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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?

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  • $\begingroup$ so you want to just confirm that d is also magnitude of vector or you want to proove? $\endgroup$ Commented Jun 6, 2013 at 20:09
  • $\begingroup$ @iostream007 I want a confirmation that i have this correct but proof is always welcome. $\endgroup$
    – Antonio
    Commented Jun 6, 2013 at 21:07

3 Answers 3

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Here is a (ugly) drawing:

enter image description here

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!

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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 $|\vec v| = d$.

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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

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