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I'm learning about planes in linear algebra, and my professor said that to define a plane two lines must intersect. However, I was thinking, what if the two lines are parallel? In that case, they would never intersect.

Can't two parallel lines still define a plane, you can think of the plane as a piece of paper and if I draw 2 parallel lines on it I have a plane. How can we define a plane with two parallel lines mathematically or do I must have a point of intersection?

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    $\begingroup$ Yes, they can. Parallel lines are defined as two lines on the same plane that never intersect, and only one distinct plane contains both of those lines. (I assume you're working with Euclidean Geometry.) $\endgroup$
    – Nate
    Mar 2 at 10:16
  • $\begingroup$ Parallel lines intersect at $\infty$ $\endgroup$
    – m-stgt
    Mar 2 at 10:21
  • $\begingroup$ @Nate -- You say "only one distinct plane contains both of those lines" while I say two planes. Why? A plane has two sides, a back and a front side, just like a coin. In contrast to coins you can't simply flip planes. Which of the two possible planes you have, you only see if you take a closer look, which side you're facing. $\endgroup$
    – m-stgt
    Mar 2 at 18:35
  • $\begingroup$ @m-stgt I think that's non-standard terminology, isn't it?  AIUI, in standard usage, planes are not oriented.  (People are welcome to define and use terms however they want, of course — but it's likely to cause confusion if they use non-standard ones without specifying.) $\endgroup$
    – gidds
    Mar 2 at 19:27
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    $\begingroup$ Strictly, I think you'd also need to specify that the two lines are distinct; two identical lines intersect at every point, but don't define a unique plane. $\endgroup$
    – gidds
    Mar 2 at 19:31

2 Answers 2

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Let the first line be $a_0 + tv$ and let the second line be $a_1 + tv$ where $t$ is a parameters.

Then your plane contain the third line $a_0+s(a_1-a_0)$ where $s$ is a parameter.

Now you have found two non-parallel lines (first line and third line) and they intersect at $a_0$. you can proceed to work on it as usual.

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A plane can be defined like this $$ E: \vec{x}=\vec{p}+r\vec{a_1}+t\vec{a_2}. $$ Here, every point $x$ can be reached via a vector $\vec{p}$ that goes from the origin to any point $P$ on the plane. Now $r,s$ are parameters (from $\mathbb R$ if you will) and the other two vectors expand the plane. They cannot be collinear (linearly dependent) or else we will get a line.

Now imagine two lines that are parallel. The first vector $\vec{a_1}$ that we can set for the plane is the vector that we get from the first line $$ L_1: \vec{x}=\vec{q}+r\vec{u}. $$ So just set $\vec{a_1}=\vec{u}$. Now we need a second vector $\vec{a_2}$ but if we take the vector $\vec{v}$ from the second line $$ L_2: \vec{x}=\vec{h}+k\vec{v}. $$ then we cannot get a plane, as $$ \vec{v}=b\cdot\vec{u} $$ for some $b$.

But we can take another clever vector, a vector that goes from the first line to the second line, for example $\vec{h}-\vec{q}$.

That vector will be collinear to our first vector that generates the plane and both lines will lie in it. Very simple.

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