Angular momentum in movement along a straight line Angular momentum is conserved when no external torque is applied, I've learned that a long time ago and know the derivation. Yet, I've now been wondering about the following case:
Let's consider a (classical) particle moving along a straight line with constant velocity $\vec{v}$, a mass $m$, and no external forces. Now, consider the angular momentum with respect to a point $O$ which is not situated along the particle's line of movement, like in the picture below.

then $\vec{r}$ is changing with time but $\vec{v}$ is not. So the angular momentum $L=\vec{r}\times \vec{v}$ will also change with time. How can this be possible, and what did I miss?
 A: $L=\vec{r}\times \vec{v} = rv\sin\theta \times\vec{n}$
(where $r$ and $v$ are the magnitudes, $\theta$ is the angle between $\vec{v}$ and $\vec{r}$ and $\vec{n}$ is a unit vector perpendicular to the plane in which $\vec{r}$ and $\vec{v}$ lie)
$v$ is constant but $r\sin\theta$ is also constant
It's the distance between $O$ and where the dotted line crosses the $y$ axis
A: You are missing the property ${\bf v}\times {\bf v}={\bf 0}$ of the vector product. So if ${\bf r}\to {\bf r}+ {\bf v}t$ the angular momentum ${\bf  L} \propto {\bf r}\times {\bf v}$does not change.
A: We have that the angular momentum is $$\vec L = m \vec r \times \vec v$$ where $\vec r = \vec r_0 + \vec v t$ (since no external force is applied, the object simply continues with its given velocity by Newton's first Law, $\vec r_0$ is some arbitrary position, $\vec v$ the velocity, and $t$ the time).
Therefore, $$\begin{align} \vec L &= m\left( \vec r_0 + \vec v t \right)\times\vec v \\  &= m \vec r_0 \times \vec v+\vec v \times \vec vt\end{align}$$
and since $\vec v \times \vec v = 0$, we have that $$\vec L = m\vec r_0\times\vec v$$ is a conserved quantity.
A: The vectors $\vec{r}$ and $\vec{v}$ span an ever changing parallelogram in the $xz$ plane (referring to the image in your question). However, the area of that parallelogram stays fixed, because the base of the parallelogram is always $|\vec{v}|$, and the height is also always constant (equal to the distance between $O$ and the line, or the absolute value of the $z$ coordinate that defines the line). This area is constant and equal to the magnitude of the cross product. It is also easy to see that the direction of the cross product is fixed as the particle moves along the line.
So the angular momentum is in fact conserved.
