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Why you can't mix x and y components of vectors?

  1. Aug 29, 2005 #1
    Hi. Does anyone know of a proof that explains why you can't mix x and y components of vectors? For example you know how if you are solving a physics problem you have to break things up into x any y components (eg: velocity). My physics teacher wanted us to find a proof online that explained why you can't combine the x and y components of a velocity vector for example. I tried looking for a proof through google but i can't find one. So i wanted to know if any of you knew of a proof like this.

    Thanks
     
    Last edited by a moderator: Jan 7, 2014
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  3. Aug 29, 2005 #2

    LeonhardEuler

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    If you are a little more exact on what you mean by "mix" and "combine", maybe I can help you. It is certainly possible to add the x and y components of a vector. You'll get a number alright, but it really won't mean much.
     
  4. Aug 29, 2005 #3
    whenever i do a physics problem i have to break things up into its components (x and y components). After you break it up if you for example add the components together that would be wrong because the components are independent of each other. I'm trying to find a proof that explains why that is true. I looked on google but i coulld not find anything.
     
  5. Aug 29, 2005 #4

    LeonhardEuler

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    There is a correct way of adding the x and y components back into a single vector, but it is not just summing the the magnitudes of the components. Are you trying to prove that some particular method of adding the components is wrong? If so what is that method?
     
  6. Aug 29, 2005 #5
    I need a proof saying that summing the the magnitudes of the components is not the correct way of adding vectors
     
  7. Aug 29, 2005 #6

    EnumaElish

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    Ha! That's a great question. The x and y components cannot be mixed (e.g. added together) because we assume that objects take the shortest route between two points, which is a straight line on the analytic (Cartesian) plane. Then we apply the Euclidian distance to the line between the two points, which is the hypothenus equality. For example, the Euclidian distance between point a = (xa,ya) and point b = (xb,yb) is [itex]\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}[/itex].

    But let's say that you need to calculate distance in a big city. The streets go in the West-East direction (the X axis) or the South-North direction (the Y axis). In this case, the shortest distance between any two points is the sum of the X component and the Y component. For example, the "street distance" between points a and b would be |xb - xa| + |yb - ya|. That's because the only way anyone can travel in a city is to cover the whole West-East distance first, and then cover the whole South-North distance (or the other way around). Any way you look at it, total distance traveled is the sum of the x component and the y component.

    In my opinion the "proof" you are looking for is the hypothenus rule, which states that the length of the hypothenus is [tex]\sqrt{\text{base}^2+\text{height}^2}.[/tex]
     
    Last edited: Aug 29, 2005
  8. Aug 29, 2005 #7
    lol, couldn't this only be done satisfactorily using linear algebra's completely general approach to vector spaces over R^3?


    everything else would just be hand-wavy, as far as i know. :/


    a safe bet would be "because they are perpindicular to each other."

    i guess you could start getting fancy and say that the projection of a vector onto another perpindicular vector is zero, etc.

    as my phys 2 prof said, "they're orthogonal--they're not talking to each other." (this was about real numbers and imaginary numbers, but the same general idea.)
     
  9. Aug 29, 2005 #8
    x and y are independent.
    Since it is 2 dimenional, this becomes parametic equation problem. X and Y are related by the relation with t, but they are independent from each other.
    but i bet it will still come out the same mathematically if you dont break it up... although this complicate the problem itself...

    my largest bet would be that the vector analysis is developed BASED ON SUCH OBSERVATION, not the nature behavior acting based on vector analysis.
     
    Last edited: Aug 29, 2005
  10. Aug 30, 2005 #9

    Gokul43201

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    Actually, separately summing the magnitudes of different components does give you the components of the sum vector ! So, there's nothing wrong with it at all !

    So, the only way to interpret your request is to show that summing the various components of a vector does not give its magnitude. This, however, is merely a result of definition. The magnitude (or norm) of a vector (in Euclidean space) is defined as the RMS value of the components. Why this definition is chosen (over say, simply adding the components) is seen from usefulness of defining the magnitude as the displacement from the origin.
     
    Last edited: Aug 30, 2005
  11. Aug 31, 2005 #10

    AKG

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    It's not really a matter of proof. We decide to model things like velocity with vectors like ax + by + cz where {x, y, z} is linearly independent. The justification for this model would have to be empirical. Also, adding the magnitudes of two vectors is certainly not the right way to add vectors since |v| + |w| is not even a vector, it is a scalar.
     
  12. Aug 31, 2005 #11

    HallsofIvy

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    Oh, PLEASE tell me that's not what you really meant to write!
     
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