Projective geometry problems

[Mondo]

Hello,

for the figure below I try to justify the reflection mapping. They say that $e_1 -> T(e_1)$ is described by a mapping $$\begin{bmatrix} cos(2\theta)\\ sin(2\theta) \end{bmatrix}$$. I can get to the same result by drawing a parpendicular from $T(e_1)$ to the X axis and then write simple trigonometry formula for cosine $$\frac{x}{r} = cos(2\theta) -> x = rcos(2\theta)$$ where 'r' is the length of the vector (here r = 1 since $e_1$ is a basis vector). Similarly for the sine function I get the expected transformation. However I can't get the tranformation for $e_2$. Since the angle is obtuse I was thinking about first writing the transformation of e2 on the X axis and lateer from X to T(e2). But since we don't know the angles I have no idea how to do this.



Thanks!

 

[LCKurtz]

I will help you with the second part. Consider the following figure:

A is the reflection of B in the line OP and you need its coordinates (x,y). Notice that angle OBA and angle COP have perpendicular sides which is why the angle at B is $\theta$ and so is the angle at A. From triangle OBP you have the length of BP is $\cos\theta$ and so is the length of PA. So the length of BA is $2\cos\theta$ and from triangle ABR you have $x=2\cos\theta \sin\theta = \sin(2\theta)$. Can you see how to get y from the same triangle?

 

[Mondo]

Thanks for the answer LCKurtz.

Can you see how to get y from the same triangle?

Yes in fact I have already solved that but in a different way than you do. I can follow the reasoning of your solution but have one questions - how are you sure that the angle at P is a right angle?

 

[pka]

I have one questions - how are you sure that the angle at P is a right angle?

If you know that $A$ is the reflection of $B$ through $P$ then $\angle OPB$ is a right angle.

A is the reflection of B in the line OP and you need its coordinates (x,y)

 

[Mondo]

Yep, that's is obvious... thanks