# Thread: Reflection and Rotation Matrices

1. ## Reflection and Rotation Matrices

calculate [R][S]

where R is the reflection matrix
[ 4/5 -3/5 ]
[-3/5 -4/5 ]

and S is the reflection matrix
[ -3/5 4/5 ]
[ 4/5 3/5 ]

i found [R][S] to be

[ -24/25 7/25 ]
[ -25/25 -24/25 ]

this is the hard part:
verify that it can be identified as the matrix of a rotation and determine the angle of rotation
try to make a picture illustrating how the composition of these reflections is a rotation

2. Originally Posted by razorfever
calculate [R][S]

where R is the reflection matrix
[ 4/5 -3/5 ]
[-3/5 -4/5 ]

and S is the reflection matrix
[ -3/5 4/5 ]
[ 4/5 3/5 ]

i found [R][S] to be

[ -24/25 7/25 ]
[ -25/25 -24/25 ] You made a typo here

this is the hard part:
verify that it can be identified as the matrix of a rotation and determine the angle of rotation
try to make a picture illustrating how the composition of these reflections is a rotation
The rotation matrix is:

$\left(\begin{array}{cc}\cos(\alpha)&-\sin(\alpha) \\\sin(\alpha) & \cos(\alpha)\end{array}\right)$

1. All your values are between -1 and 1

2. The pattern of your values correspond to the rotation matrix

3. Determine the angle of rotation.

4. Make a sketch of the two reflections. You easily can prove that you get a rotation if the two axes intersect. The angle of rotation is twice as large as the angle between the two axes. Use isosceles triangles to prove this.

3. alright i found the angle of rotation to be 16.26 and i drew the graph and showed that the lines intersect
but how do i use the fact that the lines intersect to show that i get a rotation
and how do you know that the angle of rotation is twice the angle between the lines
and finally what do you mean by use the isoceles triangle to prove this
the only method i know to find the angle between two lines is the dot-product method with cosine

4. Originally Posted by razorfever
alright i found the angle of rotation to be 16.26 and i drew the graph and showed that the lines intersect
but how do i use the fact that the lines intersect to show that i get a rotation
and how do you know that the angle of rotation is twice the angle between the lines
and finally what do you mean by use the isoceles triangle to prove this
the only method i know to find the angle between two lines is the dot-product method with cosine
I've attached the sketch.

1. $|\overline{CP}| = |\overline{CP'}| = |\overline{CP''}|$

Therefore P, P', P'' are located on a circle around C.

2. The triangles PP'C and P'P''C are isoceles.

3. Since P' is the image of P the black axis bisect the angle $\angle(PCP')$. Since P'' is the image of P' the blue axis bisect the angle $\angle(P'CP'')$.

4. I'll leave the rest for you.

5. how did you get the circle on your graph?
my graph has the two lines of reflection which pass through the origin
one line has a positive slope of 1/2
and the other line has a negative slope of 2
how and where do i add the circle?

6. Originally Posted by razorfever
how did you get the circle on your graph?
I took an arbitrary point P of the plane and did the two reflections to demonstrate where the magnitudes of the angles in question came from.

According to #1. of my previous post the original point P and its images P', P'' have the property:
$
|\overline{CP}| = |\overline{CP'}| = |\overline{CP''}|
$

and that's the definition of a circle as a locus of points which have equal distance to a fixed point C.

my graph has the two lines of reflection which pass through the origin
one line has a positive slope of 1/2
and the other line has a negative slope of 2
how and where do i add the circle?
The slopes of the axes are $m_1=\frac12$ and $m_2=-2$

Since $m_1 \cdot m_2 = -1$ the axes are perpendicular and that means the angle of rotation must be 180°. A rotation by an angle of 180° is called reflection over a point(?).

And obviously there must be a mistake in your calculations.