Is it true a plane consists of an infinite set of lines?

Theorem

A plane contains an infinite number of distinct lines.

Proof

A plane contains an infinite number of points.

Not all these points are collinear.

Let $A$, $B$ and $C$ be points in a plane $P$.

From Propositions of Incidence: Line in Plane, any two of these points determine a line.

Consider the lines $AB$, $AC$ and $BC$, all of which are distinct.

Let $X$ be one of the infinite number of points on $BC$ which is not $B$ or $C$.

Then $AX$ is a line in $P$ which is distinct from both $AB$ and $AC$.

As there is are infinite number of points on $BC$, there are an infinite number of lines incident to $A$ and $BC$.

All these lines are in $P$.

Hence the result.

$\blacksquare$

Sources

• 1952: T. Ewan Faulkner: Projective Geometry (2nd ed.) ... (previous) ... (next): Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The propositions of incidence

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A plane (formed by a linear equation) contains a line in 3D space. The linear equation has different multiples that produce the same exact plane. Does that then mean that there are an infinite number of planes that contain that line, or would that be the same plane given that it covers identical lines in 3D space?

I initially thought that there were an infinite number of planes containing that line, but I'm not so sure all of a sudden. To preface, I'm doing this to get a better understanding on my college homework.

Sorry if I didn't explain myself well. Thanks for any help given.

asked Mar 29, 2020 at 8:37

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I think it would be useful to point out here what is the difference between the linear equation and the plane itself.

Any non-trivial linear equation in three variables with coefficients in $\mathbb{R}$ defines a plane in $\mathbb{R}^3$ (i.e. in $3D$ space); such a plane is nothing but the solution set (in $\mathbb{R}^3$) to this linear equation. Now note that when you scale your linear equation by any non-zero real number, the solution set for this new scaled equation is exactly the same as the previous one, so the scaled equation defines the same plane , i.e. the same solution set.

On the other hand, given any line in $\mathbb{R}^3$ there are always infinitely many (in fact, uncountably many!) distinct planes containing that line. The intuition for this is easily represented in this picture:

One obtains infinitely many planes passing through the green line by rotating any plane containing the line along the axis given by this same line on an angle $\theta$ (for $0 < \theta \leq 2\pi$).

answered Mar 29, 2020 at 9:10

RickRick

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Not the answer you're looking for? Browse other questions tagged geometry 3d or ask your own question.

I assume the question asked by Marios was about the real plane, but planes over finite fields might also lead to interesting statements. First, let's take an example. Let $P$ be the affine plane over the two-element field $\mathbb{F}_2$. There are 4 points in this plane, but there are 6 lines (i.e. one-dimensional affine spaces) in this plane.

The proof for $\mathbb{R}$

We have a one-to-one mapping from $\mathbb{R}^2 \setminus\{(0,0)\}$ to $\mathcal{L}$ (I couldn't find simpler) : take any element $(a,b)$ and form the line $d_{a,b}$ which is tangent at the circle of center $0$ and radius $\sqrt{a^2 + b^2}$ at the point $(a,b)$. This is injective. It is not surjective, as the lines going through $(0,0)$ (let's denote it by $\mathbb{P}_1(\mathbb{R})$ ) do not have the form $d_{a,b}$.

Therefore, the set $\mathcal{L}$ have the same cardinality as $\mathbb{P}_1(\mathbb{R}) \cup \mathbb{R}^2 \setminus\{(0,0)\}$. But it is known that there is a bijection $\phi$ between $\mathbb{P}_1(\mathbb{R})$ and $\mathbb{R} \cup \infty$ : if $d \in \mathbb{P}_1(\mathbb{R})$ , let $\phi(d)$ be $x$, where $(x,1)$ is the intersection between $d$ and the horizontal line $y=1$. If $d$ is the line $y=0$, then $\phi(d) = \infty$. This is a bijection (it is clear on a drawing).

To conclude, we have a bijection of $\mathcal{L}$ onto $\mathbb{R}^3$. It can be shown that there is a bijection of $\mathbb{R}^3$ to $\mathbb{R}$.

The general case

In the general case, we might expect the number of lines to be "greater" than the plane.

First, note that there is a bijection of the projective space $\mathbb{P}_1(\Bbbk)$ onto $\Bbbk \cup \infty$, where $\infty$ is another point added to $\Bbbk$. When the field $\Bbbk$ is finite, it means that there is exactly as much one-dimensional vector spaces than elements in $\Bbbk$, plus one. When $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ they have the same cardinality.

Now note $\mathcal{L}_A$ the set of lines going through some point $x$ belonging to the set $A$. For instance, $\mathcal{L}_x$ is the set of lines going through $x$ and $\mathcal{L}_P$ is the set of all lines.

Note that $\mathcal{L}_x$ has the same cardinality as $\mathbb{P}_1(\Bbbk)$. Moreover, if $x \neq y$, then $\mathcal{L}_x \cap \mathcal{L}_y$ is a singleton (there is only one line going through $x$ and $y$ when $x \neq y$).

Using this, we show that the set of lines $\mathcal{L}$ is strictly greater than $P$, whatever be $\Bbbk$. In fact, take two points $x,y$ in $P$, we have a bijection of $\mathcal{L}_{\{x,y\}}$ onto $(\Bbbk)^2$ (we have removed an element counted twice).

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