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Math Help - Explanation of solution (parallelness)

  1. #1
    Senior Member OReilly's Avatar
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    Arrow Explanation of solution (parallelness)

    Problem that I need to solve is:

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

    This is solution from book:
    If a \subset \beta then there are two cases (look at attached picture from book).
    1) Planes \alpha and \beta have common only line a so then is \alpha  \cap \beta  = \{ a\} , so \alpha  \cap \beta \parallel a.
    2) If there is at least one more point not belonging to line a which is common to planes \alpha and \alpha ', then is \alpha  = \alpha ' so then is \alpha \parallel \beta . If it's a \cap \beta  = \emptyset , then it's either \alpha \parallel \beta , that is \alpha  \cap \beta  = \emptyset , or it's \alpha  \cap \beta  = a'. All common points of planes \alpha ' and \beta belongs to line a', so since it's \alpha  \cap \beta  = \emptyset then lines a and 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 a \subset \beta and rest of text is reffered to case when a \cap \beta  = \emptyset (I think because author didn't specify that clearly). I don't understand why author introduced plane \alpha ' at all. I don't understand red text at all. Isn't solution just to say that if planes \alpha  and \beta have one more common point not belonging to line a then it's \alpha \parallel \beta ?
    The rest of text in 2) is quite confusing me.
    My solution to when is a \cap \beta  = \emptyset is that if \alpha  \cap \beta  = \emptyset then is \alpha \parallel \beta . If is \alpha  \cap \beta  = a' then it must be a\parallel a' because if a isn't parallel to a' then it would intersect plane \beta which is contradiction. Also a' can't intersect a because then it would have only one common point with plane \beta which is contradictory to \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?
    Attached Thumbnails Attached Thumbnails Explanation of solution (parallelness)-image.gif  
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  2. #2
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    First, I totally agree with you. I see no reason to introduce \alpha'.
    On the other hand, I am not sure that I follow your proof.
    Here are the four cases that I would use.
    \begin{array}{l}<br />
 1)\;a \subset \beta \quad \& \quad \alpha  = \beta  \\ <br />
 2)\;a \subset \beta \quad \& \quad \alpha  \not= \beta  \\ <br />
 3)\;a \not\subset \beta \quad \& \quad \alpha  \cap \beta  = \emptyset  \\ <br />
 4)\;a \not\subset \beta \quad \& \quad \alpha  \cap \beta  \not= \emptyset  \\ <br />
 \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.
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Plato View Post
    First, I totally agree with you. I see no reason to introduce \alpha'.
    On the other hand, I am not sure that I follow your proof.
    Here are the four cases that I would use.
    \begin{array}{l}<br />
 1)\;a \subset \beta \quad \& \quad \alpha  = \beta  \\ <br />
 2)\;a \subset \beta \quad \& \quad \alpha  \not= \beta  \\ <br />
 3)\;a \not\subset \beta \quad \& \quad \alpha  \cap \beta  = \emptyset  \\ <br />
 4)\;a \not\subset \beta \quad \& \quad \alpha  \cap \beta  \not= \emptyset  \\ <br />
 \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?
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by OReilly View Post
    What specificaly you don't follow in my proof?
    The answer to that is just about everything.
    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.
    I just cannot help you with another proof.
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  6. #6
    Senior Member OReilly's Avatar
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    Quote Originally Posted by The Pondermatic View Post
    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.

    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.
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  7. #7
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Plato View Post
    The answer to that is just about everything.
    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.
    I just cannot help you with another proof.
    My proof isn't modification of solution given from book.

    I can't reply if you don't give me specifics what you didn't understand from my post.
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