Answer
\[
\quad k=0 \quad \text { or } \vec{v}=\overrightarrow{0}
\]
Work Step by Step
Let $\vec{v}$ be a vector in $n$ -space
\[
\left\langle a_{1}, a_{2}, \ldots, a_{n}\right\rangle=\vec{v}
\]
The standard of this vector is
\[
\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}=\|\vec{v}\|
\]
Since $a_{i}^{2} \geq 0$ for all $a_{i} \in \mathbb{R}$, we get that
\[
\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}=0 \text { iff } a_{i}=0 \text { for } i=1,2, \cdots, n
\]
and then
\[
\|\vec{v}\|=0 \Longleftrightarrow \vec{v}=\overrightarrow{0}=\langle 0,0, \ldots, 0\rangle
\]
So:
\[
\|k \vec{v}\|=|k|\|\vec{v}\|=0 \Rightarrow \quad k=0 \quad \text { or } \vec{v}=\overrightarrow{0}
\]
