Answer
a) $1=z^{2}+y^{2}+x^{2}$
b) $z^{2}+y^{2}+x^{2} \leq 1$
c) $z^{2}+y^{2}+x^{2}>1$
Work Step by Step
Let $\vec{r}=\langle x, y, z\rangle$ be an arbitrary vector in 3d-space.
a)
If $1=\|\vec{r}\|,$ then
\[
\|\vec{r}\|=1 \Rightarrow \sqrt{y^{2}+x^{2}+x^{2}}=1 \Rightarrow z^{2}+y^{2}+x^{2}=1
\]
This equation corresponds to a sphere with a radius of $1=r$, originally concentrated $ \ {O} (0,0,0) $.
b) In the same way
\[
\|\vec{r}\| \leqslant 1 \Rightarrow z^{2}+y^{2}+x^{2} \leq 1
\]
Because $\|\vec{r}\| \geq 0$ for each vector $\vec{r}$, this inequality corresponds to the set of all points inside the sphere $1=z^{2}+y^{2}+x^{2}$
c)
\[
\|\vec{r}\|>1 \Rightarrow z^{2}+y^{2}+x^{2} \geq 1
\]
This inequality corresponds to the set of all points in 3d space outside of the sphere.
$1=z^{2}+y^{2}+x^{2} .$ In other words, it corresponds to the set of all points that are at a distance $d>1$ from the original.
