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Finding the last corner of a square

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that there is a C point to these points A(-2|2|3), B(2|10|4) and D(5|-2|7), so that the quadrangle ABCD a square is. Determine the coordinates of C.

    3. The attempt at a solution
    All I have done is measure AC, CD and BC, which are 9, 9 and 12.78. But how do I get the C point from this data?

    Thanks in advance
     
  2. jcsd
  3. Aug 30, 2012 #2

    LCKurtz

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    The length of AB = 9 by calculation and the length of AD = 9 by calculation, so AD and AB are adjacent sides, with BC a diagonal of length 12.78. That means AC must be a diagonal and its length can't be 9. Something is wrong with your figures.
     
  4. Aug 30, 2012 #3
    Oh Sorry, I meant AC = 12.78, since it's a square, BC and CD must be 9 too.
     
  5. Aug 30, 2012 #4

    LCKurtz

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    Usually when you give a quadrilateral ABCD that would give the vertices in order as you go around the square. In this case with AB and AD adjacent sides, the order around the square can't be ABCD. Is that what you meant to have or do you still have typos? It might be good to state carefully exactly what you are given.

    Have you studied vectors and how to manipulate them?
     
  6. Aug 30, 2012 #5
    m21Ca.png
    This is what I mean

    I have also missed some classes, and that's why I require help from you
     
  7. Aug 30, 2012 #6

    LCKurtz

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    You didn't answer my question whether you have studied vectors. Think about what you would get if you add the vectors AB and AD. What direction would it point?
     
  8. Aug 30, 2012 #7
    AB+AD would point in the direction of AC, but I don't know those vektors, I only know the points ABD

    I know how to perform certain operations with vectors, but as I said, I missed some classes, and probably the key to resolve this problem
     
  9. Aug 30, 2012 #8

    LCKurtz

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    To get a vector between points P and Q you subtract the coordinates of P from the coordinates of Q to get the components of the vector PQ.
     
  10. Aug 30, 2012 #9
    Oh man, thanks a lot, then I suppose I have to do the following:

    AB: (2-(-2)|10-2|4-3) AD: (5-(-2)|-2-2|7-3)

    Then AC: (4+7|8-4|1+4)

    Thus the C point is (11|4|5), right?

    But this is somehow wrong because when doing calculating CD or BC I get don't get 9!

    What am I doing wrong?
     
  11. Aug 30, 2012 #10

    LCKurtz

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    No, that is not the point C. It is a vector pointing towards C. You have a vector in the right direction, but its length is whatever it came out. How long is it? How long does it need to be to fit the diagonal? How do you change its length?
     
  12. Aug 30, 2012 #11
    It's lenght is as far as I know is |C|, 12.72, which is the measure of the diagonal
     
  13. Aug 31, 2012 #12

    LCKurtz

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    OK, so it is the right length. Now, if you are at the point A, how do you use that vector with the coordinates of A to figure out the coordinates of C? And, by the way, it would be best to use exact values instead of decimal approximations if you want the exact coordinates of C.
     
  14. Aug 31, 2012 #13
    Thank you very much, I got it

    I did it like this: AC(11|4|5) = (cx-(-2)|cy-2|cz-3), then separated it in 3 simple equations: cx+2=11; cy-2=4; cy-3=5. And got C(9|6|8).

    Also, you kind of implied with your last post, that it could be measured with help of the vector's lenght, how?
     
    Last edited: Aug 31, 2012
  15. Aug 31, 2012 #14

    LCKurtz

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    When you added the two vectors from the sides to get a vector in the direction of the diagonal, you needed to check its length to make sure it is the right length for the diagonal. In this case it checked OK. Had it not been the right length, when you added its components to the coordinates of A, you wouldn't have gotten the coordinates of C.
     
  16. Aug 31, 2012 #15

    Ray Vickson

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    You are making things much more complicated than they need to be. Just look at the displacement from A to B; that is, starting from A, how do you reach B? You start from x = -2 at A and get to x = +2 at B, so you go 4 units in the x-direction. You start from y = 2 at A and get to y = 10 at B, so go 8 units in the y-direction. You start at z = 3 and end at z = 4, so go 1 unit in the z-direction. So, AB = (4,8,1). Thus, in order that ABCD be a parallelogram you need DC = AB --- draw a picture to see this --- so DC (4,8,1), hence C = D + DC = (5,-2,7) + (4,8,1) = (9,6,8).

    RGV
     
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