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Homework Help: Simple problem on vector

  1. Sep 5, 2008 #1
    If |vector A + vector B| (magnitude of sum of vectors A and B) = |vector A + vector B| (magnitude of difference between vectors A and B), how can we show that theta is equal to pi/2?

    I know that vector A - vector B can be equal to the sum of vectors A and B if vector B is a null vector, but with magnitudes, angles, I'm all confused.

    Please help.....
    Last edited: Sep 5, 2008
  2. jcsd
  3. Sep 5, 2008 #2
    If you are allowed to use dot products you can use the fact the magnitude of a vector (such as the resultant vectors for the sum and difference of A and B) equals the square root of the dot product.

    Provide the vectors with components:

    A = <a1, a2>
    B = <b1, b2>

    So the magnitudes of the sum and diff resultant vectors would be

    mag_sum = Sqrt[(a1+b1)^2 + (a2+b2)^2]
    mag_diff = Sqrt[(a1-b1)^2 + (a2-b2)^2]

    You can then use the fact that mag_sum = mag_diff and expand the squared terms, etc... and arrive at a very simple equation of the form,

    X+Y = -(X+Y)

    From there, [assuming you recognize (X+Y) :wink:...] and using the definition of "dot product" as |A||B|*Cos(theta), it's easy to see what Cos(theta), and therefor theta itself, must be for the equality to be true.

    Probably a little more involved than what's necessary... but hey - it works :rofl:

  4. Sep 5, 2008 #3


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    Welcome to PF!

    Hi spandan! Welcome to PF! :smile:

    Yes it does! :biggrin: … but the joy of vectors is that you can often prove things without using coordinates …

    in this case, |A + B|2 = (A + B).(A + B) (by definition of || :wink:),

    and |A - B|2 = (A - B).(A - B) …

    so |A + B| = |A - B| if … ? :smile:
  5. Sep 7, 2008 #4
    No, how do we showw that theta is equal to pi/2?

    Of course, pi = 3.14159..........
  6. Sep 8, 2008 #5


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    ........265358979........ :rolleyes:

    π/2 = 90º :biggrin:
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