# Transforation geometry -- using matrices!

• Feb 22nd 2008, 04:52 AM
struck
Transforation geometry -- using matrices!
This is a question from GCSE level, so please offer a simple explanation if possible. :)

I have been given the following:

To rotate a point (x,y) 90deg anti-clockwise, pre-multiply the x y matrix with:

Code:

( 0 -1 )   1  0
Similarly, for a an anti-clockwise rotation of 180deg, it's:
Code:

-1  0 0  -1
p.s., how can I create a matrix in this forum in the math tag?

It's tough to remember so memorize so many matrices when there are very similar looking matrices for reflection (on x axis, y axis and y = -x). I think perhaps, an explanation of how these matrices are created will help me. Or maybe some pointers on how to memorize at least 8-9 such matrices.

Thanks ...
• Feb 22nd 2008, 05:11 AM
mr fantastic
Quote:

Originally Posted by struck
This is a question from GCSE level, so please offer a simple explanation if possible. :)

I have been given the following:

To rotate a point (x,y) 90deg anti-clockwise, pre-multiply the x y matrix with:

Code:

( 0 -1 )   1  0
Similarly, for a an anti-clockwise rotation of 180deg, it's:
Code:

-1  0 0  -1
p.s., how can I create a matrix in this forum in the math tag?

It's tough to remember so memorize so many matrices when there are very similar looking matrices for reflection (on x axis, y axis and y = -x). I think perhaps, an explanation of how these matrices are created will help me. Or maybe some pointers on how to memorize at least 8-9 such matrices.

Thanks ...

The matrix that rotates by an anti-clockwise angle of $\, \theta \,$ is

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

p.s. The latex code I used for generating this matrix is

$$\left( \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right)$$
• Mar 18th 2008, 02:41 AM
max
I have found plenty to help me with counter clockwise rotations but when its clockwise the values i get for theta when i do arc cos and arc sin are both different, can you tell me why please?

Max
• Mar 18th 2008, 03:00 AM
mr fantastic
Quote:

Originally Posted by max
I have found plenty to help me with counter clockwise rotations but when its clockwise the values i get for theta when i do arc cos and arc sin are both different, can you tell me why please?

Max

Give an example.

To rotate clockwise, all you do is replace $\theta$ with $- \theta$ in the rotation matrix.
• Mar 18th 2008, 03:30 AM
max
Find theta and hence the angle of rotation and the direction of

top row of matrix (-1/2) (1/2)
bottom row (-1/2) (-1/2)

Clockwise direction but
when you calculate arccos (-1/2) you get 135 deg which is correct but
arcsin (-1/2) is -45 deg

What am I missing?

Max

matrix attached
• Mar 18th 2008, 05:12 AM
mr fantastic
Quote:

Originally Posted by max
Find theta and hence the angle of rotation and the direction of

top row of matrix (-1/2) (1/2)
bottom row (-1/2) (-1/2)

Clockwise direction but
when you calculate arccos (-1/2) you get 135 deg which is correct but
arcsin (-1/2) is -45 deg

What am I missing?

Max

matrix attached

$\cos \theta$ and $\sin \theta$ are both negative in the third quadrant.

So $\theta = 225^0$ or $\theta = - 135^0$.
• Mar 18th 2008, 05:51 AM
max
Thank you for your reply and I understand where the -135 deg comes from. Its negative due to being a clockwise rotation about the origin (anti clockwise it is 225 deg as you say), where I am getting lost is why do I get -45 deg for calculating arcsin (-1/√2) ?
• Mar 18th 2008, 05:59 AM
mr fantastic
Quote:

Originally Posted by max
Thank you for your reply and I understand where the -135 deg comes from. Its negative due to being a clockwise rotation about the origin (anti clockwise it is 225 deg as you say), where I am getting lost is why do I get -45 deg for calculating arcsin (-1/√2) ?

That's the correct answer, but to the wrong question!

The correct question is:

What value(s) of $\theta$ simultaneously satisfy

$\cos \theta = -\frac{1}{\sqrt{2}}$ .... (1)

$\sin \theta = -\frac{1}{\sqrt{2}}$ .... (2)
• Mar 18th 2008, 06:03 AM
max
Ahh okay thank you that actually makes it so much easier to understand.(Yes)

Max