# Explanation of solution (parallelness)

• Nov 20th 2006, 10:12 AM
OReilly
Explanation of solution (parallelness)
Problem that I need to solve is:

Let $\displaystyle a$ be line parallel to plane $\displaystyle \beta$ and let $\displaystyle \alpha$ be some plane that contains line $\displaystyle a$.
Then it's $\displaystyle \alpha \parallel \beta$ or intersection of these two planes is line parallel to $\displaystyle a$.

This is solution from book:
If $\displaystyle a \subset \beta$ then there are two cases (look at attached picture from book).
1) Planes $\displaystyle \alpha$ and $\displaystyle \beta$ have common only line $\displaystyle a$ so then is $\displaystyle \alpha \cap \beta = \{ a\}$, so $\displaystyle \alpha \cap \beta \parallel a$.
2) If there is at least one more point not belonging to line $\displaystyle a$ which is common to planes $\displaystyle \alpha$ and $\displaystyle \alpha '$, then is $\displaystyle \alpha = \alpha '$ so then is $\displaystyle \alpha \parallel \beta$. If it's $\displaystyle a \cap \beta = \emptyset$, then it's either $\displaystyle \alpha \parallel \beta$, that is $\displaystyle \alpha \cap \beta = \emptyset$, or it's $\displaystyle \alpha \cap \beta = a'$. All common points of planes $\displaystyle \alpha '$ and $\displaystyle \beta$ belongs to line $\displaystyle a'$, so since it's $\displaystyle \alpha \cap \beta = \emptyset$ then lines $\displaystyle a$ and $\displaystyle a'$ are not intersecting which means that they are parallel.

Solution from book is quite confusing to me.
Number 1) case is understable, nothing there to confuse.
Number 2) is not understable to me. First I will explain red text. Red text is reffering to case when $\displaystyle a \subset \beta$ and rest of text is reffered to case when $\displaystyle a \cap \beta = \emptyset$ (I think because author didn't specify that clearly). I don't understand why author introduced plane $\displaystyle \alpha '$ at all. I don't understand red text at all. Isn't solution just to say that if planes $\displaystyle \alpha$ and $\displaystyle \beta$ have one more common point not belonging to line $\displaystyle a$ then it's $\displaystyle \alpha \parallel \beta$?
The rest of text in 2) is quite confusing me.
My solution to when is $\displaystyle a \cap \beta = \emptyset$ is that if $\displaystyle \alpha \cap \beta = \emptyset$ then is $\displaystyle \alpha \parallel \beta$. If is $\displaystyle \alpha \cap \beta = a'$ then it must be $\displaystyle a\parallel a'$ because if $\displaystyle a$ isn't parallel to $\displaystyle a'$ then it would intersect plane $\displaystyle \beta$ which is contradiction. Also $\displaystyle a'$ can't intersect $\displaystyle a$ because then it would have only one common point with plane $\displaystyle \beta$ which is contradictory to $\displaystyle \alpha \cap \beta = a'$.

Can someone look at solution from book and my solution and give me explanation, comment or something that would clarify me that?:confused:
• Nov 20th 2006, 01:56 PM
Plato
First, I totally agree with you. I see no reason to introduce $\displaystyle \alpha'$.
On the other hand, I am not sure that I follow your proof.
Here are the four cases that I would use.
$\displaystyle \begin{array}{l} 1)\;a \subset \beta \quad \& \quad \alpha = \beta \\ 2)\;a \subset \beta \quad \& \quad \alpha \not= \beta \\ 3)\;a \not\subset \beta \quad \& \quad \alpha \cap \beta = \emptyset \\ 4)\;a \not\subset \beta \quad \& \quad \alpha \cap \beta \not= \emptyset \\ \end{array}$

I understand that this is not what you asked and therefore it may not help.
However, in each case the given conclusion does follow.
• Nov 20th 2006, 03:09 PM
OReilly
Quote:

Originally Posted by Plato
First, I totally agree with you. I see no reason to introduce $\displaystyle \alpha'$.
On the other hand, I am not sure that I follow your proof.
Here are the four cases that I would use.
$\displaystyle \begin{array}{l} 1)\;a \subset \beta \quad \& \quad \alpha = \beta \\ 2)\;a \subset \beta \quad \& \quad \alpha \not= \beta \\ 3)\;a \not\subset \beta \quad \& \quad \alpha \cap \beta = \emptyset \\ 4)\;a \not\subset \beta \quad \& \quad \alpha \cap \beta \not= \emptyset \\ \end{array}$

I understand that this is not what you asked and therefore it may not help.
However, in each case the given conclusion does follow.

What specificaly you don't follow in my proof?
• Nov 20th 2006, 03:23 PM
The Pondermatic
I think I see how it works. At least, I didn't understand the explination, but I see where A' comes in. (As I am too lazy to find the fancy characters, I will use A for the plane and a for the line.)

Given: line a
plane B
plane A
plane B || line a
line a is within plane A
not(plane A || plane B) | already proven that this can happen

Prove: a || the intersection of A and B

Proof:
Let there be a plane A', which is parallel to B and contains a.
Since A and B intersect, there must be a line b that is contained by both A and B.
So now there are parallel planes A' and B, each containing a line (a and b, repectively).
Since a and b are both within plane A, they must either intersect or be parallel. However, they are also the intersections of A and B and A and A', and B and A' are defined to be parallel, a and b must also be parallel.
Moreover, since b is defined as the intersection of A and B, a is parallel the intersection of A and B. And there is your proof.

Sorry if that was sketchy. I don't remeber my theorums from last year. :p
• Nov 20th 2006, 03:35 PM
Plato
Quote:

Originally Posted by OReilly
What specificaly you don't follow in my proof?

It seems to me as if you are extending or modifying the proof in the text.
I thought we agreed that the text is off the mark on its proof.
I also understand that my approach may not be helpful to you.
• Nov 20th 2006, 03:37 PM
OReilly
Quote:

Originally Posted by The Pondermatic
I think I see how it works. At least, I didn't understand the explination, but I see where A' comes in. (As I am too lazy to find the fancy characters, I will use A for the plane and a for the line.)

Given: line a
plane B
plane A
plane B || line a
line a is within plane A
not(plane A || plane B) | already proven that this can happen

Prove: a || the intersection of A and B

Proof:
Let there be a plane A', which is parallel to B and contains a.
Since A and B intersect, there must be a line b that is contained by both A and B.
So now there are parallel planes A' and B, each containing a line (a and b, repectively).
Since a and b are both within plane A, they must either intersect or be parallel. However, they are also the intersections of A and B and A and A', and B and A' are defined to be parallel, a and b must also be parallel.
Moreover, since b is defined as the intersection of A and B, a is parallel the intersection of A and B. And there is your proof.

Sorry if that was sketchy. I don't remeber my theorums from last year. :p

Even without plane A' it can be proven that a is parallel to b.
If a wouldn't be parallel to b then it would intersect b and plane B so it must be parallel (since inital statement was that a is parallel to B). Also if b wouldn't be parallel to a then it would have only one common point with plane B which is again against inital assumption that b is common line to planes A and B.
• Nov 20th 2006, 03:40 PM
OReilly
Quote:

Originally Posted by Plato