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How to calculate the rotation matrix in 3D in terms of an arbitrary axis of rotation? Given a unit vector $V=V_{x}e_{x}+V_{y}e_{y}+V_{z}e_{z}$ How to calculate the rotation matrix about that axis?

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migrated from physics.stackexchange.com Dec 11 '12 at 10:18

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See e.g. Wikipedia – Qmechanic Dec 9 '12 at 8:40
    
But the Wikipedia page just tells the formula . I want to know how to derive this – vbvbcxv Dec 9 '12 at 8:41
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It's just a conjugation of the simple matrix for a rotation around the $z$-axis, which is effectively just 2 x 2 matrix, by another rotational matrix that rotates the North pole to the point $V$, which is a product of rotation in the theta-direction and the phi-direction to get where you need to get. Conjugation by $U$ is $URU^{-1}$ where the product is matrix product. It's possible ineffective to write these things without matrices so if you don't know matrices, this is a reason to learn them. At any rate, it's not really physics, it's linear algebra and geometry and a basic one. – Luboš Motl Dec 9 '12 at 9:15
    
thanks a lot. It's pretty simple . I think it can also be solved by considering an arbitrary vector ,taking the projection of that vector into the plane perpendicular to the axis of rotation and rotate that vector . – vbvbcxv Dec 9 '12 at 11:05
    
This is a math question and belongs on Mathematics – Sklivvz Dec 9 '12 at 14:17

I think you need the Rodrigue's rotation matrix composition.

If your unit rotation axis is $\vec{v} = (V_x,V_y,V_z)$ and the rotation angle $\theta$ then the rotation matrix is

$$ R = \boldsymbol{1}_{3\times3} + \vec{v}\times\,(\sin\theta) + \vec{v}\times\vec{v}\times\,(1-\cos\theta) $$

where $\vec{v}\times = \begin{pmatrix} 0 & -V_z & V_y \\ V_z & 0 & -V_x \\ -V_y & V_x & 0 \end{pmatrix}$ is the $3\times 3$ cross product operator matrix.

For example a rotation about the unit $\vec{v}=(\frac{\sqrt{3}}{3},0,\text{-}\frac{\sqrt{6}}{3})$ the rotation matrix is

$$ R = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix} +\begin{pmatrix} 0&\frac{\sqrt{6}}{3}&0\\ \text{-}\frac{\sqrt{6}}{3}&0&\frac{\sqrt{3}}{3}\\0&\frac{\sqrt{3}}{3}&0\end{pmatrix} \sin\theta + \begin{pmatrix} \text{-}\frac{2}{3}&0&\frac{\sqrt{2}}{3}\\0&\text{-}1&0\\\text{-}\frac{\sqrt{2}}{3}&0&\text{-}\frac{1}{3}\end{pmatrix} (1-\cos\theta) $$

which collects to:

$$ R = \frac{1}{3} \begin{pmatrix} 1\cos\theta+1 & \sqrt{6}\sin\theta & \sqrt{2}\cos\theta-\sqrt{2} \\ -\sqrt{6}\sin\theta& 3 \cos\theta & -\sqrt{3}\sin\theta \\ \sqrt{2}\cos\theta - \sqrt{2} & \sqrt{3}\sin\theta& \cos\theta+2 \end{pmatrix} $$

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Noting that $C(u)^2=uu^T-I$, Rodrigue's formula is the same as $R(u,\theta)$ in my answer. However, on MSE, it is preferable to answer with more than a link. Consider adding a bit of what Rodrigue's rotation matrix composition says, if not a proof. – robjohn Dec 11 '12 at 15:34
    
Thanks for the pointer, and since my answer was first hopefully it will get accepted. – ja72 Dec 11 '12 at 20:20
    
Thanks for completing the answer. (+1) Unfortunately, there is no author for this question on math.SE, so no answer will be accepted unless the author registers on physics.SE and associates their account on math.SE. – robjohn Dec 11 '12 at 20:27
    
Yeah hand typing 5 3x3 matrices in $LaTeX$ is painful, but I agree, also necessary for a complete answer. – ja72 Dec 11 '12 at 20:31
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I have asked the author from physics.SE to register there and join here so that they can get reputation and so that they can accept here, too. – robjohn Dec 12 '12 at 1:00

Suppose $|u|=1$ and $C(u)x=u\times x$, that is $$ C(u)=\begin{bmatrix}0&-u_3&u_2\\u_3&0&-u_1\\-u_2&u_1&0\end{bmatrix} $$ Then, for the rotation matrix $R(u,\theta)$, we have $$ \frac{\mathrm{d}}{\mathrm{d}\theta}R(u,\theta)x=C(u)R(u,\theta)x $$ That is, the direction of motion of rotation is perpendicular to the axis of rotation and the position of the point rotated.

Thus, $$ R(u,\theta)=e^{\theta C(u)} $$ One property of the cross product matrix, $C(u)$, is that $$ C(u)^3=-C(u) $$ which aids in computing the exponential via power series: $$ R(u,\theta)=e^{\theta C(u)}=I+C(u)\sin(\theta)+C(u)^2(1-\cos(\theta)) $$ where $$ C(u)=\begin{bmatrix}0&-u_3&u_2\\u_3&0&-u_1\\-u_2&u_1&0\end{bmatrix}\quad\text{and}\quad C(u)^2=\begin{bmatrix}u_1^2-1&u_1u_2&u_1u_3\\u_1u_2&u_2^2-1&u_2u_3\\u_1u_3&u_2u_3&u_3^2-1\end{bmatrix} $$

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