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Hopelessly confused with vectors

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data
    In the diagram ABCD is a square, P and Q are the midpoints of DC and BC respectively. Show that PA + QA = 1.5 CA
    check attachment for diagram


    2. Relevant equations
    none (i think?)


    3. The attempt at a solution
    PA+QA=3/2 CA
    =PC+CA+QC+CA
    =PA+AC+QC+2CA
    =PA+AC+QA+AC+2AC
    =PA+QA+4AC

    and I'm stuck here. I also would like to know if there is a difference between prove
    and show

    Secondly is anyone could give some tips how to deal with these problems with proofs (also in trigonometry) it would be greatly appreciated. I am pretty much running through the dark with these problems.

    P.S I don't know how to get the little arrows above the letters to show they are vectors, if anyone could tell me how...
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2008 #2

    danago

    User Avatar
    Gold Member

    The way i would do it would be to let the length AD be represented by the vector a, and let DC be represented by b. Then try writing PA, QA and CA all in terms of only a and b, and see if you can see what to do next.
     
  4. Sep 7, 2008 #3
    I assume the notation means for example PA is a vector directed from point P toward point A. Start by writing vector sums for PA and QA in terms of CA and other vectors. So (again, these are vector equations)

    PA = PC + CA and similarly QA = QC + CA

    Adding these two equations yields

    PA + QA = PC + QC + CA + CA = PC + QC + 2CA

    If we are correct so far, we suspect that PC + QC must someow be equal to -0.5 CA.

    Let O be the center of the square. Then consider the square OQCP. The diagonal from lower left to upper right of this square is the vector 0.5AC (note AC, not CA). Also from this square one can see that

    PC + QC = 0.5AC

    Now recall that, for example, AC = - CA, so that 0.5AC = - 0.5 CA, so

    PC + QC = 0.5AC = -0.5CA

    and we have what we need. There's probably a more direct proof, but this one works, and it makes use of vectors.
     
  5. Sep 7, 2008 #4
    See this is as much as I suspected but I did not know for sure how PC + QC = 0.5AC. I know how to add a vector say QC + CR which would be equal to QR, but how do you add vectors like that i.e PC + QC?

    Secondly, how can we assume the origin, O, is at the centre of the square?
     
  6. Sep 7, 2008 #5
    Secondly, how can we assume the origin, O, is at the centre of the square?

    We don't need a coordinate system for this problem. Point O is just a way to label the center of the square.

    See this is as much as I suspected but I did not know for sure how PC + QC = 0.5AC. I know how to add a vector say QC + CR which would be equal to QR, but how do you add vectors like that i.e PC + QC?

    Slide PC to position OQ. Slide QC to position PO. The two vectors are now two sides of a parallelogram whose diagonal is their sum and = 0.5AC.
     
  7. Sep 7, 2008 #6
    But wouldn't it be QC to OP? I know that the diagonal is the sum, but I still don't understand how the sum could be 0.5AC, or how we even obtain the value for the diagonal.
     
  8. Sep 7, 2008 #7
    But wouldn't it be QC to OP? I know that the diagonal is the sum, but I still don't understand how the sum could be 0.5AC, or how we even obtain the value for the diagonal.

    Yes it would. The diagonal goes from point O, the midpoint of AC (since O is at the center of the square) to point C, and its length is 1/2 the length of the diagonal (AC) of the original square. Thus PC + QC = 0.5 AC
     
  9. Sep 8, 2008 #8
    OH RIGHT..now I see. Wow vectors are really interesting once you understand.

    Another question I have is about position vectors. Sometimes the question asks you about a position vector A eg. 2a - 3b relative to the origin, O, so then it's like OA = 2a - 3b. But then some ask you questions about the position vector alone

    Eg. the position vector of A and B are a - 3b and 2a + 5b respectively. Find the position of the mid pt. of AB.

    So if you were to draw a vector triangle how does it look? How do you work a question like this?
     
  10. Sep 8, 2008 #9
    See attached document.
     

    Attached Files:

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