Confusion regarding equational form of plane

I am confused on the equational definition of a plane which states that

The collection of points $(x, y, z)$ in $\mathbb{R}^{3}$ satisfying an equation of the form $$ a x+b y+c z=d, $$ with at least one of the constants $a, b$, or $c$ nonzero, is a plane in $\mathbb{R}^{3}$.

I am confused on the $(x,y,z)$ part. Is this $3$-tuple representing one singular point given by three scalars? Or is it referring to 3 seperate points that define the plane? If so, would these points always be the x, y, and z-intercepts respectively?

Solution 1:

$(x,y,z)$ denotes a single variable point. The definition gives us a necessary and sufficient condition for an arbitrary point $(x,y,z)$ to be in that plane. This becomes clearer with examples.

Consider the equation $2x - 3y + 5z = 0$ (here, we are considering $a = 2$, $b = -3$, $c = 5$). This equation defines one plane. In order to know which points belong to this plane, we evaluate the variables $x$, $y$, $z$ with the coordinates of every point we want to use. Examples:

• Does the point $(1, -1, 2)$ lie in the plane $2x - 3y + 5z = 0$? No, because $2\cdot 1 - 3\cdot(-1) + 5\cdot 2 = 15 \neq 0$. Note that we have used $x = 1$, $y = -1$, and $z = 2$. That is, in this example, we have used $(x,y,z) = (1, -1, 2)$.
• Does the point $(1, -1, -1)$ lie in the plane $2x - 3y + 5z = 0$? Yes, because $2\cdot 1 - 3\cdot(-1) + 5\cdot(-1) = 0$. Here, we have used $x = 1$, $y = -1$, and $z = -1$. That is, in this example, we have used $(x,y,z) = (1, -1, -1)$.

So, $(x,y,z)$ represent every choice you can do for a single point, with coordinates $x$, $y$, $z$. For each choice of such a point, the equation $ax + by + cz = 0$ says wether or not that point lies in the corresponding plane.