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I know how to normalize a 2 component vector, but I need to normalize a 3-component vector? If it's the same formula as 2-component vector normalization, then how do I figure out the magnitude of a 3-component vector?

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2 Answers 2

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It's exactly the same idea: given a non-zero vector $\mathbf{x} = [x_1, x_2, x_3]$, its magnitude is $$\Vert\mathbf{x}\Vert = \sqrt{x_1^2 + x_2^2 + x_3^2}.$$

You normalize $\mathbf{x}$ by dividing by this magnitude. The idea extends to any number of dimensions.


Side note: to see why this would be true, imagine the projection $\mathbf{p}$ of $\mathbf{x}$ onto the $x_1,x_2$-plane. The length of that vector is $\Vert \mathbf{p} \Vert = \sqrt{x_1^2+x_2^2}$. Now consider the triangle from the origin with $\mathbf{p}$ as a base and extending up to $\mathbf{x}$: its hypotenuse is $\sqrt{\Vert \mathbf{p} \Vert^2 + x_3^2} = \sqrt{x_1^2 + x_2^2 + x_3^2}$.

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  • $\begingroup$ Quick tip: \Vert looks nicer than ||. $\endgroup$
    – K.defaoite
    Apr 23, 2021 at 19:30
  • $\begingroup$ @K.defaoite I appreciate it! I knew there was a symbol but couldn't remember off the top of my head. $\endgroup$
    – Théophile
    Apr 23, 2021 at 19:38
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In 3-dimesnions, $\mathbf{u} = (u_1, u_2, u_3)$, so its $\frac{\mathbb{u}}{\|\mathbb{u}\|}$ where $\|\mathbf{u}\|$ is $\sqrt{u_{1}^1 + u_{2}^2 + u_{3}^2}$

And more generally for a vector $\mathbf{u} \in \mathbb{R}^n$: $\|\mathbf{u}\|$ is just $\sqrt{u_{1}^1 + \dots + u_{n}^2}$

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