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I was reading that the vectors must have magnitude of $1$ to be called normalized; however, none of the problems I solved gave me a vector magnitude of $1$ after normalizing. I also thought that taking the norm of a vector produced the unit vector on the same line with that vector which also equals to $1$. What am I getting wrong here?

Here’s the source

Note: I‘ve checked the other answers, but it didn‘t seem to click. Hope you can enlighten me.

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2 Answers

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Indeed, by definition, a normalized vector is one with a magnitude of 1. Also by definition, taking the norm of a vector is actually taking the root of its scalar product (for a vector $v$, $|v| := \sqrt{\langle v,v\rangle}$). Thus, it is false to say that "taking the norm of a vector produc[e] the unit vector on the same line with that vector" : we clearly see that by definition, taking the norm of a vector $v$ output a scalar and not an other vector.

To obtain the vector $v_n$ of magnitude 1 that is "on the same line" as $v$ (the "normalized vector of $v$") we can simply take $\frac{v}{|v|}$ wich is what is done one the piece of mathematics that you have circled in red.

(you could easily prove that fact in $\mathbb{R^3}$ by letting $v := (a,b,c) \in \mathbb{R^3}$, calculating $\frac{v}{|v|}$ and taking the norm of $\frac{v}{|v|}$. You would see that it gives 1 for any value of $a$, $b$ and $c$)

Best of luck in your Linear Algebra journey,

LS

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Why, all of the vectors are normalized. The magnitude of a vector is given by

$$ \|\textbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$

The definition of a normalized vector is so that $\|\textbf{v}\|$, its magnitude is $1$.

Also, the way to normalize a vector is to scale the vector down by its magnitude. Let $\textbf{v}^\prime$ (bad notation, I don't know how to make the caret symbol in top) be the normalized vector. Then

$$ \textbf{v}^\prime = \frac{\textbf{v}}{\|\textbf{v}\|} $$

Reason why they are normalized:

$$ \sqrt{(1/\sqrt{2})^2 + 0^2 +(1/\sqrt{2})^2} = 1$$

$$ \sqrt{(1/2)^2 + (\sqrt{2}/2)^2 +(1/2)^2} = 1$$

$$ \sqrt{(1/2)^2 + (-\sqrt{2}/2)^2 +(1/2)^2} = 1$$

There is nothing fancy here. You can easily check that it is equal to $1$.

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