Magnitude and unit vectors

The magnitude of a vector $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ or $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ is its length:

2D: $|\mathbf{v}| = \sqrt{x^2 + y^2}$

3D: $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$

In i/j/k notation: $|x\mathbf{i} + y\mathbf{j} + z\mathbf{k}| = \sqrt{x^2 + y^2 + z^2}$.

Why Pythagoras works in 3D. Take $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$. First stay on the floor: going 3 along $x$ and 4 along $y$ covers a floor distance of $\sqrt{3^2 + 4^2} = 5$. Now go 12 straight up. "Floor distance" and "up" are at right angles, so use Pythagoras again: $\sqrt{5^2 + 12^2} = \sqrt{3^2 + 4^2 + 12^2} = 13$. Two right-angled triangles, one after the other, give the 3D formula.

The magnitude is a length, so it is never negative, and it is 0 only for the zero vector $\mathbf{0}$ (no move at all).