Carlos Ureña / Varios

Rotations in 2D and 3D

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Perpendicular vectors in 2D

We consider $\mathcal{B}=\{\mathbf{x},\mathbf{y}\}$ as an orthonormal basis in a 2D vector space. The coordinates in that basis of a vector $\mathbf{v}$ are $(a,b)$, and we want to obtain another vector ${\mathbf{v}}_{\perp }$, perpendicular to $\mathbf{v}$, and with the same length. The coordinates of ${\mathbf{v}}_{\perp }$ are $(a_{\perp },b_{\perp })=(-b,a)$ (in what follows we identify vectors with its coordinates in $\mathcal{B}$).

This perpendicular vector is obtained by rotating $\mathbf{v}$ ${90}^o$ towards its left side (counterclockwise, as long as we imagine $\mathbf{y}$ to the left of $\mathbf{x}$). We could use the reverse convention (to rotate clockwise), and use $(b,-a)$, but the first option is far more usual.

It is easy to show that the dot product of both vectors is cero, and that they are also of equal length $$\mathbf{v}\cdot {\mathbf{v}}_{\perp }=a(-b)+ba=0\Vert \mathbf{v}\Vert =\Vert {\mathbf{v}}_{\perp }\Vert =\sqrt{a^2+b^2}$$ The (lineal) relation between $\mathbf{v}$ and ${\mathbf{v}}_{\perp }$ can be expressed by using a 2x2 matrix $\mathbf{P}$ acting on coordinate column vectors (relative to $\mathcal{B}$): $${\mathbf{v}}_{\perp }=\left(\begin{array}{c}{a}_{\perp }\\ b_{\perp }\end{array}\right)=\left(\begin{array}{c}-b\\ a\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\mathbf{P}\mathbf{v}$$

Rotations in 2D

In a 2D vector space we want to define the rotation transformation $R$. This rotation is a rotation around the origin of the vector space, with angle $\theta$ (in radians).

If we want to rotate vector $\mathbf{v}$, we begin by using the previous definition of ${\mathbf{v}}_{\perp }=\mathbf{P}\mathbf{v}$ and define the rotated vector ${\mathbf{v}}^{\prime }$ as $${\mathbf{v}}^{\prime }=(\cos \theta )\mathbf{v}+(\sin \theta ){\mathbf{v}}_{\perp }$$ note that when $\theta =0$, we have ${\mathbf{v}}^{\prime }=\mathbf{v}$.

Let us assume that the coordinates of $\mathbf{v}$ in the orthonormal basis $\{\mathbf{x},\mathbf{y}\}$ are $(a,b)$ (that is, $\mathbf{v}=a\mathbf{x}+b\mathbf{y}$). This allows us to compute the coordinates of ${\mathbf{v}}^{\prime }$, by substituting: $$\begin{array}{rl}{\mathbf{v}}^{\prime }& =(\cos \theta )\mathbf{v}+(\sin \theta ){\mathbf{v}}_{\perp }\\ & =(\cos \theta )(a\mathbf{x}+b\mathbf{y})+(\sin \theta )(-b\mathbf{x}+a\mathbf{y})\\ & =(a\cos \theta -b\sin \theta )\mathbf{x}+(a\sin \theta +b\cos \theta )\mathbf{y}\\ & =a^{\prime }\mathbf{x}+b^{\prime }\mathbf{y}\end{array}$$ where $$\begin{array}{rl}{a}^{\prime }& =a\cos \theta -b\sin \theta \\ b^{\prime }& =a\sin \theta +b\cos \theta \end{array}$$ values $(a^{\prime },b^{\prime })$ are the coordinates of the rotated point ${\mathbf{p}}^{\prime }$.

Matrix form for 2D rotations

The rotation transformation can we written in homogeneous coordinates by using a 3x3 matrix ${\mathbf{R}}_{\theta }$: $${\mathbf{v}}^{\prime }=\left(\begin{array}{c}{a}^{\prime }\\ b^{\prime }\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)={\mathbf{R}}_{\theta }\mathbf{v}$$ Matrix ${\mathbf{R}}_{\theta }$ can also be obtained by writing $\mathbf{v}$ as $\mathbf{I}\mathbf{v}$, where $\mathbf{I}$ is the identity matrix, and ${\mathbf{v}}_{\perp }$ as $\mathbf{P}\mathbf{v}$, then: $$\begin{array}{rl}{\mathbf{v}}^{\prime }& =(\cos \theta )\mathbf{v}+(\sin \theta ){\mathbf{v}}_{\perp }\\ & =(\cos \theta )\mathbf{I}\mathbf{v}+(\sin \theta )\mathbf{P}\mathbf{v}\\ & =[(\cos \theta )\mathbf{I}+(\sin \theta )\mathbf{P}]\mathbf{v}\end{array}$$ this implies we can write ${\mathbf{R}}_{\theta }$ as a combination of $\mathbf{I}$ and $\mathbf{P}$: $${\mathbf{R}}_{\theta }=(\cos \theta )\mathbf{I}+(\sin \theta )\mathbf{P}=(\cos \theta )\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+(\sin \theta )\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$$

Perpendicular vectors in 3D. Cross product.

In a 3D vector space, for any given vector $\mathbf{w}=(b_0,b_1,b_2)$, there are infinitelly many vectors perpendicular to $\mathbf{w}$ and with the same length as $\mathbf{w}$ (all those form a circunference in the plane perpendicular to $\mathbf{w}$).

To select one of those perpendicular vectors, we must use a third vector $\mathbf{k}$ (non colinear with $\mathbf{w}$), and compute the cross product of $\mathbf{k}$ and $\mathbf{w}$. Then, vector $\mathbf{k}\times \mathbf{w}$ is perpendicular to $\mathbf{w}$ (but not neccesarly with the same length as $\mathbf{w}$). If $(a_0,a_1,a_2)$ are the coordinates of $\mathbf{k}$, then: $$\mathbf{k}\times \mathbf{w}=\left(\begin{array}{c}{a}_0\\ a_1\\ a_2\end{array}\right)\times \left(\begin{array}{c}{b}_0\\ b_1\\ b_2\end{array}\right)=\left(\begin{array}{c}{a}_1{b}_2-a_2{b}_1\\ a_2{b}_0-a_0{b}_2\\ a_0{b}_1-a_1{b}_0\end{array}\right)$$ This cross-product operation (pre-multiplying by $\mathbf{k}$), can also be written as a linear matrix operation by using matrix ${\mathbf{P}}_{\mathbf{k}}$ defined as follows: $${\mathbf{P}}_{\mathbf{k}}=\left(\begin{array}{ccc}0& -a_2& a_1\\ a_2& 0& -a_0\\ -a_1& a_0& 0\end{array}\right)$$ It is easy to check that pre-multiplying $\mathbf{k}$ by $\mathbf{w}$ is the same as to apply ${\mathbf{P}}_{\mathbf{k}}$ to $\mathbf{w}$ $${\mathbf{P}}_{\mathbf{k}}\mathbf{w}=\left(\begin{array}{ccc}0& -a_2& a_1\\ a_2& 0& -a_0\\ -a_1& a_0& 0\end{array}\right)\left(\begin{array}{c}{b}_0\\ b_1\\ b_2\end{array}\right)=\left(\begin{array}{c}{a}_1{b}_2-a_2{b}_1\\ a_2{b}_0-a_0{b}_2\\ a_0{b}_1-a_1{b}_0\end{array}\right)=\mathbf{k}\times \mathbf{w}$$

Rotations in 3D and Rodrigues formula

In 3D, the rotation transform uses an unit-length vector $\mathbf{u}=(u_0,u_1,u_2)$ as the rotation axis, and an angle $\theta$. In order to rotate $\mathbf{v}$, we decompose it in two components, ${\mathbf{v}}_{\parallel }$ (parallel to $\mathbf{u}$) and ${\mathbf{v}}_{⊢}$ (perpendicular to $\mathbf{u}$, but in the plane spawned by $\mathbf{u}$ and $\mathbf{v}$) $$\mathbf{v}={\mathbf{v}}_{\parallel }+{\mathbf{v}}_{⊢}$$ where ${\mathbf{v}}_{\parallel }$ is obtained by projecting $\mathbf{v}$ onto rotation axis, that is: $${\mathbf{v}}_{\parallel }=(\mathbf{v}\cdot \mathbf{u})\mathbf{u}$$ and ${\mathbf{v}}_{⊢}$ is obtained as: $${\mathbf{v}}_{⊢}=\mathbf{v}-{\mathbf{v}}_{\parallel }$$ in the case $\mathbf{v}$ and $\mathbf{u}$ are collinear, vector ${\mathbf{v}}_{⊢}$ is the null vector and ${\mathbf{v}}_{\parallel }=\mathbf{v}$. We define an additional perpendicular vector ${\mathbf{v}}_{\perp }$, which is perpendicular to both ${\mathbf{v}}_{\parallel }$ and to ${\mathbf{v}}_{⊢}$, and has the same length as ${\mathbf{v}}_{⊢}$: $${\mathbf{v}}_{\perp }=\mathbf{u}\times {\mathbf{v}}_{⊢}$$ Note that, as $\mathbf{u}$ is unit-length and perpendicular to ${\mathbf{v}}_{⊢}$, then ${\mathbf{v}}_{\perp }$ has the same length as ${\mathbf{v}}_{⊢}$.

With these definitions in mind, we can define the rotated vector ${\mathbf{v}}^{\prime }$ by observing that its component parallel to the axis does not changes, while the component perpendicular to the axis is rotated as in 2D (in the plane perpendicular to the axis), that is, we define $${\mathbf{v}}^{\prime }={\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}+(\sin \theta ){\mathbf{v}}_{\perp }$$

We rewrite ${\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}$ as follows: $$\begin{array}{rl}{\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}& ={\mathbf{v}}_{\parallel }+(\cos \theta )(\mathbf{v}-{\mathbf{v}}_{\parallel })\\ & =(\cos \theta )\mathbf{v}+(1-\cos \theta ){\mathbf{v}}_{\parallel }\\ & =(\cos \theta )\mathbf{v}+(1-\cos \theta )(\mathbf{v}\cdot \mathbf{u})\mathbf{u}\end{array}$$ Vector ${\mathbf{v}}_{\perp }$ can we rewritten as: $$\begin{array}{rl}{\mathbf{v}}_{\perp }& =\mathbf{u}\times {\mathbf{v}}_{⊢}\\ & =\mathbf{u}\times (\mathbf{v}-{\mathbf{v}}_{\parallel })\\ & =\mathbf{u}\times \mathbf{v}-\mathbf{u}\times {\mathbf{v}}_{\parallel }\\ & =\mathbf{u}\times \mathbf{v}\end{array}$$ here, as $\mathbf{u}$ and ${\mathbf{v}}_{\parallel }$ are parallel, we have substituted their cross product by $0$. Now we can write the expression for ${\mathbf{v}}^{\prime }$ which is called Rodrigues' Rotation formula: $$\begin{gathered} \boxed{{\displaystyle \\ {\mathbf{v}}^{\prime }=(\cos \theta )\mathbf{v}+(1-\cos \theta )(\mathbf{v}\cdot \mathbf{u})\mathbf{u}+(\sin \theta )\mathbf{u}\times \mathbf{v}}} \end{gathered}$$

Matrix form for 3D Rotations

We can express Rodrigues formula in terms of cross-products, this leads to the general rotation matrix. We begin by rewriting ${\mathbf{v}}_{⊢}$ as: $${\mathbf{v}}_{⊢}={\mathbf{v}}_{\perp }\times \mathbf{u}=-\mathbf{u}\times {\mathbf{v}}_{\perp }=-\mathbf{u}\times (\mathbf{u}\times \mathbf{v})$$ This allows to write ${\mathbf{v}}_{\parallel }$ by using cross-products: $${\mathbf{v}}_{\parallel }=\mathbf{v}-{\mathbf{v}}_{⊢}=\mathbf{v}+\mathbf{u}\times (\mathbf{u}\times \mathbf{v})$$

Vector ${\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}$ is expressed as follows: $$\begin{array}{rl}{\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}& ={\mathbf{v}}_{\parallel }+(\cos \theta )(\mathbf{v}-{\mathbf{v}}_{\parallel })\\ & =(\cos \theta )\mathbf{v}+(1-\cos \theta ){\mathbf{v}}_{\parallel }\\ & =(\cos \theta )\mathbf{v}+(1-\cos \theta )[\mathbf{v}+\mathbf{u}\times (\mathbf{u}\times \mathbf{v})]\\ & =\mathbf{v}+(1-\cos \theta )\mathbf{u}\times (\mathbf{u}\times \mathbf{v})\end{array}$$

So we end up with this expression for ${\mathbf{v}}^{\prime }$: $$\begin{array}{rl}{\mathbf{v}}^{\prime }& ={\mathbf{v}}_{\parallel }+(\cos \theta ){\mathbf{v}}_{⊢}+(\sin \theta ){\mathbf{v}}_{\perp }\\ & =\mathbf{v}+(1-\cos \theta )\mathbf{u}\times (\mathbf{u}\times \mathbf{v})+(\sin \theta )\mathbf{u}\times \mathbf{v}\end{array}$$

This last expression is written just in terms of $\mathbf{v},\mathbf{u}$ and $\theta$, uses just cross-products, and is equivalent to Rodrigues' formula. From that we obtain the matrix ${\mathbf{R}}_{\theta ,\mathbf{u}}$ for rotation in 3D, which is expressed in terms of the powers of matrix ${\mathbf{P}}_{\mathbf{u}}$: $$\begin{gathered} \boxed{{\displaystyle \\ {\mathbf{R}}_{\theta ,\mathbf{u}}=\mathbf{I}+(\sin \theta ){\mathbf{P}}_{\mathbf{u}}+(1-\cos \theta ){\mathbf{P}}_{\mathbf{u}}^2}} \end{gathered}$$ where $\mathbf{I}$ is the identity matrix (that is, $\mathbf{I}={\mathbf{P}}_u^0$), and $${\mathbf{P}}_{\mathbf{u}}=\left(\begin{array}{ccc}0& -u_2& u_1\\ u_2& 0& -u_0\\ -u_1& u_0& 0\end{array}\right)$$ Thus: $${\mathbf{P}}_{\mathbf{u}}^2=\left(\begin{array}{ccc}0& -u_2& u_1\\ u_2& 0& -u_0\\ -u_1& u_0& 0\end{array}\right)\left(\begin{array}{ccc}0& -u_2& u_1\\ u_2& 0& -u_0\\ -u_1& u_0& 0\end{array}\right)=\left(\begin{array}{ccc}{u}_0^2-1& u_0{u}_1& u_0{u}_2\\ u_0{u}_1& u_1^2-1& u_1{u}_2\\ u_0{u}_2& u_1{u}_2& u_2^2-1\end{array}\right)$$ where we have used $u_0^2+u_1^2+u_2^2=1$. By expanding the matrixes:

$${\mathbf{R}}_{\theta ,\mathbf{u}}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+(\sin \theta )\left(\begin{array}{ccc}0& -u_2& u_1\\ u_2& 0& -u_0\\ -u_1& u_0& 0\end{array}\right)+(1-\cos \theta )\left(\begin{array}{ccc}{u}_0^2-1& u_0{u}_1& u_0{u}_2\\ u_0{u}_1& u_1^2-1& u_1{u}_2\\ u_0{u}_2& u_1{u}_2& u_2^2-1\end{array}\right)$$ And finally: $${\mathbf{R}}_{\theta ,\mathbf{u}}=\left(\begin{array}{ccc}(1-c)u_0^2+c& (1-c)u_0{u}_1-s{u}_2& (1-c)u_0{u}_2+su1\\ (1-c)u_0{u}_1+s{u}_2& (1-c)u_1^2+c& (1-c)u_1{u}_2-s{u}_0\\ (1-c)u_0{u}_2-s{u}_1& (1-c)u_1{u}_2+s{u}_0& (1-c)u_2^2+c\end{array}\right)\text{where:}\{\begin{array}{l}s=\sin \theta \\ c=\cos \theta \end{array}$$ More compactly: $$\begin{gathered} \boxed{{\displaystyle \\ {\mathbf{R}}_{\theta ,\mathbf{u}}=\left(\begin{array}{ccc}{a}_{00}+c& a_{01}-s{u}_2& a_{02}+su1\\ a_{10}+s{u}_2& a_{11}+c& a_{12}-s{u}_0\\ a_{20}-s{u}_1& a_{21}+s{u}_0& a_{22}+c\end{array}\right)\text{where:}\{\begin{array}{l}s=\sin \theta \\ c=\cos \theta \\ a_{ij}=(1-c)u_i{u}_j\end{array} \\ }} \end{gathered}$$ (this page uses MathJax)