Are the difference of two vectors orthogonal if the angle between the two vectors approaches 0? (Attempted proof) Suppose that $\vec{a}=(x,y), \vec{a`}=(x', y'), \Delta \vec{a} = (x'-x, y'-y), \theta \rightarrow 0$ where $\theta$ is the angle between $\vec{a}$ and $\vec{a'},$ and the magnitudes are equal,  $a=a'$
Is it true that $\vec{a}$ is orthogonal to $\vec{a'}$?  I have been using this fact for nearly a year in several physics based courses, but I have no idea how I would prove it.  Below is my attempt at a proof (dot product):
$\vec{a} \cdot \Delta\vec{a} = x(x'-x) + y(y'-y) \rightarrow 0$  Since $\theta \rightarrow 0 \implies \vec{a'} \rightarrow \vec{a}$
But this seems slightly unsatisfactory, because the dot product doesn't say that two vectors are orthogonal if the dot product is near 0, but only equal to 0 exactly. ??
My second attempt uses the cosine law, for what I would assume be a more vigorous proof, taking $\alpha$ for the angle between $\vec{a}$ and $\Delta\vec{a}$
$a'^2=a^2+\Delta a^2 - 2a\Delta a\cos\alpha \implies \Delta a^2 = 2a\Delta a\cos\alpha \implies \Delta a = 2a\cos \alpha \implies \frac{\Delta a}{2a} = \cos \alpha$ 
However, since $\theta \rightarrow 0 \implies$ $\Delta a \rightarrow 0$ then $\alpha = \arccos(0) = \frac{\pi}{2}$ and therefore the vectors are orthogonal.
Please point out any flaws in my proofs, and give me advice.  Thanks!  Another thing I'm sort of iffy about is the statement $\theta \to 0 \implies \Delta \vec{a} \to 0$. Is this correct? It feels weird to use and wrong, because of it being a vector and all
 A: There are a lot of imprecision in the result you cited. A correct result would be 

Given a vector $\vec a$ in $\mathbb R^2$, let $\vec a'(\theta)$ be a function such that the angle between  $\vec a'(\theta)$ and $\vec a$ is $\theta$, and $\|\vec a'(\theta)\|=\|\vec a\|$ (notice that for every $\theta$ there's a unique vector satisfying these conditions). If you have a succession $\theta_n$ that converges monotonically to $0$, then the angle between $\Delta \vec a(\theta_n)=\vec a - \vec a'(\theta_n)$ and $\vec a$ tends to $\pi/2$ or $-\pi/2$.

In particular it is  false that $\Delta\vec a$ is orthogonal to $\vec a$. It is true if and only if $\theta=0$.
Moreover, is correct to say that $\Delta\vec a\to 0$, but is formally more correct to write $\|\Delta \vec a\|\to 0$.
A: If I try to understand your question before trying to answer it.
You have a vector $\vec a$ that you rotate with an angle $\theta$ to get the vector $\vec a^\prime$. You then denote $\Delta \vec a$. Your question is: is the limit of the angle $\widehat{(\vec a, \Delta \vec a)}$ equal to $\pi/2$ when $\theta \to 0$? Is that your question?
If yes... Then the answer is yes! 
To prove it, just compute the angles of the isoscele triangle built on $\vec a$ and $\vec a^\prime$.
