Geometric consequence of absolute vectors If $| \mathbf{a} + \mathbf{b} | = | \mathbf{a} | + | \mathbf{b} |$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors, what is the geometrical significance of this? 
My first thought was that the vectors $\mathbf{a}$ and $\mathbf{b}$ must be in the first quadrant, only having positive components. 
 A: Let's see what $a+b$ is, geometrically. The dark blue arrows are $a$ and $b$. The orange vector is $a+b$.

Essentially, you just stick one on top of the other. Now, we know that the absolute value of a vector just represents the length; can you see what happens when $|a+b|=|a|+|b|$?
Essentially, they would have to be pointing in the same direction! It has nothing to do with what quadrant they're in, it only says they're the same direction. The converse is also true; if $a$ and $b$ point in the same direction, then $|a+b|=|a|+|b|$ (do you see why?).
A: Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$.  Using properties of the dot product, 
$$|\mathbf a+\mathbf b|=|\mathbf a|+|\mathbf b|$$
$$\implies|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2+2|\mathbf a||\mathbf b|$$
$$\implies(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)=\mathbf a\cdot \mathbf a+\mathbf b\cdot \mathbf b+2|\mathbf a||\mathbf b|.$$ 
But the left side is $\mathbf a\cdot \mathbf a + \mathbf b\cdot \mathbf b + 2\mathbf a\cdot \mathbf b = \mathbf a\cdot \mathbf a +\mathbf b \cdot \mathbf b+2|\mathbf a||\mathbf b|\cos\theta$, so this means $\cos\theta=1$.
A: Squaring of the both sides gives:
$$\vec{a}\vec{b}=|\vec{a}||\vec{b}|.$$
Now, can you understand what happens here? 
A: A norm for which is valid  $$|a+b|=|a|+|b|\Rightarrow \text{$a$ and $b$ are parallel}$$ 
is called strict.  (All norms deriving from an inner product are strict, e.g.)
An example for a non-strict norm is the Manhattan-norm, see https://en.wikipedia.org/wiki/Taxicab_geometry.  Here $$|(1,0)|=|(0,1)|=1\text{ and } |(1,1)|=2,$$ hence $$|(1,0)+(0,1)|=|(1,0)|+|(0,1)|,$$ but the vectors aren't parallel.
