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Vector proof of generalized parallelogram law for quadrilaterals

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The following theorem in geometry suggests a vector identity involving three vectors A, B, and C. Guess the identity and prove that it holds for vectors in Vn. This provides a proof of the theorem by vector methods.
    "The sum of the squares of the sides of any quadrilateral exceeds the sum of the squares of the diagonals by four times the square of the length of the line segment which connects the midpoints of the diagonals."


    2. Relevant equations
    I'm not completely sure what's relevant here besides various properties of the dot product and properties of vectors themselves, but the previous problem was similar and in fact is a special case of this problem for parallelograms:

    [tex]\left\|A + B\right\|^2 + \left\|A - B\right\|^2 = 2 \left\|A\right\|^2 + 2\left\|B\right\|^2 [/tex]

    I also have the identities resulting from expanding [itex]\left\|A + B\right\|^2[/itex] and [itex] \left\|A - B\right\|^2 [/itex]. I also know the Cauchy-Schwarz inequality. And by the way, ||A|| represents the magnitude of a vector A. Vn is a vector space of n-tuples of real numbers.

    3. The attempt at a solution
    The problem statement implies that only three vectors are necessary to prove this, but I don't see how since a quadrilateral could have four unequal, nonparallel sides. In addition, I don't know how to represent the diagonals as a sum of two vectors which represent two adjacent sides since the diagonal would not end at the vertex of the parallelogram created by those two vectors.

    Obviously, I can't guess the identity until I can find a way to translate the english into vectors, and so I can't move very far with this problem. Strangely enough, I could not find this proof after googling and searching here for a good two hours...

    Edit: Okay, I've made some more progress. By drawing it out, I found that I could assign a name to each side and diagonal of a quadrilateral with just three vectors. So far, the identity I'm supposed to figure out in terms of those three vectors is:

    [tex]\left\|A\right\|^2 + \left\|B\right\|^2 + \left\|A + C\right\|^2 + \left\|B - C\right\|^2 = \left\|C\right\|^2 + \left\|A + B\right\|^2 + 4\left\|.5B - .5A - .5C\right\|^2[/tex]

    I'm still working out what that midpoint segment could be in terms of the three vectors I have already, hence the question mark where that term goes. I made A and B adjacent sides of a quadrilateral and C the diagonal connecting the vertex where vectors A and B meet to the opposite vertex. If it matters, I have B pointing in an upward direction, A pointing towards the tail of B, and C pointing away from A and B. Thus, the last two terms on the left side are the lengths of two sides of the quadrilateral, and the next-to-last term on the right side is the other diagonal's length.

    Edit 2: Okay, I've found out what the identity is. I've edited the above equation with that final term. I found the last term from the drawing I made.
     
    Last edited: Sep 2, 2009
  2. jcsd
  3. Sep 2, 2009 #2

    Dick

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    That's one way to do it, probably not the least complicated, but it should work. To find the midpoint of a diagonal, figure out the vector representation of each endpoint of the diagonal, add them and divide by 2.
     
  4. Sep 2, 2009 #3

    Dick

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    Are you sure about the last term? I think you might have a sign wrong.
     
  5. Sep 2, 2009 #4
    Yeah., the sign is wrong... As I was trying to prove it, I found out that there should be a minus sign before the ".5C" term. I'm still trying to figure out why right now. Thanks for checking the earlier part, btw.

    What I did was:

    [tex]4\left\|(B - .5(A + B)) + .5C\right\|^2 = 4\left\|(B - .5A - .5B) + .5C\right\|^2 = 4\left\|.5B - .5A + .5C\right\|^2[/tex]
     
  6. Sep 2, 2009 #5

    Dick

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    Look at the C diagonal. One endpoint is A the other endpoint is A+C. Add and divide by two. Look at the other diagonal. One endpoint is 0 and the other endpoint is A+B. Add and divide by two. Now take the difference.
     
  7. Sep 2, 2009 #6
    Ahh. I see what I did wrong now. I had flipped one of the vectors the wrong way while subtracting. However, I understand your method now and like it much better ("converting" the vectors to points). That clears up the identity properly if I subtract right. Though, looking at it, it doesn't matter which way I subtract the two points.

    The rest of the proof is just a simple matter of expanding everything out. Thanks for the help!
     
  8. Sep 2, 2009 #7

    Dick

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    Right. There's lots of other ways to do this. For example, I set the quadrilateral points to be 0, A, A+B and A+B+C. You're welcome!
     
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