Vector Analysis & Linear Algebra

Question

I'm given a positive number, a unit vector $u \in \mathbb{R} ^n $ and a sequence of vectors $ \{ b_k \} _{ k \geq 1} $ such that $|b_k - ku| \leq d $ for every $ k=1,2,...$.

This obviously implies $ |b_k| \to \infty $ . But why does this imply $\angle (u,b_k) \to 0 $ ? I've tried proving it using some inner-product calculations, but without any success.

In addition, why the given data implies that there must exist $i<j$ such that $|b_i| \leq \frac{\delta}{4} |b_j| $ ?

Thanks a lot !

Answer 1

You have $|b_k| \leq |b_k -k u| + k=d+k$, and $|b_k| \geq k-|b_k -k u|=k-d$. This gives: $$\lim_{k \to \infty} \frac{\langle u, b_k-ku\rangle}{|b_k|} = 0, \;\; \lim_{k \to \infty} \frac{\langle u, ku\rangle}{|b_k|} = 1,$$ so it follows that $$\lim_{k \to \infty} \frac{\langle u, b_k \rangle}{|b_k|} = \lim_{k \to \infty} \frac{\langle u, b_k -ku\rangle + \langle u, ku\rangle}{|b_k|} = 1.$$

Answer 2

From $|b_k-k\,u|\le d$ you get that for all sufficiently large $k$ $$ k-d\le |b_k|\le k+d $$ and $$ k-d\le u\cdot b_k\le k+d. $$ Then $$ \frac{k-d}{k+d}\le\cos(\angle(u,b_k))\le\frac{k+d}{k-d}. $$ It follows that $\cos(\angle(u,b_k))\to1$ and $\angle(u,b_k)\to0$