The 3-D Coordinate System — Question 4

Question 4 – Solution

Strategy Translate each geometric distance into coordinates: distance to the $xy$-plane is $|z|$, while distance to the $z$-axis is the radial distance $\sqrt{x^2+y^2}$.

Inequality and boundary We require $$|z|\le\sqrt{x^2+y^2}\quad\Longleftrightarrow\quad \boxed{z^2\le x^2+y^2}.$$ The boundary $z^2=x^2+y^2$ is a double cone with vertex at the origin. The set $S$ is the region on or outside that cone in the radial direction.

Membership For $(1,2,3)$, $z^2=9>1^2+2^2=5$, so it is not in $S$. For $(2,-2,1)$, $z^2=1\le 8$, so it is in $S$.

Symmetry and verification The inequality uses only squares, so changing any coordinate’s sign preserves membership; rotations about the $z$-axis also preserve $x^2+y^2$. On the $xy$-plane, every point satisfies it, while nonzero points on the $z$-axis do not, matching the distance interpretation.