equation for cube?

Question

this question is quite obvious i was reading about writing the equation for cube on wikipedia its given by

In analytic geometry, a cube's surface with center $(x_0, y_0, z_0)$ and edge length of $2a$ is the locus of all points $(x, y, z)$ such that

$$ \max \lbrace \lvert x-x_0\rvert,\lvert y-y_0\rvert,\lvert z-z_0\rvert\rbrace =a.$$

what does $\max\{|x-x_{0}|,|y-y_{0}|,|z-z_{0}|\}=a.$ means does it mean theat maximum value among all of the three $x,y,z$

Answer 1

For real numbers $u$, $v$, $w$ the function $(u,v,w)\mapsto \max\{u,v,w\}$ gives the largest occurring value among the three. A condition of the form $\max\{u,v,w\}=a$ means that at least one of the three is $=a$, and that none is $>a$. If a point $P_0=(x_0,y_0,z_0)\in{\mathbb R}^3$ and a number $a>0$ are given the condition $$\max\bigl\{|x-x_0|, \>|y-y_0|, \>|z-z_0|\bigr\}=a$$ then says that a point $(x,y,z)\in{\mathbb R}^3$ belongs to the set in question (namely the surface $\partial C$ of the cube $C$ with center $P_0$ and side length $2a$) if at least one of the unsigned coordinate differences $|x-x_0|$, $|y-y_0|$, $|z-z_0|$ is $=a$, and none is $>a$. Geometric intuition should tell you that this is indeed the description of the set $\partial C$.

Answer 2

$\max\{\lvert x-x_0\rvert,\lvert y-y_0\rvert,\lvert z-z_0\rvert\}$ does indeed mean that you take the greatest value among the three values $\lvert x-x_0\rvert$, $\lvert y-y_0\rvert$, and $\lvert z-z_0\rvert$.

The set of points such that $\max\{\lvert x-x_0\rvert,\lvert y-y_0\rvert,\lvert z-z_0\rvert\} = a$ includes the eight vertices \begin{align} &(x_0 - a, y_0 - a, z_0 - a),\\ &(x_0 - a, y_0 - a, z_0 + a),\\ &(x_0 - a, y_0 + a, z_0 - a),\\ &(x_0 - a, y_0 + a, z_0 + a),\\ &(x_0 + a, y_0 - a, z_0 - a),\\ &(x_0 + a, y_0 - a, z_0 + a),\\ &(x_0 + a, y_0 + a, z_0 - a),\\ \text{and }&(x_0 + a, y_0 + a, z_0 + a). \end{align}

It also includes the faces and edges of the cube with those vertices.

For example, all the points $(x,y,z)$ where $x = x_0 + a$, $y_0 - a < y < y_0 + a$, and $z_0 - a < z < z_0 + a$ satisfy $\max\{\lvert x-x_0\rvert,\lvert y-y_0\rvert,\lvert z-z_0\rvert\} = a$ because $\lvert x-x_0\rvert = a$, $\lvert y-y_0\rvert < a$, and $\lvert z-z_0\rvert < a$; that is, $a$ is the greatest of the three values. Those points are the points on the face with vertices $(x_0 + a, y_0 - a, z_0 - a)$, $(x_0 + a, y_0 - a, z_0 + a)$, $(x_0 + a, y_0 + a, z_0 - a)$, and $(x_0 + a, y_0 + a, z_0 + a)$.

The points $(x,y,z)$ such that $x = x_0 + a$, $y = y_0 - a$, and $z_0 - a < z < z_0 + a$ are the points along the edge connecting the vertices $(x_0 + a, y_0 - a, z_0 - a)$ and $(x_0 + a, y_0 - a, z_0 + a)$.

Other cases ($x=x_0 - a$, $y=y_0 + a$, $z = z_0 - a$, etc.) give the other faces and edges of the cube.