A 4x4 homogeneous matrix for a 90 degree rotation about Y axis?

According to my book

Rotations through an angle $\theta$ about the $x$, $y$, and $z$ axes are performed using the following transformation matrices.

For example, $$R_x(\theta) = \left[ \begin{matrix} 1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right]$$

Ry0= | cos0 0 sin0 0|
| 0 1 0 0|
| -sin0 0 cos0 0|
| 0 0 0 1|

Rz0= | cos0 -sin0 0 0 |
| sin0 cos0 0  0|
| 0 0 1 0|
| 0 0 0 1|


And I just need to put the angle in it no matter which rotate about axis? After that, how can I get a $4\times 4$ matrix?

Thank you

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To express ordinary $\mathbb{R}^n \to \mathbb{R}^n$ linear transformation into homogeneous coordinates just add another row and column where every term is equal to $0$ but the diagonal, which should be $1$.

For example if $A$ is your transformation matrix, then the new matrix would be $$A_H = \left[\begin{matrix}A& 0 \\\ 0& 1 \end{matrix}\right].$$ In your example $$A = \left[\begin{matrix}1& 0 & 0 \\\ 0& \cos\theta &-\sin\theta\\\ 0& \sin\theta &\cos\theta \end{matrix}\right],$$ so $$A_H = \left[\begin{matrix}1& 0 & 0 & 0 \\\ 0& \cos\theta &-\sin\theta & 0 \\\ 0& \sin\theta &\cos\theta & 0 \\\ 0 & 0 & 0 & 1 \end{matrix}\right].$$

Hope that helps ;-)

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Thanks , it helps, how about y and z? – Leo Chan Apr 21 '12 at 10:39
Is my try correct ? Thank you........... – Leo Chan Apr 21 '12 at 10:42
Yes, it is, as long you know the difference between $0$ (zero) and $\theta$ (theta). For more information try en.wikipedia.org/wiki/Rotation_matrix and en.wikipedia.org/wiki/Transformation_matrix, actually there is a lot on this on the web, GIYF. – dtldarek Apr 21 '12 at 10:44
Thanks for your help – Leo Chan Apr 21 '12 at 11:06

Let the rotation axis be a line through a fixed point $(x_0,y_0,z_0,1)^T$ and an infinite point (direction of the line) $(a,b,c,0)^T$, without loss of generality, we assume $a^2+b^2+c^2=1$. Use right-handed rule for rotation, then a general homogeneous rotation with angle $\theta$ can be obtained as:

$$\boldsymbol{R}^{3D}\left(x_0,y_0,z_0,a,b,c,\theta\right)=\mathscr{C}_1+\left(\sin\theta\cdot\mathscr{A}_2- \left(1-\cos\theta\right)\cdot\mathscr{O}_3\right)\cdot \mathscr{T}_4$$

where: $$\mathscr{C}_1= \left[\begin{array}{*{20}{c}} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&2-\cos\theta \end{array}\right]$$

$$\mathscr{A}_2 = {\begin{array}{c} \underbrace{ \begin{array}{c} {{\left[ {\begin{array}{*{20}{c}} {\color{blue}0}&{\color{blue}-c}&{\color{blue}b}&0\\ {\color{blue}c}&{\color{blue}0}&{\color{blue}-a}&0\\ {\color{blue}-b}&{\color{blue}a}&{\color{blue}0}&0\\ 0&0&0&0 \end{array}} \right]}} \end{array} } \\ \text{Antisymmetric matrix} \end{array}}$$

$$\quad\mathscr{O}_3= \begin{array}{c} \underbrace{ {I - {\left[ {\begin{array}{*{20}{c}} a\\ b\\ c\\ 0 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} a&b&c&0 \end{array}} \right]}} }\\ \text{Orthographic parallel projection} \end{array}$$

$$\mathscr{T}_4=\begin{array}{c} \underbrace{\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - {x_0}}\\ 0&1&0&{ - {y_0}}\\ 0&0&1&{ - {z_0}}\\ 0&0&0&1 \end{array}} \right]}\\ \text{Translation}\end{array}$$

The L-C formulation of homogeneous 3D rotation is similar to Rodrigues' but they are not the same in essence. Details are available here: A submission on homogeneous rotation to arXiv.org

For your case, substitute $x_0=y_0=z_0=0$, $a=b=0$, $c=1$ and $\theta=\dfrac{\pi}{2}$ into the L-C formulation then you can obtain the desired homogeneous rotation matrix.

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