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Homework Help: Quick Question

  1. Sep 21, 2005 #1
    I am doing a assingment for my classical mechanics class that requires the proof of:
    The dot product of |A dot B| <= (less than or equeal to) |A| |B| .

    I did the algebraic proof fine but we are required to do a geometic proof as well. This leaves me with the question what is the geometic view of the Dot Product?

    At first I brushed it off thinking that it was a prjection of some sort, but I have no idea what |A||B|Cos(Angle) would be geometirally?

    Thanks for the help
    Derek
     
  2. jcsd
  3. Sep 21, 2005 #2

    Fermat

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    |B|Cos(Angle) is the projection of B upon A.

    Lat A and B be two vectors coming out from the origin. They have an angle of Φ between them.
    If you draw a line from the tip of B to the vector A such that the line is perdindicular to A, and intersects it at P, say, then |B|cosΦ is the distance |OP| which is the projection of B onto A.
     
  4. Sep 21, 2005 #3
    I know that |B|Cos(angle) is the projection on A but what is |A|B|Cos(angle)???
    This is the question
     
  5. Sep 21, 2005 #4

    Fermat

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    |A||B|Cos(Φ) is simply the product of two scalars and is usually interpreted as two co-linear lines, |A| and |B|Cos(Φ).

    Hmmm.

    Can you do this ?

    Let |A||B|Cos(Φ) be the product of two scalars, S1 and S2, where S1 = |A| and S2 = |B|cos(Φ).
    The product of any two scalars is an area: A1 = S1*S2.

    So now sketch a little rectangle with sides labelled S1 and S2.

    Then |A||B| is also the product of two scalars S1 and S3, where S1 = |A| and S3 = |B|.
    The product of these two scalars is A2 = S1*S3.

    So now sketch another little rectangle with sides labelled S1 and S3.

    Can you do something like this to (geometrically) show that A1 <= A2 ?
     
  6. Sep 21, 2005 #5

    Integral

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    |A||B|cos ([itex] \Theta[/itex]) is a scaler and represents the magnitude of the vector, the direction is that of A,
     
  7. Sep 21, 2005 #6

    HallsofIvy

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    You want a "geometric" proof of |A dot B|<= |A||B| and you know that, geometrically, A dot B= |A||B|cos(&theta;)?

    Okay how large or how small can cos(&theta;) be?
     
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