# Mechanics rotation Question

1. Jan 11, 2014

### Genericcoder

Consider two cartesian coordinate system xyz and x y z that initally concide. The x y z undergoes three successive counterclockwise 45 rotations about the following axes: first, about the fixed z-axis;second, about its own x-axis( which has been now rotated); finally, about its own z-axis ( which has now been rotated);finally, about its own z-axis (which has also been rotated). Find the components of a unit vector X in the xyz coordinate system that points along the direction of the x-axis in rotated x y z system.

Attempted solution

Let matrix A,B,& C denote the 3 rotation such that A denote the rotation 45 about z-axis,B rotation 45 about its x-axis,& C denote the matrix which has been rotated about its own z axis.

A = [cos 45 sin 45 0]
[-sin 45 cos45 0]
[0 0 1 ]

B = [ 1 0 0 ]
[ 0 cos45 -sin45]
[ 0 sin45 cos45]

C = [cos45 sin45 0]
[-sin45 cos45 0]
[0 0 1]

I have them everything right but in my book it has in second B matrix it has
j . k as -sin45 but why is that!!?

2. Jan 11, 2014

### D H

Staff Emeritus
I'm confused about what you are asking. Are you asking why it's then second row in the B matrix that has the negated value rather than the third?

3. Jan 11, 2014

### Genericcoder

Yes in the book its the answer for j.k = -sin(45) instead of sin(45) even though its cos(90 - theta) = sin(theta)
or is their something I am missing?

4. Jan 11, 2014

### haruspex

There's an isomorphism consisting of rotating the three axes, z to x, x to y, y to z. This should map A to B.
Applying that to your A matrix gives
B = [ 1 0 0 ]
[ 0 cos45 sin45]
[ 0 -sin45 cos45]

Edit: on second thoughts, that's a rotation about the original x axis.

Last edited: Jan 12, 2014
5. Jan 11, 2014

### Genericcoder

Ye I can see that coz cos(90 - theta) = sin(theta) thank you haru
And cos(90 + theta) = -sin(theta) so I was wondering do I have the other matrices right for A B and C? @ haru?

6. Jan 11, 2014

### D H

Staff Emeritus
Talking about isomorphisms is perhaps a bit advanced for this Introductory Physics question.

The answer lies in the right hand rule. Set up a right handed coordinate system, with positive rotations indicated by the right hand rule. Now describe what happens with rotations about the various axes.

I'll start with rotation about the +z axis because that's equivalent to the canonical 2D rotation in the x-y plane. Here's a rotation about the z axis looks:

Notice that a positive rotation about the +z axis rotates the +x axis toward the +y axis and rotates the +y axis toward the -x axis:
\begin{aligned} \hat x' &= \phantom{-}\cos \theta\,\hat x + \sin \theta\,\hat y \\ \hat y' &= -\sin\theta\,\hat x + \cos\theta\, \hat y \\ \hat z' &= \phantom{-\sin\theta\,\hat x + \cos\theta\,\hat y} + \hat z \end{aligned}
Writing the above in matrix form,
$$\begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} = \begin{pmatrix} \phantom{-}\cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \,\, \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}$$
That matrix on the right? That's your matrix C.

Next let's look at a rotation about the +x axis:

Here a positive rotation about the +x axis rotates the +y axis toward the +z axis and rotates the +z axis toward the -y axis:
\begin{aligned} \hat y' &= \phantom{-}\cos \theta\,\hat y + \sin \theta\,\hat z \\ \hat z' &= -\sin\theta\,\hat y + \cos\theta\, \hat z \\ \hat x' &= \phantom{-\sin\theta\,\hat y + \cos\theta\,\hat z} + \hat x \end{aligned}
Writing the above in matrix form,
$$\begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \phantom{-}\cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{pmatrix} \,\, \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}$$
That matrix on the right? That's your matrix A.

Finally, let's look at a rotation about the +y axis:

Here a positive rotation about the +y axis rotates the +z axis toward the +x axis and rotates the +x axis toward the -z axis:
\begin{aligned} \hat z' &= \phantom{-}\cos \theta\,\hat z + \sin \theta\,\hat x \\ \hat x' &= -\sin\theta\,\hat z + \cos\theta\, \hat x \\ \hat y' &= \phantom{-\sin\theta\,\hat z + \cos\theta\,\hat x} + \hat y \end{aligned}
Writing the above in matrix form,
$$\begin{pmatrix} \hat x' \\ \hat y' \\ \hat z' \end{pmatrix} = \begin{pmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos \theta \end{pmatrix} \,\, \begin{pmatrix} \hat x \\ \hat y \\ \hat z \end{pmatrix}$$
That matrix on the right? That's your matrix B.

I took the above images from slide 63 of a set of tutorial slides presented at Siggraph 98:
http://www.sdsc.edu/~moreland/courses/Siggraph98/vrml97/slides/mt0000.htm

7. Jan 11, 2014

### Genericcoder

Thanks alot DH that makes sense.

8. Jan 12, 2014

### Genericcoder

I have a question why do we have to multiply the matrix right to left instead of left to right to get the product

for example in my example it is :

CBA to get the total rotated components.