Addition/Subtraction of vectors regarding norm If we subtract a vector with a certain norm from another one with the same norm what's the norm of the resulting vector?
Since norm is just a length indicator it should remain the same, but I cannot prove that formally.
 A: Note that $\Vert a-b\Vert\neq\Vert a\Vert - \Vert b \Vert$ in general. The difference between two vectors with the same norm is not necessarily the zero vector.
For instance, in the real vector space $(\mathbb{R},+,\cdot)$ with the Euclidean norm (the absolute value), we have
$$4=|2-(-2)|\neq|2|-\vert-2|=0$$
Also, it can be shown that (this is the triangle inequality)
$$\Big\vert\Vert a\Vert-\Vert b\Vert\Big\vert\leqslant\Vert a-b\Vert\leqslant\Vert a\Vert+\Vert b\Vert$$
A: In Euclidean geometry, the norms of vectors is related by the triangle inequality:
$\Vert a + b \Vert \le \Vert a \Vert + \Vert b \Vert$
and
$\Vert a - b \Vert \le \Vert a \Vert + \Vert b \Vert$
and
$\Vert a \Vert - \Vert b \Vert \le \Vert a - b \Vert$.
A: OK, let's take $u,v$ with equal Euclidean norm and let's look whether it's possible that $u-v$ has the same norm as well. I’ll denote by $\langle x,y\rangle$ the inner product. We can compute
$$
\|u-v\|^2=\langle u-v,u-v\rangle=\langle u,u\rangle-2\langle u,v\rangle+\langle v,v\rangle=\|u\|^2-2\langle u,v\rangle+\|v\|^2
$$
If we want $\|u\|=\|v\|=\|u-v\|=a$, the condition becomes
$$
2\langle u,v\rangle=a^2
$$
So you see that this is a very strict condition on the two vectors. For instance, if the two vectors are orthogonal, then $\|u-v\|=\|u\|=\|v\|$ if and only if $u=v=0$.
More generally, if you consider that $\langle u,v\rangle=\|u\|\,\|v\|\cos\alpha$, you see that, for $a\ne0$, the condition becomes $2\cos\alpha=1$: the angle between the two vectors has to be $\pi/3$ or $-\pi/3$, which means an equilateral triangle is involved.
