- #1

ForceBoy

- 47

- 6

## Homework Statement

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The problem consists of deriving the matrix for a 3 dimensional rotation.

My approach consisted of constructing an arbitrary vector and rewriting this vector in terms of its magnitude and the angles which define it. Then I increased the angles by some amount each. I would expand the equations then simplify.

My problem comes towards the end of the derivation where I encounter an expression which I don't know how to simplify in a way that will make everything come out nicely and in a way that is useful.

## Homework Equations

The arbitrary vector is as follows:

## \vec v = \begin{pmatrix}x \\ y \\ z \end{pmatrix} = r \begin{pmatrix} \cos \theta \cos \alpha \\ \cos \theta \sin \alpha \\ \sin \theta \end{pmatrix}##

where ## r ## is the magnitude of the vector, ## \theta## the angle the vector forms with the ##\text{xy-plane}## and ## \alpha## the angle the xy projection of the vector forms with the x axis.

The angle addition identities will also be used

## The Attempt at a Solution

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Before the rotation transformation the vector is:

## \vec v = \begin{pmatrix}x \\ y \\ z \end{pmatrix} = r \begin{pmatrix} \cos \theta \cos \alpha \\ \cos \theta \sin \alpha \\ \sin \theta \end{pmatrix}##

After the transformation has been applied:

## \vec v' = \begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} = r \begin{pmatrix} \cos (\theta+ \phi) \cos (\alpha + \beta) \\ \cos (\theta + \phi) \sin (\alpha + \beta) \\ \sin (\theta+\phi) \end{pmatrix}##

Expanding the trig functions:

## \vec v' = r \begin{pmatrix} (\cos \theta \cos \phi - \sin \theta \sin \phi)(\cos \alpha \cos \beta - \sin \alpha \sin\beta) \\ (\cos \theta \cos \phi - \sin \theta \sin \phi)(\sin \alpha \cos \beta + \sin \beta \cos \alpha) \\ \sin \theta \cos \phi +\sin \phi \cos \theta \end{pmatrix} ##

Foiling out the binomials:

##\vec v' = r \begin{pmatrix} \cos \theta \cos \phi \cos \alpha \cos \beta - \cos \theta \cos \phi \sin \alpha \sin \beta - \sin \theta \sin \phi \cos \alpha \cos \beta + \sin \theta \sin \phi \sin \alpha \sin\beta \\ \cos \theta \cos \phi \sin \alpha \cos \beta + \cos \theta \cos \phi \sin \beta \cos \alpha - \sin \theta \sin \phi \sin \alpha \cos \beta - \sin \theta \sin \phi \sin \beta \cos \alpha \\ \sin \theta \cos \phi + \sin \phi \cos \theta \end{pmatrix} ##

Distributing the ## r ## and simplifying:

##\vec v' = \begin{pmatrix} x \cos \phi \cos \beta - y \cos \phi \sin\beta - z \sin \phi (cos \alpha cos \beta + \sin \alpha \sin\beta) \\ y \cos \phi \cos \beta + x \cos \phi \sin \beta - z \sin \phi (\sin \alpha \cos \beta - \sin \beta \cos \alpha) \\ z \cos \phi +r \sin \phi \cos \theta \end{pmatrix} ##

##\vec v' = \begin{pmatrix} x \cos \phi \cos \beta - y \cos \phi \sin\beta - z \sin \phi \cos(\alpha-\beta) \\ x \cos \phi \sin \beta + y \cos \phi \cos \beta + - z \sin \phi \sin(\alpha-\beta) \\ z \cos \phi +r \sin \phi \cos \theta \end{pmatrix} ##

Things would work out nicely here and I could represent ## \vec v'## as a product of ##\vec v## and a matrix. My problem is getting rid of the ugly ## r\sin\phi \cos\theta##. Could you provide a hint as to how I would proceed? Also please point out any errors I might've made with the simplifications. Thanks!

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