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Vector notation. just an explanation

  1. Sep 9, 2008 #1
    i dont have a problem jsut a little confusion. what exactly does the i hat and j hat notation mean when given a vector problem.

    such as [tex]\vec{}a[/tex]= 0.023 [tex]\hat{}i[/tex]m/s[tex]^{}2[/tex] + 0.046 [tex]\hat{}j[/tex]m/s[tex]^{}2[/tex]

    what is this statement actually saying. i know what i need to do with it, but dont quite grasp what this statement means


    thanks
     
  2. jcsd
  3. Sep 9, 2008 #2

    Hootenanny

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    [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex] are orthogonal unit vectors, the former is parallel to the x-axis and the latter is parallel to the y-axis. Explicitly, in Cartesian coordinates [itex]\hat{i} = \left(1,0\right)[/itex] and [itex]\hat{j} = \left(0,1\right)[/itex].

    Does that make sense?
     
  4. Sep 9, 2008 #3
    It is convenient (it make the math easier) to be able to express a vector in terms of components which related to the coordinate system in use. For the Cartesian coordinate system, we write vectors in terms of components which are parallel to the coordinate axes (x,y,z). i hat is by definition a unit vector (magnitude 1 unit) parallel to the x axis. We use a unit vector because we can construct the x component of any vector by multiplying i hat by the magnitude of the x component of the vector.

    The same is true for j hat and (when necessary) k hat, except that they are parallel to the y and z axes, repspectively.
     
  5. Sep 9, 2008 #4
    so basically x and y. what exactly is meant by the unit vector. i think i get the it means it has a magnitude of one unit. but you lost me on the mulipltying them. im using them deal with time, displacment initial and final velocity. I have the questions answered as i had someone try to explain it a little last night. so i have the answer to what we want but not sure how we arrived at it.

    bc the initial velocity is 3.8 [tex]\hat{}i[tex] the velocity after ten seconds is 4.08 m/s[tex]\hat{}i[tex]+ .46 m/s [tex]\hat{}j[tex]
     
  6. Sep 9, 2008 #5

    ok here is what i know

    i-hat = going to the right
    j-hat= going up
    k-hat= is mostly used as hypotenuse which is usulay find by using Pythagorean equation.

    so (i-hat)^2 + (j-hat)^2 = (k-hat)^2
     
  7. Sep 9, 2008 #6
    so if im looking at it just in a generic way, that is just the length of the line/ magnitude if its a vector
     
  8. Sep 9, 2008 #7

    Hootenanny

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    This is an extremely unorthodox use of notation and I have never seen it used.

    As I said in my previous post [itex]\hat{i}[/itex], [itex]\hat{j}[/itex] and [itex]\hat{k}[/itex] are mutually orthogonal (i.e. all perpendicular to each other) unit vectors. [itex]\hat{i}[/itex] is parallel to the x-axis, [itex]\hat{j}[/itex] is parallel to the y-axis and [itex]\hat{k}[/itex] is parallel to the z-axis.
     
  9. Sep 9, 2008 #8
    yes....... it just all about two dimensional motion.the only thing is you have to remember the vertical and horizontal components are separated.using i-hat and j-hat.i think they call this independent of vertical and horizontal motion.
     
  10. Sep 9, 2008 #9
    damn, i know this stuff should be easy, i think im thinking about it way to hard and confusing myself.
     
  11. Sep 9, 2008 #10
    ok then late say you have i-hat and j-hat and the question asked you to find k hat which is a magnitude ..........how do you gone find and what formula you gone use.
     
  12. Sep 9, 2008 #11

    Hootenanny

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    The easiest way I can put it is this. The unit vector gives the direction and the scalar coefficient gives the magnitude. For example, suppose we have a vector:

    [tex]\bold{v} = 2\hat{i} + 3\hat{j}[/tex]

    This means that to draw the vector (v) you would move 2 units along the x-axis and 3 units up the y-axis.
     
  13. Sep 9, 2008 #12

    Hootenanny

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    The point is the question wouldn't ask that because the standard definition of k-hat is for the unit vector parallel to the z-axis. k-hat is already in standard use and so wouldn't be used for such a quantity.
     
  14. Sep 9, 2008 #13
    sorry this is such a big picture, but this is what i have to work with.

    [​IMG]
    [​IMG]
    [​IMG]
     
  15. Sep 9, 2008 #14
    I will agree on that .
     
  16. Sep 9, 2008 #15

    Hootenanny

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    Jbright1406: So I'm guessing that your having trouble with part (c). Could you post your working?
     
  17. Sep 9, 2008 #16
    im at the library an ddont have it on the comp. im tryin to work it on paper but i think im plugging it it wrong.
    try this
    (.23*10+.5*.023*10^2)ihat + (0.46*10+.5*.046*10^2)jhat
    3.45 i hat + 6.9 jhat
     
  18. Sep 10, 2008 #17

    Hootenanny

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    I'm note sure where your pulling those numbers from, but note that the values in red should be the initial velocity of the boat before the acceleration.
     
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