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NotTim Group Title

Consider u=[2,3] and v=[-4,-6]. Find 2 unit vectors parallel to u.

  • 2 years ago
  • 2 years ago

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  1. NotTim Group Title
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    I THINKING OF treating it as a parallel thing, but that's not correct, is it?

    • 2 years ago
  2. TuringTest Group Title
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    \[{\vec u\over||\vec u||}\]makes it a unit vector, did you know that?

    • 2 years ago
  3. NotTim Group Title
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    what's that.

    • 2 years ago
  4. TuringTest Group Title
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    dividing by the norm/magnitude of the vector

    • 2 years ago
  5. FoolForMath Group Title
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    Unit vector parallel to \(\vec{u}\) is : \( \frac{1}{\sqrt{13}} <2,3> \)

    • 2 years ago
  6. NotTim Group Title
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    ok... I don't understand, but I don't think I really need further explaination @TuringTest

    • 2 years ago
  7. NotTim Group Title
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    Also, what did you type FFM? That looks...strange

    • 2 years ago
  8. FoolForMath Group Title
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    But there is one one unique vector (parallel) isn't ?

    • 2 years ago
  9. NotTim Group Title
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    uhhh...i dunno. Was just gunna treat this as collinear. NEver saw how to deal with parallels in my txtbook

    • 2 years ago
  10. TuringTest Group Title
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    FFM I think is pointing out that the other, \(\vec v\), is antiparallel I suppose... I just see it as i.e. magnitude is the length of the vector\[\vec u=<2,3>\] and the magnitude is\[||\vec u||=\sqrt{2^2+3^2}\]\(\vec v\) is parallel to \(\vec u\) so you should be able to do the same trick there

    • 2 years ago
  11. NotTim Group Title
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    I think the equation editor is messing around with us. I see squares everywhere. Also, maybe I should skip to the chase. How would I get [0.55,0,83],[-0.55,-0.83]?

    • 2 years ago
  12. TuringTest Group Title
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    what is the magnitude of u ?

    • 2 years ago
  13. NotTim Group Title
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    [2,3]

    • 2 years ago
  14. TuringTest Group Title
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    no, read what I wrote above magnitude is the \(length\) of the vector, which we can find with pythagorus\[||\vec u||=\sqrt{2^2+3^2}\] and the magnitude is

    • 2 years ago
  15. TuringTest Group Title
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    to amke a vector into a unit vector, divide by its magnitude

    • 2 years ago
  16. TuringTest Group Title
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    *make

    • 2 years ago
  17. NotTim Group Title
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    aww...what exactly is a uni vector. I though the point [2,3] was a vector itself...

    • 2 years ago
  18. TuringTest Group Title
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    A unit vector is a vector that has a length (i.e. "magnitude) of 1

    • 2 years ago
  19. NotTim Group Title
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    ok. um. gimme a moment to do the math here

    • 2 years ago
  20. TuringTest Group Title
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    so a vector \(\vec u=<x,y>\) is a unit vector iff\[||\vec u||=\sqrt{x^2+y^2}=1\]after dividing a vector by its magnitude, it should make sense that it will become a unit vector (you are essentially dividing a vector by it's length, which will logically give 1)

    • 2 years ago
  21. NotTim Group Title
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    wait, how do i divide [2,3] by sqrt 13?

    • 2 years ago
  22. TuringTest Group Title
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    do each component individually: it is a scalar multiplication

    • 2 years ago
  23. NotTim Group Title
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    im really fuzzy on this. I divide 2 by sqrt 13, and then 3 by sqrt 13 right?

    • 2 years ago
  24. TuringTest Group Title
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    yep, and those will be you new components

    • 2 years ago
  25. NotTim Group Title
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    umm.srry. pencil is breaking on me here...

    • 2 years ago
  26. TuringTest Group Title
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    try a calculator ;)

    • 2 years ago
  27. NotTim Group Title
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    im going to get a new pencil. bak in a flash

    • 2 years ago
  28. TuringTest Group Title
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    \[{2\over\sqrt{13}}\]just requires a calculator!!!!

    • 2 years ago
  29. NotTim Group Title
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    argh. srry. wat nxt now

    • 2 years ago
  30. NotTim Group Title
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    got [0.55,0.83]

    • 2 years ago
  31. TuringTest Group Title
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    hooray

    • 2 years ago
  32. TuringTest Group Title
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    now for the next one, do the exact same trick ...or use a shortcut (which takes a little explanation) and get the answer in 2 seconds whichever you prefer

    • 2 years ago
  33. NotTim Group Title
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    I guess you don't wanna explain both methods?

    • 2 years ago
  34. TuringTest Group Title
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    well they are really based on the same idea: multiplying a vector be a scalar will leave it parallel to the original...

    • 2 years ago
  35. NotTim Group Title
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    wat do yu mean "exact same trick"

    • 2 years ago
  36. TuringTest Group Title
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    actually that was a poor choice of phrasing; upi need to regognize what I'm about to say either way: let's look at \(\vec u\) and \(\vec v\)|dw:1333605852213:dw|let's say that \[\vec v=c\vec u\]then what does \(\vec v\) look like?

    • 2 years ago
  37. TuringTest Group Title
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    let \(c=2\) and we double the length of \(\vec u\):|dw:1333606018155:dw|notice that they remain parallel

    • 2 years ago
  38. TuringTest Group Title
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    here we are doing the same thing: \(\vec v\) is a scalar multiple of \(\vec u\) with \(c=-1\) this will change the direction of the vector (hence the term :anti-parallel")

    • 2 years ago
  39. NotTim Group Title
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    so wait, v and u are actually related? im suppose to use them both? i was just thinking of using u and random other vectrs

    • 2 years ago
  40. NotTim Group Title
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    oohhh...i see. scalar multitplier of -2.

    • 2 years ago
  41. TuringTest Group Title
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    |dw:1333606190135:dw|yeah, sorry my drawing is wrong, it should be -2 but note: that won't change the \(unit\) \(vector\) because all unit vectors have length 1 so all we need to do is notice that \(\vec v\) is a scalar multiple of \(\vec u\), so it has the same unit vector, but the - sign changes the direction it is still parallel though, in the sense that they never intersect

    • 2 years ago
  42. NotTim Group Title
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    ok.

    • 2 years ago
  43. NotTim Group Title
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    so, v is parallel to u. actually. wait. we already got the answer. thats magical.

    • 2 years ago
  44. TuringTest Group Title
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    ...another way to say it is that a unit vector has only one parallel vector to it: it's negative therefor no calculation is really needed. it's better to understand it the first way though, in my opinion

    • 2 years ago
  45. TuringTest Group Title
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    yeah you got it :)

    • 2 years ago
  46. NotTim Group Title
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    wait...what's the "first" way.

    • 2 years ago
  47. TuringTest Group Title
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    that \(\vec v\) is a (negative) scalar multiple of \(\vec u\), and therefor has the same unit vector in the opposite direction

    • 2 years ago
  48. NotTim Group Title
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    does that mean its parallel to u?

    • 2 years ago
  49. TuringTest Group Title
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    though sorry, now that I reread the question v isn't even important

    • 2 years ago
  50. TuringTest Group Title
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    but yes, as I said v is a scalar multiple of u when one vector is a scalar multiple of the other, they are parallel

    • 2 years ago
  51. NotTim Group Title
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    yeah. i kidna just included every thing related to the question. given the chance, i would include the background info by the text too

    • 2 years ago
  52. TuringTest Group Title
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    scalar multiples only change the length of the vector

    • 2 years ago
  53. TuringTest Group Title
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    hence they do not change whether or not they are parallel

    • 2 years ago
  54. TuringTest Group Title
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    but it only asks for two unit vectors perpendicular to \(\vec u\), so \(\vec v\) has nothing to do with it; it's coincidentally parallel but is not needed to find the answer

    • 2 years ago
  55. NotTim Group Title
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    so vector, wit ha scalar multiple of the original vector, is not actually parallel to it? man. we've taken up quite some time with this.

    • 2 years ago
  56. TuringTest Group Title
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    unit vectors parallel to*

    • 2 years ago
  57. TuringTest Group Title
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    I repeat: scalar multiples only change the length of the vector, hence they do not change whether or not they are parallel so in\[c\vec u=\vec v\] where \(c\) is a scalar, \(\vec u\) and \(\vec v\) are parallel

    • 2 years ago
  58. NotTim Group Title
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    is there a situation when the vectors are not parallel, but are only different due to a scalar multiplier. many question i give

    • 2 years ago
  59. TuringTest Group Title
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    no, unless there is some huge exception that is slipping my mind...

    • 2 years ago
  60. TuringTest Group Title
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    scalar multiplication of a vector changes only \(direction\) and/or \(magnitude\) (i.e. length)

    • 2 years ago
  61. NotTim Group Title
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    so i pick, say [3,2,6]. Then it is parallel to [6,4,12]?

    • 2 years ago
  62. TuringTest Group Title
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    yes. in higher math, like linear algebra, you will learn that vectors that can be written as scalar multiples of each other are called "linearly dependent" that means they are parallel, which is very important in setting up coordinate systems and such, which is what you do in linear algebra

    • 2 years ago
  63. TuringTest Group Title
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    look at it like slope

    • 2 years ago
  64. NotTim Group Title
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    ok. thank you very much. all my questions have been cleared up, im not done my review, and i wish to take a nappy.

    • 2 years ago
  65. TuringTest Group Title
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    ok then, glad I could help rest well, a weary mind is not terribly useful in mathematics ;)

    • 2 years ago
  66. NotTim Group Title
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    should i sleep now, and risk not studying later, or study, but not sleep ever?

    • 2 years ago
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