Suppose we have a closed ball $S$ of radius $r$ centered at the origin $O$ of $\mathbb{R}^3$. Suppose further that we are given a point $v$ on the ball's boundary (i.e., $\|v\|=r$), and a point $u$ in the ball's interior (i.e., $\|u\|<r$). We assume that $v$ and $u$ are not too far from each other, so $\|v-u\| \leq r$. Let $\theta$ denote the angle between the vectors $\overrightarrow{vO}$ and $\overrightarrow{vu}$.
Is it true that the ball $S'$ of radius $\|v-u\| \cos \theta$ centered at $u$ is contained in $S$?
And if not, is there any (interesting) radius of the form $\|v-u\|f(\theta)$ which works?
This arose in the context of an (elementary) differential geometry problem. We're asked to prove that the osculating circles of a curve $\mathbf{r}(t)$ (with certain hypotheses on the curve) on some sphere $S$ centered at the origin must be contained within sphere. The above claim is one avenue of attack I'm attempting, but it also makes for an interesting geometric problem in and of itself.
The intuition seems fairly clear. If the angle $\theta$ is large, there is much little wiggle room for the ball containing $u$, and $\cos \theta$ comes off as a natural factor. Also, the extreme cases $\theta=0$ or $\theta=\frac{\pi}{2}$ seem to be okay.
No, this is not true, since $\|v-u\|\cos\theta$ doesn’t go to zero as $u$ approaches the boundary on any given ray from the origin (except for the extreme ones that you checked).