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Find the geometric relation between vectors A and B

  1. Sep 2, 2008 #1
    Vectors A and B each lie in the x-y plane. The magnitude of A + B equals the magnitude of A - B. Find the geometric relation between vectors A and B.
    (Hint: Express the vectors in unit-vector notation. Use your knowledge of dot products.)

    I really don't even know what this problem is asking. What is meant by the geometric relation? I'm not sure how dot products relate. What is my final answer going to be in terms of? Is it going to be an equation? A sentence? I'm really lost. Can someone break down what the objective of this problem is or how to start it?
     
  2. jcsd
  3. Sep 2, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi pf09! Welcome to PF! :smile:

    The square of the magnitude of a vector is its dot-product with itself: |A|2 = A.A :smile:
    Stop philosophising! :rolleyes:

    If you actually solve the equation, you should see what the point is. :wink:

    So what equation do you get? :smile:
     
  4. Sep 2, 2008 #3
    Re: Welcome to PF!

    hi, thank you for the welcome.

    i know that about the dot product; i just dont know how it relates.
    solve which equation? what am i solving for?
    i'm not philosophizing. i just dont know what i'm supposed to do.
     
  5. Sep 2, 2008 #4
    i can add vectors, find magnitudes, do dot products, find the angle between vectors,.. i can do all that. i just dont know what this problem is asking. if someone could tell me what i'm supposed to do, i could probably do it. please, i actually dreamt about this problem between my first and last post.
     
  6. Sep 2, 2008 #5

    HallsofIvy

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    The "magnitude" of A is the square root of [itex]A\cdot A[/itex] so if the magnitudes of A+ B and A- B are the same you know that [itex](A+B)\cdot(A+B)= (A-B)\cdot(A-B)[/itex] multiply that out and see what it tells you.

    One way to do this problem purely geometrically would be to realize, from the "parallelogram" definition of the sum of two vectors, that A+ B is one diagonal of the parallelogram having sides A and B, while A- B is the other diagonal. If those have the same length, what does that tell you about the parallelogram?
     
  7. Sep 2, 2008 #6

    tiny-tim

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    The equation is |A + B| = |A - B|.

    Square it, and then use the general forumula |A|2 = A.A
     
  8. Sep 2, 2008 #7
    thank you for these starting points.

    ok so after i square both sides of |A + B| = |A - B|, i get what hallsofivy gave me - (A+B).(A+B) = (A-B).(A-B). am i doing this next part right? from there, i said that (A+B).(A+B) = <(a1 + b1)^2, (a2 + b2)^2, (a3 + b3)^2> and (A-B).(A-B) = <(a1 - b1)^2, (a2 - b2)^2, (a3 - b3)^2>. i'm not sure where to go from here.

    also, to get back to hallsofivy's help, with a parallelogram having diagonals of the same length, i get a rectangle. so would it be enough for a final answer (after i've shown work) to say that vectors A and B are perpendicular?
     
  9. Sep 2, 2008 #8

    tiny-tim

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    oooh! :cry:

    Never use components if you don't have to … that's the joy of vector notation! :wink:

    Use the distributive law … (P + Q).R = P.Q + P.R.

    Try again! :smile:
     
  10. Sep 2, 2008 #9
    haha ok my book still uses components for all the proofs. i have to leave for school now, ill try to get back to this in between classes today. thanks for your help.
     
  11. Sep 2, 2008 #10

    atyy

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    Is the dot product of two vectors is a vector or a number?

    What is the dot product of two perpendicular vectors?
     
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