Linear Algebra. Norm. If $\|v\| = 2$ and $\|w\| = 3$  , what are the largest and smallest values possible for $\|v-w\|$  ? Give a geometric explanation of your results.
 A: Triangular and inverse triangular inequalities are fundamental inequalities that extend to normed vector spaces a simple properties of triangles: 

given a triangle with sides $a,b,c$ than any side is less than the sum
  of the other two.

It is not difficult to see that this can be expressed as $|a-b|\le c\le a+b$. 
In a vector space, given two vectors $\vec v$ and $\vec w$ we have a ''triangle'' with sides $\vec v$, $\vec w$ and $\vec w-\vec v$ (or $\vec v-\vec w$) and, if it is a normed space, the length of its sides is $||\vec v||$, $||\vec w||$ and $||\vec v-\vec w||$.
So, for $a=||\vec v||$, $b=||\vec w||$ and $c=||\vec v-\vec w||$ we have:
$$
\left|\,||\vec v||-||\vec w||\;\right|\le ||\vec v -\vec w|| \le ||\vec v||+||\vec w||
$$
This is the intuition behind the formal definition of a norm, and  the statement can be formally deduced from the axioms that define a norm function. ( see, e.g. : http://en.wikipedia.org/wiki/Triangle_inequality#Reverse_triangle_inequality.
So, in your case you have $1\le ||\vec v -\vec w|| \le 5 $.
