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

The answer is yes to both. 


*

*If you have a vector $(v_{1},...,v_{n})\in\mathbb{R}^{n}$ and a real number $a\in \mathbb{R}$, then $(av_{1},...,av_{n})=a(v_{1},...,v_{n})$. This is the definition for multiplication of a vector by a real number.

*If $v\in\mathbb{R}^{n}$ and $a\in \mathbb{R}$, then it always holds that $\|av\|=|a|\cdot \|v\|$. In your case of $\mathbb{R}^{3}$ and Euclidean norm, you can see it for example as following. If $v=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$ and $a\in\mathbb{R}$, then 
\begin{equation*}
\|av\|=\sqrt{(av_{1})^{2}+(av_{2})^{2}+(av_{3})^{2}}=\sqrt{a^{2}(v_{1}^{2}+v_{2}^{2}+v_{3}^{2})}=|a|\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}=|a|\cdot\|v\|.
\end{equation*}
