Norm of vector in $\mathbb{R}^3$ with multiple If we have a vector in $\mathbb{R}^3$ (or any Euclidian space I suppose), say $v = (-3,-6,-9)$, then:


*

*May I always "factor" out a constant from a vector, as in this example like $(-3,-6,-9) = -3(1,2,3) \implies (1,2,3)$ or does the constant always go along with the vector?

*If yes on question 1, then if I want to compute the norm, is the correct computation the following: $||v|| = |-3|\sqrt{14} = 3\sqrt{14}$ ? If so, is the only reason that we take the absolute value of -3 because we don't want a negative length?


I'm sorry if things are obvious but I just want to make sure I actually get this correctly.
Best regards

As a rule, this all falls from the distributive rule. If $v=(ax,ay,az)$, then $$\begin{align}||v|| &= \sqrt{(ax)^2 + (ay)^2+(az)^2} = \sqrt{a^2(x^2+y^2+z^2)}\\
&=\sqrt{a^2}\sqrt{x^2+y^2+z^2}\end{align}$$
And $\sqrt{a^2}=|a|$
And yes, ultimately, this is all because we want distances to be positive, but we also want the norm to be well-defined.If $w=(1,-2,3)$, is $|w|=\sqrt{14}$ or $-\sqrt{14}$?  There is no guidance here to choose one or the other.
