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NotTim

  • 2 years ago

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

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  1. NotTim
    • 2 years ago
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    I THINKING OF treating it as a parallel thing, but that's not correct, is it?

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

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

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

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

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

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

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

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

  10. TuringTest
    • 2 years ago
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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

  11. NotTim
    • 2 years ago
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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]?

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

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

  14. TuringTest
    • 2 years ago
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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

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

  16. TuringTest
    • 2 years ago
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    *make

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

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

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

  20. TuringTest
    • 2 years ago
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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)

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

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

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

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

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

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

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

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

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

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

  31. TuringTest
    • 2 years ago
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    hooray

  32. TuringTest
    • 2 years ago
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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

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

  34. TuringTest
    • 2 years ago
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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...

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

  36. TuringTest
    • 2 years ago
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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?

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

  38. TuringTest
    • 2 years ago
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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")

  39. NotTim
    • 2 years ago
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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

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

  41. TuringTest
    • 2 years ago
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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

  42. NotTim
    • 2 years ago
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    ok.

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

  44. TuringTest
    • 2 years ago
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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

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

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

  47. TuringTest
    • 2 years ago
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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

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

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

  50. TuringTest
    • 2 years ago
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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

  51. NotTim
    • 2 years ago
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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

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

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

  54. TuringTest
    • 2 years ago
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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

  55. NotTim
    • 2 years ago
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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.

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

  57. TuringTest
    • 2 years ago
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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

  58. NotTim
    • 2 years ago
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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

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

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

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

  62. TuringTest
    • 2 years ago
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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

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

  64. NotTim
    • 2 years ago
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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.

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

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

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