) + (x * sin(a/2))i + (y * sin(a/2))j + ( z * sin(a/2))k. Rotation quaternions are unitary. $\endgroup$
    – Jeff
    Jul 19, 2013 at 2:33

Here is a Matlab function that can be used to calculated the rotation from one vector to another.

Example 1:

>> v1=[1 2 3]';
>> v2=[4 5 6]';
>> fcn_RotationFromTwoVectors(v1, v2)

ans =

0.9789 0.0829 0.1870
 -0.0998 0.9915 0.0829
 -0.1785 -0.0998 0.9789

Example 2:

>> v1=[1 2 0]';
>> v2=[3 4 0]';
>> fcn_RotationFromTwoVectors(v1, v2)

ans =

0.9839 0.1789 0
 -0.1789 0.9839 0
 0   0     1.0000


function R=fcn_RotationFromTwoVectors(v1, v2)
% R*v1=v2
% v1 and v2 should be column vectors and 3x1

% 1. rotation vector
% 2. rotation angle
% 3. rotation matrix, using Rodrigues' formula

function x_skew=fcn_GetSkew(x)
x_skew=[0 -x(3) x(2);
 x(3) 0 -x(1);
 -x(2) x(1) 0];

You could say you are looking for a transformation between two orthonormal bases:

$$ M*[\vec{i},\vec{j},\vec{k}]=[\vec{i}',\vec{j}',\vec{k}'] $$ where

  • $\vec{i}$ and $\vec{i}'$ are the two vectors in question ("from" and "to")

and the missing parts can be picked in a way which suits the purpose:

  • $\vec{j}=\vec{j}'=\frac{\vec{i}\times\vec{i}'}{||\vec{i}\times\vec{i}'||}$, the axis of rotation, which is left in place and is orthogonal to the $\vec{i},\vec{i}'$ plane (and has to be normalized)
  • $\vec{k}=\vec{i}\times\vec{j}, \vec{k}'=\vec{i}'\times\vec{j}'$, because you need a pair of 3rd vectors for completing the bases.

Since $[\vec{i},\vec{j},\vec{k}]$ is an orthogonal matrix, there is no need for 'real' inversion and the transformation is

$$ M=[\vec{i}',\vec{j}',\vec{k}']*[\vec{i},\vec{j},\vec{k}]^T $$ Except when $\vec{i},\vec{i}'$ do not stretch a plane (and thus $||\vec{i}\times\vec{i}'||=0$). This calculation will not produce $I$, e.g. dies on both $\vec{i}=\vec{i}'$ and $\vec{i}=-\vec{i}'$

Disclaimer: Sorry about the necro, and especially for the partial repeat: I see Kjetil's answer, but I simply do not understand what and why that skewed matrix is doing there, and while Kuba's answer says it builds on Kjetil's, it introduces trigonometry on top of that, slightly defeating the idea (of course I understand that the trigonometric part is expressed with dot/cross products at the end)

Plus this was my first time with LaTeX, I feel $\vec{i}~\vec{i}'$ look a bit too similar, but $\vec{i}~\vec{i'}$ is just plain ugly. And writing that fraction properly is way above me.


Given two unit vectors $\hat a$ and $\hat c$, reflecting a vector $x$ across the orthogonal complement of $\hat a$ and then for $\hat c$ will rotate the part of $x$ in the span of $\hat a$ and $\hat c$ by twice the angle from $\hat a$ to $\hat c$. Letting $\hat c = \frac{\hat a + \hat b}{|\hat a+\hat b|}$ be the unit vector that bisects $\hat a$ and $\hat b$, the composition of the two reflections $R_{\hat c} \circ R_{\hat a}$ will rotate $\hat a$ to $\hat b$, and any vector $x$ by the angle from $\hat a$ to $\hat b$. Recall that the reflection transformation is $R_n(x) = x - 2 \frac{x \cdot n}{n \cdot n} n$.


Sadly I don't have enough points to comment on the accepted answer but as others have noted, the formula doesn't work when a == -b.

To solve this edge case you have to create a normal vector of a by for example using the formula found here (a,b and c being the components of the vector):

function (a,b,c)  
    return  c<a  ? (b,-a,0) : (0,-c,b)  

then make the rotation matrix by rotating vector a around this normal by Pi.


General solution for n dimensions in matlab / octave:

%% Build input data n = 4; a = randn(n,1); b = randn(n,1);

%% Compute Q = rotation matrix A = a*b'; [V,D] = eig(A'+A); [~,idx] = min(diag(D)); v = V(:,idx); Q = eye(n) - 2*(v*v');

%% Validate Q is correct b_hat = Q'*a*norm(b)/norm(a); disp(['norm of error = ' num2str(norm(b_hat-b))]) disp(['eigenvalues of Q = ' num2str(eig(Q)')])


You can easily do all this operation using the Vector3 library.

The following four steps worked for me.

Vector3D axis = Vector3D.CrossProduct(v1, v2);

if (axis.Magnitude != 0)
    double AngleFromZaxis = Vector3D.AngleBetween(new Vector3D(0, 0, 1), vAxis);
    Vector3D vAxis = new Vector3D(axis.X, axis.Y, axis.Z);
    Matrix3D m = Matrix3D.Identity;
    Quaternion q = new Quaternion(vAxis, AngleFromZaxis);

    m.RotateAt(q, centerPoint);
    MatrixTransform3D mT = new MatrixTransform3D(m);


    myModel.Transform = group;
  • 3
    $\begingroup$ You use an undeclared variable AngleFromZaxis in your example there. And group. Hmm this code is a bit of a mess unfortunately. $\endgroup$ Apr 30, 2015 at 21:39
  • $\begingroup$ Now this works native with WPF, without the Vector3 Library. $\endgroup$
    – Sen Jacob
    Oct 1, 2019 at 13:43

This is an interesting problem. It guarantees that you rotate the unit vector $a$ to coincide with $b$. Still it would not be enough to rotate a rigid body to make it coincide with another. I post an example of this below but first let me point a few important references:

  1. @glennr correctly pointed out the Rodrigues rotation formula . Note that this formula is the same as the one shown by @Jur var den vec here. The scaling factors are due to normalization of the cross and dot products since the vector $v$ in the Wikipedia website (here $a$) is arbitrary and here $a$ is unit. The Wikipedia page only requires vector manipulations (need to be good at taking cross products).
  2. Here is an elegant proof that does not depend too much on geometrical constructions but it requires simple linear ordinary differential equations for matrices, in addition to the cross product representation as a matrix operator.

  3. The final reference is the most complete document that I have found on this topic. It is titled ROTATION: A review of useful theorems involving proper orthogonal matrices referenced to three dimensional physical space.

Now, for an example. This stack exchange link is about a question I posted on the rotation of a tetrahedron. I found and answer to my question by two consecutive rotations of matrices. The apex of the tetrahedron was at $(0,0,\sqrt(3))$ and I wanted it to be at $(-1,-1,1)$. Of course I wanted to verify that all vertices were rotated to their correct place and the product of matrices: \begin{eqnarray} Y = \left ( \begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array} \right ) = \left ( \begin{array}{ccc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{array} \right ) \end{eqnarray}

and \begin{eqnarray} P = \left ( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{array} \right ) = \left ( \begin{array}{ccc} \frac{\sqrt{3}}{3} & 0 & -\frac{\sqrt{2}}{\sqrt{3}} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{\sqrt{3}} & 0 & \frac{\sqrt{3}}{3} \\ \end{array} \right ) \end{eqnarray}

That is the matrix: \begin{eqnarray*} \frac{1}{6} \left ( \begin{array}{ccc} \sqrt{6} & -3\sqrt{2} & - 2 \sqrt{3} \\ \sqrt{6} & 3 \sqrt{2} & -2 \sqrt{3} \\ 2 \sqrt{6} & 0 & 2 \sqrt{3} \end{array} \right ) \end{eqnarray*}

did the job.

I thought that the Rodrigues rotation (the one being considered here) would do the job but it did not. I computed the Rodrigues rotation for this problem. Here is the result:

\begin{eqnarray*} R = \left ( \begin{array}{ccc} \frac{\sqrt{3}+1}{2 \sqrt{3}} & -\frac{\sqrt{3}-1}{2 \sqrt{3}} & -\frac{1}{\sqrt{3}} \\ -\frac{\sqrt{3}-1}{2 \sqrt{3}} & \frac{\sqrt{3}+1}{2 \sqrt{3}}& -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3} } \end{array} \right ) \end{eqnarray*}

This matrix correctly maps the vector $(0,0,1)$ into the vector $\frac{1}{\sqrt{3}} (-1,-1,1)$ as promised, but the other three vectors on the base of the pyramid (tetrahedron) are not correctly mapped into their positions.

The meaning of this is that, to map rigid bodies through rotations, there is not a magic formula where only one matrix could be found and instead a multiplication of elementary (or non elmentary) matrices is required to do the job, unless we are lucky. Right?


Here's how to find the transformation from one triangle to another.

First triangle has vertices $a,b,c$ and normal $n$ and second triangle has vertices $a',b',c'$ and normal $n'$.

First we will find transformation $f(x)=Mx+t$ from reference triangle to the first triangle and another transformation $g(x)=M'x+t'$ from reference triangle to the second triangle. Then the transformation mapping first triangle to the second is $$T(x) = g(f^{-1}(x)) = M'(M^{-1}(x-t)) + t' = Rx + s$$ where $R = M'M^{-1}$ and $s = -M'M^{-1}t + t'$.

As the reference triangle we will use triangle with vertices $(0,0,0),(1,0,0),(0,1,0)$ and normal $(0,0,1)$

Then: $$M = [ b-a, c-a, n]$$ $$t = a $$ $$M' = [ b'-a', c'-a', n' ] $$ $$t' = a' $$

You have to be cautious and give vertices $a,b,c$ and $a',b',c'$ in right order. This has to satisfy $((b-a)\times (c-a))\cdot n > 0$ and the same for second triangle.

If those two triangles are not the same then matrix $R$ will not be orthogonal. But we can find isometry which maps one triangle to the another as close as possible in some sense. For this you can use Kabsch algorithm which is well explained here.


One way to proceed is as following:

Start by constructing one orthonormal basis for each of the vectors $\vec{n_1}$ and $\vec{n_2}$. This can be done by the trick given in an answer to another question [1]. This will result in two transformation matrices

$$R_1=\begin{bmatrix}\vec{u_1} & \vec{v_1} & \vec{w_1}\end{bmatrix}$$

and $$R_2=\begin{bmatrix}\vec{u_2} & \vec{v_2} & \vec{w_2}\end{bmatrix}$$

Then the rotation matrix for aligning $\vec{n_1}$ onto $\vec{n_2}$ becomes


[1] https://math.stackexchange.com/q/712065


I have a simpler method comes from Erigen's "Mechanics of Continua". R is rotational matrix that rotate vector "a" align with vector "b" Matlab Code:

%%%%%% Rotate vector a align with vector b%%%%%%%%%% 

syms ax ay az bx by bz k real

a=[ax ay az]'

b=[bx by bz]'

R=[bu(1)*au(1) bu(1)*au(2) bu(1)*au(3);

   bu(2)*au(1) bu(2)*au(2) bu(2)*au(3);

   bu(3)*au(1) bu(3)*au(2) bu(3)*au(3)]

To verify:


A zero result means that $c$ (rotated $a$) and $b$ are aligned with each other.


A zero result means that $c$ (rotated $a$) and $a$ are of the same length.

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ Jul 25, 2014 at 17:50
  • $\begingroup$ Note that R = (au*bu')'. Or, simply, $R_{ij} = \hat{b}_i \hat{a}_j$ $\endgroup$ Aug 14, 2014 at 18:49
  • $\begingroup$ Of course this doesn't really work. Say let a=[1 0 0], b=[0 1 0], we should get a simple $90^\circ$ rotation around the z axis, with R=[0 1 0; 1 0 0; 0 0 1]. $\endgroup$ Aug 14, 2014 at 18:54
  • $\begingroup$ In all cases the $R$ squishes $a \times b$ to zero. $\endgroup$ Aug 14, 2014 at 19:36

This paper shows detail how to solve this problem. Assume we want to rotate vector f to vector t,

  1. Let v = f x t, u = v/||v||, c = f . t, h = 1-c/1-c^2
  2. The formula of the rotation is \begin{bmatrix} c + hv_x^2 & hv_xv_y-v_z & hv_xv_z+v_y \\ hv_xv_y+v_z & c+hv_y^2 & hv_yv_z-v_x \\ hv_xv_z-v_y & hv_yv_z+v_x & c+hv_z^2 \end{bmatrix}

Python code:

import numpy as np
v = np.cross(f, t)
u = v/np.linalg.norm(v)
c = np.dot(f, t)
h = (1 - c)/(1 - c**2)

vx, vy, vz = v
rot =[[c + h*vx**2, h*vx*vy - vz, h*vx*vz + vy],
      [h*vx*vy+vz, c+h*vy**2, h*vy*vz-vx],
      [h*vx*vz - vy, h*vy*vz + vx, c+h*vz**2]]
  • $\begingroup$ Note that the definition u = v/||v|| in bullet 1 and the corresponding statement in the Python code u = v/np.linalg.norm(v) are redundant as $u$ is never used. It is defined in the original paper as preparation for some special-case handling later on. $\endgroup$
    – Pedro
    Nov 7, 2021 at 13:23

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