Find the equations of $x-y$, $x-z$ and $z-y$ planes.

Question

Do the $y-x$ and $x-y$ planes have the same equations? I think that the equation of the $x-y$ plane can be $x+y+z=0$ or $x+y+z=4$ or $ax+by+cz=$ any real number and $a,b,c$ are arbitrary real numbers. Am I wrong?

Answer 1

x-y plane is the plane containing the line $x=0;z=0$ and $y=0;z=0$ The plane equation is $$z=0$$ x-y plane and y-x plane have the same equation. $z=0$

Answer 2

Every plane in three-dimensional Cartesian space has an equation of the form $$ ax + by + cz + d = 0,$$ where $a$, $b$, $c$, and $d$ are real numbers. The choice of which numbers to use as the values of $a$, $b$, $c$, and $d$ determines what plane you get.

The $x,y$ plane is the plane that contains the $x$ axis and the $y$ axis. If you let each point $(x,y,z)$ in Cartesian space correspond to a vector $\small\begin{bmatrix} x \\[-.7ex] y \\[-.7ex] z \end{bmatrix}$, the $x,y$ plane consists of any linear combination of the vectors $e_1 = \small\begin{bmatrix} 1 \\[-.7ex] 0 \\[-.7ex] 0 \end{bmatrix}$ and $e_2 = \small\begin{bmatrix} 0 \\[-.7ex] 1 \\[-.7ex] 0 \end{bmatrix}$. In that sense the plane is described by two vectors.

The $x,y$ plane also has an equation of the form $ ax + by + cz + d = 0,$ but for this plane $a = b = d = 0$ and $c$ can be any non-zero real number. We can set $c= 1$, for example, in which case the equation $ ax + by + cz + d = 0$ becomes $$ 0\cdot x + 0 \cdot y + 1 \cdot z + 0 = 0,$$ that is, $$ z = 0 .$$

Another fact about a plane with the equation $ ax + by + cz = d,$ in general, is that the vector $\small\begin{bmatrix} a \\[-.7ex] b \\[-.7ex] c \end{bmatrix}$ is perpendicular to that plane. The plane consists of points corresponding to all vectors $\small\begin{bmatrix} x \\[-.7ex] y \\[-.7ex] z \end{bmatrix}$ such that $$\begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} x \\[-.7ex] y \\[-.7ex] z \end{bmatrix} + d = 0.$$ In the case of the $x,y$ plane, recalling that $a = b = d = 0$ for that plane, the vector $\small\begin{bmatrix} 0 \\[-.7ex] 0 \\[-.7ex] 1 \end{bmatrix}$ is perpendicular to the plane. In that sense there is a single vector defining the plane.