On the magnitude of vectors Imagine a vector a in three dimensional Cartesian coordinates, the vector's endpoint coordinates are ($X_a,Y_a,Z_a$). Now let's assume this vector is in standard form (the vector "begins" at the origin).
The formula for the distance between the endpoint and the origin in Cartesian three-space is as follows where d represents distance and X,Y,Z are the coordinates.
$d^2=X^2+Y^2+Z^2$
With that said is "d" equal to the magnitude of the vector, since the magnitude of a vector is represented by the length of the line segment?
 A: Yes, the magnitude of a vector $\vec v$ is often called the norm of a vector: $|\vec v|$, (more technically, the $L_2$ norm), and for a vector $\vec v \in \mathbb R^3$, $\vec v = \langle x, y, z\rangle$, $\;|\vec v| = \sqrt{x^2 + y^2 + z^2}.\;$ 
This corresponds with the Euclidean distance $d$ of a point $(x, y, z)$ in $\mathbb R^3$, measured from the origin, where $d = \sqrt{x^2 + y^2 + z^2}$
So indeed, the magnitude of such a vector $|\vec v| = d$. 
A: Here is a (ugly) drawing:

The $\color{blue}{base}$ of the triangle can be viewed as in the $xy$-plane and using Pythagoras' Theorem we get the length  $L_B = \sqrt{x^2 + y^2}$. The $\color{green}{height}$ of the triangle has length $L_H = \sqrt{z^2}$. Therefore, using Pythagoras' theorem, the square of the length of $\color{red}{A = (x,y,z)}$ is $$L_A^2= L_B^2 + L_H^2 = \left(\sqrt{x^2 + y^2}\right)^2 + \left(\sqrt{z^2}\right)^2 = x^2 + y^2 + z^2 = D^2.$$
Hope this helps!
A: Yes you've got it right, d is essentially the distance from beginning of the vector to its end point, making it the magnitude of the vector in question
