Why the norm of this vector is 1?

Question

let $f\in C^1(B\subseteq\mathbb{R}^n,\mathbb{R}^m)$, $x,x_0$ in the open B and $v\in\mathbb{R}^m$ if we choose $v=\frac{f(x)-f(x_0)}{||f(x)-f(x_0)||}$ then $||v||=1$ why?

Answer 1

Because$$\left\lVert\frac{f(x)-f(x_0)}{\lVert f(x)-f(x_0)\rVert}\right\rVert=\frac1{\lVert f(x)-f(x_0)\rVert}\lVert f(x)-f(x_0)\rVert=1.$$