General

Orthogonal transformations a, b and c of the group of orthogonal transformations in 3 dimensions O(3) must uphold the following laws:

a,b∈O(3)\Rightarrow a⋅b∈O(3) completeness.
a⋅(b⋅c)=(a⋅b)⋅c assocativity.
∃e:a⋅e=e⋅a unit.
∃a^{-1} inverse.

Specific

The page tries to calculate what happens to a point after rotation around an arbitrary axis through the origin.

The following variables are used:

R The distance of the point to the origin (the origin is also assumed to be where the axis goes through).

\begin{pmatrix}x_ω \\ y_ω \\ z_ω \end{pmatrix} The axis (direction and angular velocity).

\begin{pmatrix}x \\ y \\ z \end{pmatrix} The location of the point relative to the origin.

Given that the point will only move on a plane perpendicular to the axis, we get:

x⋅x_ω+y⋅y_ω+z⋅z_ω=0

Given that the distance of the point to the origin will stay the same, we get:

x²+y²+z²=R²

Solve the latter equation for z:

z=√{R²-x²-y²}

Put into the first equation:

x⋅x_ω+y⋅y_ω+{√{R²-x²-y²}}⋅z_ω=0
{√{R²-x²-y²}}=÷{-x⋅x_ω-y⋅y_ω}{z_ω}
\pm (R²-x²-y²)=÷{x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω}{z_ω²}
\pm (R²⋅z_ω²-x²⋅z_ω²-y²⋅z_ω²)=x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω

For the positive branch, that is:

R²⋅z_ω²-x²⋅z_ω²-y²⋅z_ω²=x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω
R²⋅z_ω²=x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω+x²⋅z_ω²+y²⋅z_ω²
R²⋅z_ω²=x²⋅(x_ω²+z_ω²)+y²⋅(y_ω²+z_ω²)+2⋅x⋅x_ω⋅y⋅y_ω
x²⋅(x_ω²+z_ω²)+2⋅x⋅x_ω⋅y⋅y_ω+y²⋅(y_ω²+z_ω²)-R²⋅z_ω²=0

FIXME: solved for x, assumed x_ω²+z_ω²≠0:

x=÷{-2⋅x_ω⋅y⋅y_ω \pm √{4⋅x_ω²⋅y²⋅y_ω²-4⋅(x_ω²+z_ω²)⋅(y²⋅(y_ω²+z_ω²)-R²⋅z_ω²)}}{2⋅(x_ω²+z_ω²)}
x=÷{-x_ω⋅y⋅y_ω \pm √{x_ω²⋅y²⋅y_ω²-(x_ω²+z_ω²)⋅(y²⋅(y_ω²+z_ω²)-R²⋅z_ω²)}}{x_ω²+z_ω²}

For the negative branch, that is:

-R²⋅z_ω²+x²⋅z_ω²+y²⋅z_ω²=x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω
-R²⋅z_ω²=x²⋅x_ω²+y²⋅y_ω²+2⋅x⋅x_ω⋅y⋅y_ω-x²⋅z_ω²-y²⋅z_ω²
-R²⋅z_ω²=x²⋅(x_ω²-z_ω²)+y²⋅(y_ω²-z_ω²)+2⋅x⋅x_ω⋅y⋅y_ω
x²⋅(x_ω²-z_ω²)+2⋅x⋅x_ω⋅y⋅y_ω+y²⋅(y_ω²-z_ω²)+R²⋅z_ω²=0

FIXME: solved for x, assumed x_ω²-z_ω²≠0:

x=÷{-2⋅x_ω⋅y⋅y_ω \pm √{4⋅x_ω²⋅y²⋅y_ω²-4⋅(x_ω²-z_ω²)⋅(y²⋅(y_ω²-z_ω²)+R²⋅z_ω²)}}{2⋅(x_ω²-z_ω²)}
x=÷{-x_ω⋅y⋅y_ω \pm √{x_ω²⋅y²⋅y_ω²-(x_ω²-z_ω²)⋅(y²⋅(y_ω²-z_ω²)+R²⋅z_ω²)}}{x_ω²-z_ω²}

Cross Product Solving

v⃗=ω⃗⨯r⃗
r⃗=?

Ooookay...

\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}=\begin{pmatrix} ω_1 \\ ω_2 \\ ω_3 \end{pmatrix}⨯\begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix}

v_1=ω_2⋅r_3-ω_3⋅r_2
v_2=ω_3⋅r_1-ω_1⋅r_3
v_3=ω_1⋅r_2-ω_2⋅r_1

Better do it with a matrix...

\begin{pmatrix} 0 & -ω_3 & ω_2 \\ ω_3 & 0 & ω_1 \\ -ω_2 & ω_1 & 0 \end{pmatrix}⋅\begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix}=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
\begin{pmatrix} 1 & 0 & ÷{ω_1}{ω_3} \\ 0 & 1 & ÷{-ω_2}{ω_3} \\ 0 & 0 & 2⋅÷{ω_2⋅ω_1}{ω_3} \end{pmatrix}⋅\begin{pmatrix} r_1 \\ r_2 \\ r_3 \end{pmatrix}⋅\begin{pmatrix} ÷{v_2}{ω_3} \\ ÷{-v_1}{ω_3} \\ v_3+÷{ω_2⋅v_2+ω_1⋅v_1}{ω_3} \end{pmatrix}
r_3=÷{ω_3}{2⋅ω_1⋅ω_2}⋅(v_3+÷{ω_1⋅v_1+ω_2⋅v_2}{ω_3})
r_2=÷{-v_1}{ω_3}+÷{1}{2⋅ω_1}⋅(v_3+÷{ω_1⋅v_1+ω_2⋅v_2}{ω_3})
r_1=÷{v_2}{ω_3}-÷{1}{2⋅ω_2}⋅(v_3+÷{ω_1⋅v_1+ω_2⋅v_2}{ω_3})

Abbreviate: α:=÷{1}{2}⋅(v_3+÷{ω_1⋅v_1+ω_2⋅v_2}{ω_3}):

r_3=÷{ω_3}{ω_1⋅ω_2}⋅α
r_2=÷{-v_1}{ω_3}+÷{1}{ω_1}⋅α
r_1=÷{v_2}{ω_3}-÷{1}{ω_2}⋅α