Change of angle between two skew lines during individual rotation Given two skew lines, $\mathbf{x}_1 + \mathbf{u}_1 t$ and $\mathbf{x}_2 + \mathbf{u}_2 t$ where $\mathbf{u}_1$ and $\mathbf{u}_2$ are unit vectors along the lines, we might be able to rotate each line with angular velocity $\mathbf{w}_1$ and $\mathbf{w}_2$. I am interested in calculating the change of angle between the two lines due to the rotations. The sine of the angle should be $| \mathbf{u}_1 \times \mathbf{u}_2 |$. For simplicity, I dealt with $(\mathbf{u}_1 \times \mathbf{u}_2)^2 = (\mathbf{u}_1 \times \mathbf{u}_2) \cdot (\mathbf{u}_1 \times \mathbf{u}_2)$. Diferentiating this expression with respect to $t$, I get
$$
\frac{d}{dt}(\mathbf{u}_1 \times \mathbf{u}_2)^2 = 2(\mathbf{u}_1 \times \mathbf{u}_2)\cdot (\frac{d\mathbf{u}_1}{dt}\times \mathbf{u}_2 + \mathbf{u}_1 \times \frac{d\mathbf{u}_2}{dt}).
$$
Using $\frac{d \mathbf{u}}{d t} = \mathbf{w} \times \mathbf{u}$, the above becomes
$$
\frac{d}{dt}(\mathbf{u}_1 \times \mathbf{u}_2)^2 = 2(\mathbf{u}_1 \times \mathbf{u}_2)\cdot ((\mathbf{w}_1\times \mathbf{u}_1)\times \mathbf{u}_2 + \mathbf{u}_1 \times (\mathbf{w}_2 \times \mathbf{u}_2)).
$$
Now, using $(\mathbf{A}\times\mathbf{B})\times\mathbf{C} = -(\mathbf{C}\cdot\mathbf{B})\mathbf{A} + (\mathbf{C}\cdot\mathbf{A})\mathbf{B}$ and $\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = (\mathbf{A}\cdot\mathbf{C})\mathbf{B} - (\mathbf{A}\cdot\mathbf{B})\mathbf{C}$, the vector triple product terms become
$$
(\mathbf{w}_1 \times \mathbf{u}_1) \times \mathbf{u}_2 = -(\mathbf{u}_2\cdot \mathbf{u}_1)\mathbf{w}_1 + (\mathbf{u}_2 \cdot \mathbf{w}_1) \mathbf{u}_1 \\
\mathbf{u}_1 \times (\mathbf{w}_2 \times \mathbf{u}_2) = (\mathbf{u}_1\cdot \mathbf{u}_2)\mathbf{w}_2 - (\mathbf{u}_1 \cdot \mathbf{w}_2) \mathbf{u}_2
$$
then, the above differentiation becomes quite a simple expression:
$$
\frac{d}{dt}(\mathbf{u}_1 \times \mathbf{u}_2)^2 = 2(\mathbf{u}_1 \cdot \mathbf{u}_2)(\mathbf{w}_2 - \mathbf{w}_1)\cdot(\mathbf{u}_1 \times \mathbf{u}_2).
$$
because $\mathbf{u}_i \cdot (\mathbf{u}_1 \times \mathbf{u}_2) = 0$ with $i = 1,2$.
The interpretation of this result is tricky for me because of $\mathbf{u}_1 \cdot \mathbf{u}_2$ term. Does it really mean that if $\mathbf{u}_1$ and $\mathbf{u}_2$ were orthogonal initially, the derivative of its cross product is zero regardless of how each line rotates. Or were there error in my calculation so far?
 A: You derivation is correct but the function you're differentiating is the square of the sine of the angle, i.e.
$f(t) = \sin^2 \theta(t) $
So when you differentiate with respect to $t$, you will get
$ f'(t) = 2 \sin \theta \cos \theta \dfrac{d \theta}{d t } $
And this will be $0$ if $\theta = 90^\circ$ because of the $\cos \theta$ term.
An alternative to your derivation is to start with
$ \cos \theta = u_1 \cdot u_2 $
Then
$ (- \sin \theta ) \dfrac{d \theta} {dt} = u_1' \cdot u_2  + u_2' \cdot u_1 $
Since $u_1' = \omega_1 \times u_1 $ and $u_2' = \omega_2 \times u_2 $ , then
$ (- \sin \theta ) \dfrac{d \theta} {dt} = (\omega_1 \times u_1) \cdot u_2  + (\omega_2 \times u_2) \cdot u_1 $
Now, using $ a \cdot (b \times c) = c \cdot ( a \times b) = b \cdot (c \times a) $, we get
$ (- \sin \theta ) \dfrac{d \theta} {dt} = (\omega_2  - \omega_1 ) \cdot (u_1 \times u_2) $
And this expression shows explicitly that even when the initial angle $\theta = 90^\circ$, then $\dfrac{d \theta}{dt} $ will be different from $0$ (as long as $\omega_2 - \omega_1 \ne 0 $).
If, on the other hand $u_1$ and $u_2$ were parallel initially, then the above equation cannot be used to calculate $\dfrac{d \theta}{dt} $, because we have $0$ on both sides of the equation ( $\sin \theta = 0 $ and $ u_1 \times u_2 = 0$ ).
In this case, we know that $u_1(t)$ and $u_2(t)$ will move on circles defined parametrcially by
$ u_1(t) = u \cos(\omega_1 t) + (\hat{\omega_1} \times u ) \sin(\omega_1 t) $
$ u_2(t) = u \cos(\omega_2 t) + ( \hat{\omega_2} \times u ) \sin(\omega_2 t ) $
Therefore
$ \cos(\theta(t) ) = u_1 \cdot u_2 = \cos(\omega_1 t) \cos(\omega_2 t) + (\hat{\omega_1} \times u ) \cdot ( \hat{\omega_2} \times u ) \sin(\omega_1 t ) \sin(\omega_2 t) $
Now,
$ (a \times c) \cdot (b \times c ) = c \cdot ( (a \times c) \times b )\\
= - c \cdot ( b \times (a \times c) ) = - c \cdot ( a (b.c) - c (a.b) )\\  = c^2 (a.b) - (a.c)(b.c) $
Therefore,
$ (\hat{\omega_1} \times u ) \cdot ( \hat{\omega_2} \times u ) = \hat{\omega_1} \cdot \hat{\omega_2} $
And we now have
$ \cos(\theta(t)) = \cos(\omega_1 t) \cos(\omega_2 t) + (\hat{\omega_1} \cdot \hat{\omega_2}) \sin(\omega_1 t) \sin(\omega_2 t) $
The above equations models the angle $\theta(t)$ as a function of time $t$.
Using Taylor series expansion, you can show that
$ | \dfrac{ d\theta}{dt } | = \| \omega_1 - \omega_2 \| = \sqrt{ (\omega_1 - \omega_2 ) \cdot (\omega_1 - \omega_2 ) } $
The above expression gives the magnitude (the absolute value) of $\dfrac{d \theta}{dt} $ at $t = 0$ in the special case that initially the two vectors $u_1 $ and $u_2$ coincide.
