Translating a Plane

Question

I am trying to understand plane equations but am finding it a bit confusing. My understanding of the plane equation says that for points that lie in the plane they will give an output of $0$ i.e. $f(x) = 0$. If I want to define a plane that gives an output of 1 then I have $f(x)=1$, but isn't this the same as $f(x) - 1 = 0$, so it would be the same as defining a plane that has an intercept that is $1$ lower.

Why is it that when I see diagrams they show that the line is shifted away from the origin i.e. with a higher intercept? Why does it not now have a lower intercept?

Answer 1

Hint:

I suggest you how to see the question for a straight line, and you can extend easily to a plane. A stright line has equation: $$ f(x)=0 \iff ax+by+c=0 $$

If $b \ne 0$ the intercept with the $y$ axis is given by:

$$ x=0 \Rightarrow by+c=0\Rightarrow y=-\frac{c}{b} $$

If you write the new equation: $$ ax+by+c=1 $$ you find: $$ x=0 \Rightarrow by+c=1\Rightarrow y=-\frac{c}{b}+\frac{1}{b} $$ and, depending on the signs of $b$ and $c$, this adds or subtract something to the first intercept.

Answer 2

Planes can be defined as $$Ax + By + Cz + D = 0$$ where $A,B,C,D\in\mathbb{R}.$ According to your definition, you would have $$f(x,y,z) = Ax + By + Cz + D.$$ For each constante value $k\in\mathbb{R}$, the equation $f(x,y,z)=k$ defines a different plane. In fact, for any function $f$, these are the so called level sets. Anyhow, as you said, you can always redefine (translate) your function so that the surfaces fulfills the homogeneous equation: $$f(x,y,z)=k \implies g(x,y,z) = f(x,y,z)-k = 0$$

Adding or multiplying the equation by constants does not change the surface that it represents, but it does change the function for which you are choosing a level set. in other words, $f(x,y,z) = k$ and $g(x,y,z)=0$ represent the same surface (a plane), but they arise from different level sets.

Note also that, for $C\ne0$ $$Ax + By + Cz + D = 0 \iff A'x + B'y + z = D',$$ where $A'=A/C$, $B'=B/C$ and $D'=-D/C$. This new equation still represents the same plane (but a level set of a different function!). But, in this case, $D'$ can be interpreted as the height, from the origin, where the plane cuts the $z$-axis. Of course, you can deduce analogous information for the other axes by transforming the equation accordingly.