Calculate Rotation Matrix to align Vector A to Vector B in 3d?

Solution 1:

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.

Solution 2:

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

Derivation

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);

Testing:

> 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 =

   0.043477
   0.036412
   0.998391
> b = ru()
b =

   0.60958
   0.73540
   0.29597
> 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)) + \
     ssc(cross(A,B))^2*(1-dot(A,B))/(norm(cross(A,B))^2)

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

Solution 3:

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 a different 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 $a = b$.