Two equivalent definitions of two equivalent vectors In the book "Introduction to linear algebra" by Serge Lang, the author gives the definition of vector equivalence as follows:

If $\vec{AB}$ and $\vec{CD}$ are two located vectors, we shall say that they are equivalent if $B-A=D-C$

To me, it is obvious that to be equivalent means they have the same norm, let's say, for example:
Let two vectors be defined as by the set of four points:
$P=(2, 3, -4); Q=(-1, 3, 5)$ and $A=(-2, 3, -1); B=(-5, 3, 8)$
Then 
$Q-P=(-3, 0, 9)$; $B-A=(-3, 0, 9).$
The norm of these vectors can be computed using the distance formula
$d=\sqrt{(x-x_1)^{2}+(y-y_1)^{2}+(z-z_1)^2}$
So the norm for $\vec{PQ}$ = $\sqrt{(-1-2)^{2}+(3-3)^2+(5+4)^2}=\sqrt{9+0+81}=\sqrt{90}$
The norm for $\vec{AB}=\sqrt{(-5+2)^2+(3-3)^2+(8+1)^2}=\sqrt{9+0+81}=\sqrt{90}$
So are these two definitions equivalent? Can you transform the distance formula into the definition of Serge Lang?
 A: This is not the same. There are many inequivalent vectors with the same norm, say the vectors
$$e_1 = \begin{pmatrix}
1 \\ 0
\end{pmatrix}, \;\; e_2=\begin{pmatrix}
0 \\ 1
\end{pmatrix}$$
in $\mathbb R^2$ have the same norm but are inequivalent, since they point in different directions. (In your setting, you might obtain these vectors by letting $A = (0, 0), B = (1, 0), C = (0, 0), D = (0, 1)$). You really need to check whether all components of $B - A$ and $D- C$ are equal, not just their length.

Edit: To the follow-up question whether Serge Lang's definition is true, I reply the following. Definitions are always true without any justification (that's kind of the definition of definition if you want), but they may not capture what you intuitively associate to the word that is being defined. I don't know what you think equivalent vectors should be, but Serge Lang's definition as far as I understand it captures the following:
Two vectors are equivalent if and only if they have the same length and point in the same direction.
You might be familiar with the idea that adding vectors is the same as taking them and putting the tail of the second vector the the head of the first, and the vector that connects them is the resulting vector. In the same way, subtracting vectors gives us the vector connecting the points indicated in the following picture:
 
Since all unlocated vectors are only ever considered up to moving them around in a parallel way (equivalently, only length and direction are relevant), we have that the located vector $\vec{AB}$ can be transformed to an unlocated vector $B-A$, because then the only information that is left is "how far are $A$ and $B$ apart" and "in which direction do we have to go to get from $A$ to $B$".
In any other books, vectors and points are actually the same. You can transfer between these points of view to think of unlocated vectors just as located vectors whose tail sits at the zero vector $0$. Maybe this helps.
