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Geometric Proof With Vectors

  1. Apr 19, 2014 #1
    1. The problem statement, all variables and given/known data

    This problem is from Mary Boas' "Mathematical Methods in the Physical Sciences" 3rd Ed.

    Capter 3 Section 4 Problem 3

    Use vectors to prove the the following theorems from geometry:

    3. The diagonals of a parallelogram bisect each other.

    2. Relevant equations

    Just the parallelogram principle with vectors.

    3. The attempt at a solution

    My drawing is here: https://www.flickr.com/photos/81639412@N00/13916871186/

    $$\frac { 1 }{ 2 } \left( A+B \right) =\frac { 1 }{ 2 } \left( -A+B \right) $$ Where A&B are vectors. A does not equal zero so I obviously failed to prove anything. Example 2 in the text seems to prove a similar theorem this way. The vector for P and P' point to the same spot if the vectors are translated, but since they are pointing different directions and are different lengths they are not equivalent. However, the book seems to make it work in example two. I am not sure what went wrong. I never took geometry. I skipped it to take trig then calculus, and I am thinking it really would come in handy in this section. I am not used to proofs and other pure math formalism. I was a physics major that graduated about 15 years ago, so please go easy on me :D

    Thanks,
    Chris Maness
     
  2. jcsd
  3. Apr 19, 2014 #2

    Mark44

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    Please fix your link. It opens to a Flickr page that says "Adult Content", and asks you to sign in.
     
  4. Apr 19, 2014 #3

    Fredrik

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    When I clicked the link, I got the message "ADULT CONTENT You must be signed in to see this content".

    Dictionary.com defines "bisect" as "to cut or divide into two equal parts". So I would interpret the problem as asking you to first find the point where those two lines meet, and then show that's it's the midpoint of the line segment from 0 to A+B, as well as the midpoint of the line segment from A to B.

    You could use that every line in a vector space is the range of a map ##t\mapsto x+tv## (with domain ##\mathbb R##), where ##x## is any point on the line, and ##v## is any tangent to the line (e.g. the difference between two points on the line).
     
  5. Apr 19, 2014 #4
    Haha, it is definitely not adult contact. I had miss clicked a setting in Flickr. It should be ok now.

    Chris
     
  6. Apr 19, 2014 #5
    Yes, I figured that is what was meant by bisect, and did make an attempt. My drawing is really relevant here.

    Could you please unpack your second line? I am not following.

    Thanks,
    Chris Maness
     
  7. Apr 19, 2014 #6

    SammyS

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    Here is the photo. It was tough getting without signing in. . . . . . SammyS

    attachment.php?attachmentid=68827&stc=1&d=1397943633.jpg
     

    Attached Files:

  8. Apr 19, 2014 #7
    I did remove the adult content flag on it. For some reason I had clicked to tag every photo I upload as adult content. Thanks for posting it.

    Chris
     
  9. Apr 19, 2014 #8

    Fredrik

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    Let x and v be vectors such that v≠0. Define a function f by f(t)=x+tv for all real numbers t. Do you see what kind of set the range of f is?

    The problem with your attempt is that you're just writing down a formula for the midpoint of the line segment from A to B, and that formula is wrong. You shouldn't just try to write down a formula. You should also explain why it holds. You got the right answer for the midpoint of the line segment from 0 to A+B. How did you do that? Can you use the same method for the line segment from A to B?
     
  10. Apr 19, 2014 #9
    I am not sure what you mean by "set" as my training in any mathematical formalism is VERY limited. The math I know is basic calculus applied to physical systems and that is why I am going through the M. Boas text to build up my frail math skills. I was sailing through fairly smoothly until I hit chapter three, and I am now gracelessly bumbling through it.

    I do see in my mind that the function where x is a fixed length/direction vector and v varies in length but not direction with all values of "t". Adding those two will form a resultant vector that varies with "t".

    I see adding A + B as forming a vector from the tail of A to the tip of B. Half of that sum would be a vector that spans to the midpoint of the parallelogram (the center of mass so to speak). I can do that for all for corners of the parallelogram. However, I am not sure how to prove that all four midway vectors point to the same location without resorting to a coordinate system, and the example number two that Mary shows in her book does not resort to a coordinate system. That seems to be what the point of the exercise was.

    It is a bit hard to describe without reproducing the text and drawing for her example here. I can post it if you don't have access to the text book.

    Thank you for helping me,
    Chris Maness
     
  11. Apr 19, 2014 #10

    Fredrik

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    Aha. You should probably take a look at the most basic stuff about sets as soon as you can, so that you at least understand terms like "function", "domain" and "range. I'll try to avoid talking about sets here.

    The point I was trying to make was that if you for each real number t draw the point f(t) in a picture like the one you drew, the result is a straight line.

    v doesn't vary, only the product tv does. You can draw x as a point somewhere in the picture. v is, like every other vector*, represented by an arrow that starts at 0. The line defined by the function f is the line through x that's parallel to v. So if you draw a copy of v starting at x, you should find it easy to locate the point f(t)=x+tv. If you can figure out how to describe your two lines in this way, the rest will be easy.

    *) All vectors can be represented in the diagram either as a point, or as an arrow from 0 to that point.

    From the tail of A to the tip of B? The tails of A and B are both at 0, so the result would be B.

    You can find A+B by drawing a copy of A starting at the tip of B. Maybe that's what you meant. Keep in mind however that the arrow that represents A+B is drawn from 0 to the tip of that copy of A.

    The four corners of the parallelogram are 0, A, B, A+B.

    I do have access to the book, but I haven't looked at it yet. The way I would do this is to define functions f and g such that

    f(0)=0, f(1)=A+B, and f(t) is on the line through 0 and A+B for all real numbers t.
    g(0)=A, g(1)=B, and g(t) is on the line through A and B for all real numbers t.

    If these lines intersect, there must exist exactly one pair of real numbers t and s such that f(t)=g(s). If you find those numbers, you're pretty much done.
     
    Last edited: Apr 20, 2014
  12. Apr 19, 2014 #11

    LCKurtz

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    A and B are two adjacent sides. You know the left side is the mid point of one diagonal. The right side isn't the mid point of the other diagonal. To get to the mid point of the other diagonal you have to move along A then half way along the other diagonal. How does that change the right side of your equation?
     
  13. Apr 20, 2014 #12
    Ok, after sleeping on it, and re-reading what Mary did in her example, I see her logic. It is a different approach than I think you guys took (not saying it is better). But here is her reasoning applied to problem number three.

    One diagonal of the parallelogram is a vector represented by (A+B), and the opposite diagonal is (-A+B) because this vector can be translated back to the original parallelogram. From what I understand about vectors -- if they are translated from one place to another without changing direction or magnitude they are completely equivalent. So with A and B, (A+B), and (-A+B) we now have a parallelogram with two lines bisecting in the middle.

    To prove this. The mid length of A+B is where the other line should cross, so 1/2*(A+B) terminates at point P. Now if we follow A from O to the tail of the other diagonal, we can arrive at the same point by adding to A our P' pointing vector 1/2*(-A+B). Since we get the same vector 1/2*(A+B) back we know that P and P' are the same point and are both the midpoint of both diagonals we have proven the diagonals in a parallelogram bisect.

    I think I have captured the spirit of example 2 in Chapter 3 section 4 in application to problem number 3. Now I know why my students groan when they talk about their geometry class :D I skipped it in junior college figuring if I tested out of it, I didn't really need it.

    Thanks for your patience guys. I feel ashamed that I passed calculus and differential equations with good marks, and I stumbled so severely on this problem.

    Sincerely,
    Chris Maness
     
  14. Apr 20, 2014 #13

    LCKurtz

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    Or, to write it out, you are observing that since$$
    \frac { 1 }{ 2 } \left( A+B \right) =A+\frac { 1 }{ 2 } \left( B-A \right)$$is an identity, the two midpoints are the same.
     
  15. Apr 20, 2014 #14

    Fredrik

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    OK, my approach is perhaps a little more complicated than it needs to be. I found the lines first, and used them to find the point of intersection, and it turned out to be the midpoint of both lines. As you have noticed, it's faster to just write down formulas for the midpoints and note that they're the same point.

    My solution:

    Define f by f(t)=t(A+B) for all t in the interval [0,1]. Define g by g(t)=A+t(B-A) for all t in the interval [0,1]. The range of f is the line segment from 0 to A+B. (Note that f(0)=0 and f(1)=A+B). The range of g is the line segment from A to B. (Note that g(0)=A and g(1)=A+B-A=B).

    If there's a point p where they intersect, then there must exist s and t in the interval [0,1] such that f(s)=p=g(t). So s(A+B)=A+t(B-A). This is equivalent to (s-1+t)A+(s-t)B=0. Since A and B are linearly independent, this implies that s-1+t=0 and s-t=0. The latter implies that s=t. So the former becomes 2s-1=0, which implies that s=1/2. Since s=t, this implies that t=1/2.

    Unless we want to appeal to a theorem that says that the lines intersect, we should also verify that they do:
    f(1/2)=(1/2)(A+B)=(A+B)/2
    g(1/2)=A+(1/2)(B-A)=(2A+B-A)/2=(A+B)/2
     
  16. Apr 20, 2014 #15
    Thanks guys. LCKutz that is a very eloquent way to put it.

    Chris
     
  17. Apr 20, 2014 #16

    LCKurtz

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    You're welcome. That's what I was hinting at in post #11.
     
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