Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have written an algorithm for solving the following problem: Given two 3-vectors, say: $a,b$, find rotation of $a$ so that its orientation matches $b$.

However, I am not sure if the following algorithm works in all cases:

1) Find axis and angle using cross product and dot product:

$$\mathbf{x}=\frac{a\times b}{||a\times b||}\\ \theta=\cos^{-1}(\frac{a\cdot b}{||a||\cdot ||b||})$$

3) Find rotation matrix using exponential map:

$$\mathbf{R}=e^{\mathbf{A}\theta} =\mathbf{I}+\sin(\theta)\cdot \mathbf{A}+\left(1-\cos(\theta)\right) \cdot \mathbf{A}^{2}$$

where $\mathbf{A}$ is a skew-symmetric matrix corresponding to $\mathbf{x}$:

$$\mathbf{A}=[\mathbf{x}]_{\times}=\begin{bmatrix}0 & -\mathbf{x}_{3} & \mathbf{x}_{2} \\ \mathbf{x}_{3} & 0 & -\mathbf{x}_{1} \\ -\mathbf{x}_{2} & \mathbf{x}_{1} & 0\end{bmatrix}$$

Notes:

The axis is computed using cross product as this gives vector perpendicular to both $a$ and $b$. Only direction of the axis is important, hence it is divided by its magnitude. However, I am not sure if $\mathbf{x}$ will always have the proper direction (the result can be $-\mathbf{x}$ instead of $\mathbf{x}$?).

The rotation matrix is computed using Rodrigues' rotation formula.

Finally, the vector $\mathbf{R}a$ should have same direction as $b$.

I have tested this numerically and it seems working, but I would like to be sure the formulas work for any two $a,b$.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

This is the right general approach, but the corner case $\|a\times b\| \approx 0$ must be handled.

If $\theta < \epsilon,$ $R=I$.

If $\pi-\theta < \epsilon$, you can choose for $\mathbf{x}$ any vector orthogonal to $\mathbf{a}$, for instance $\mathbf{x} = \frac{\mathbf{a} \times e_i}{\|\mathbf{a}\times e_i\|}$, where $i$ is the index of the component of $\mathbf{a}$ with least magnitude.

share|improve this answer
    
I was aware of the first case (very small angle), but you have shown the angle approaching 180° is also important to handle. Thanks. –  Libor Feb 2 '13 at 21:53
add comment

I'm not clear on why you have a factor of $A^2$ in your expression for $R$. In particular, wikipedia lists the matrix form for the Rodrigues formula as

$$R = I \cos \theta + A \sin \theta + (1-\cos \theta) x x^T$$

share|improve this answer
    
Note that for $k$ a unit vector, $A^2 = kk^T - I$. –  user7530 Feb 2 '13 at 21:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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