# How to extract an equation from transformation matrix multiplication?

I am trying to rotate a point in a 3D space in the 3 axis together around a specific origin point.

Unfortunately I can't use matrices in my application,All I can do is just the basic math operations (including the trigonometrical functions).

So Instead of the normal transformation matrix,I want a normal function that has 3 angles (x,y,z) and the original location of the vertex and the origin point location as inputs and return the new point location.

Is it possible to do that without using matrices?

• Can't you just write out the matrices, apply them to an arbitrary vector $(x,y)^T$ and then use the resulting equation? Jan 15 '16 at 19:38
• I can do this,But I want to see if there was an existed function to do it,as I don't know how to implement the origin point portion on my matrix. Jan 15 '16 at 19:42
• Any specific reason why matrices are out? Are you programming in a language that does not have arrays? Jan 15 '16 at 19:47
• No,I am an artist,I work with node based system,call it a visual programming language,We can't do arrays there. Jan 15 '16 at 19:49
• To expand on my previous comment, it is possible to change the origin with matrices, in which case you can obtain a general formula by multiplying all the matrices out with a general vector. I have posted an answer that elaborates on this. Jan 15 '16 at 20:06

The trick is to write out your matrices in homogeneous coordinates (with an extra column to allow for affine transformations i.e., translation matrices). Letting the point you want to rotate about equal $(a,b,c)$ gives:

$$R_x(\theta) = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos\theta & -\sin\theta & 0 \\ 0 & \sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$R_y(\theta) = \begin{bmatrix} \cos\theta & 0 & \sin\theta & 0\\ 0 & 1 & 0 & 0 \\ -\sin\theta & 0 & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$R_z(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta & 0 & 0\\ \sin\theta & \cos\theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$T_1(a,b,c) = \begin{bmatrix} 1 & 0 & 0 & -a\\ 0 & 1 & 0 & -b\\ 0 & 0 & 1 & -c\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

$$T_2(a,b,c) = \begin{bmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 0 & b\\ 0 & 0 & 1 & c\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Now applying these matrices as follows to a general vector $v = (x,y,z,1)^T$ gives a general formula:

$$T_2(a,b,c)R_z(\theta)R_y(\theta)R_x(\theta)T_1(a,b,c)v.$$

Afterwards, the fourth entry of the vector can be dropped. The logic behind this is that you first translate $(a,b,c)$ to be at the origin, then you rotate, then you translate $(a,b,c)$ to be back to where it was originally.