Re: Help to find a solution

One way to do this:

1) translate so that R is the origin by subtracting R from each vector.

P becomes $\displaystyle (1- (7^{1/2}- 5), 0- 2)= (6- \sqrt{7}, -2)$ and Q becomes $\displaystyle (7- (7^{1/2}- 5),-2)= (12- \sqrt{7}, -2)$

2) Rotate through by 2.3 radians by multiplying those vectors by the matrix $\displaystyle \begin{pmatrix}cos(2.3) & -sin(2.3) \\ sin(2.3) & cos(2.3)\end{pmatrix}$.

3) Translate **back** by adding R to each vector.

As for "homogeneous coordinates" do you know what they are? We represent the point (x, y) as the three-vector (x, y, 1) with the understanding that if, after some transformation, the last component is not 1, we divide through by it. That is, (x, y, a) represents the same point as (x/a, y/a, 1). In homogeneous coordinates, a translation of (x, y) by (a, b) is given by the matrix multiplication $\displaystyle \begin{pmatrix}1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ 1\end{pmatrix}= \begin{pmatrix}x+ a \\ y+ b\\ 1\end{pmatrix}$ while rotation through angle $\displaystyle \theta$ is given by the matrix multiplication $\displaystyle \begin{pmatrix}cos(\theta) & -sin(\theta) & 0 \\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ 1\end{pmatrix}= \begin{pmatrix}x cos(\theta)- y sin(\theta) \\ x sin(\theta)+ y cos(\theta) \\ 1\end{pmatrix}$.

Find the matrices corresponding to the translation of R to the origin, rotation by 2.3 radians, and translation back and multiply them (in the correct order).

Re: Help to find a solution

Hei, tanks for replying. I tried to continue the prosess in har first part and i multiples These vectors to the rotation matrices and Then added R to them, but i did not git the right answer. Can you show how we can do this

The same for the homogenous coordinates. What would be x , y , a and b in this question