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anonymous

  • 5 years ago

Does anyone know Unit vectors and length stuff really well? I'm boggled..

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  1. amistre64
    • 5 years ago
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    somewhat really well, but kinda sorta maybe like.... sure :)

  2. amistre64
    • 5 years ago
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    i and j are unit vectors.... i = <1,0>

  3. anonymous
    • 5 years ago
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    Hahaha, okay. I think I understand how to find length of a vector- that's pretty straight forward. How can I use that kind of math to find a vector parallel to another?

  4. amistre64
    • 5 years ago
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    since vectors arent tied to any specific spot, any vector with the same direction is parallel; if they are in the same direction and of the same magnitude they are equal vectors

  5. amistre64
    • 5 years ago
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    if the "slope" of the vectors are equal, they are parallel right?

  6. anonymous
    • 5 years ago
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    Yeah, okay. So if I am given a vector in R^4, I solve for it's length, and then what?

  7. anonymous
    • 5 years ago
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    Yeah, totally got that.

  8. amistre64
    • 5 years ago
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    <4,7> is parallel to <12,21>

  9. anonymous
    • 5 years ago
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    Agreed.

  10. amistre64
    • 5 years ago
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    r^4?.... 4d stuff eh..

  11. anonymous
    • 5 years ago
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    Yeah. That doesn't really matter though, does it? length is length.

  12. amistre64
    • 5 years ago
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    distance follows the same rules into other dimensions; you just keep tacking on the extra stuff under the radical.

  13. anonymous
    • 5 years ago
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    I'm given a solution here, and they just took 1/(length)*(my original vector)

  14. amistre64
    • 5 years ago
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    \[distance of R^n = \sqrt{x^2 + y^2 + z^2 + ......n^2}\]

  15. anonymous
    • 5 years ago
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    Yep.

  16. anonymous
    • 5 years ago
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    So them multiplying by the reciprocal of my length... Is that just them essentially 'rescaling' my vector, giving me one that is parallel (since it is basically underneath my original? It's on the same spot, but just not as long)?

  17. amistre64
    • 5 years ago
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    not to sure about the answer to that.... its been awhile since I got down and dirty with vectors...

  18. anonymous
    • 5 years ago
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    Does that make sense, though? If I took the vector <1, 0, 0, 0> and multiplied it by two, I would get <2, 0, 0, 0>, which is parallel.

  19. amistre64
    • 5 years ago
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    that makes sense yes, its magnitude has increased but its direction is the same. it is straight in the air if i see it correctly

  20. anonymous
    • 5 years ago
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    Right. But then why would they choose to multiply it by 1/(length)? there must be some significance..

  21. amistre64
    • 5 years ago
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    must be, but I would have to review the material that you are getting it from to understand it better :/

  22. anonymous
    • 5 years ago
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    Darn. Thanks though!

  23. amistre64
    • 5 years ago
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    youre welcome..wish i could have been more helpful :)

  24. anonymous
    • 5 years ago
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    It's strange too, since they are telling me to check whether the vector I got had the length of 1...

  25. amistre64
    • 5 years ago
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    all unit vectors are of length "1". I recall that much :)

  26. anonymous
    • 5 years ago
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    Maybe that's why.. Haha, thanks. Stay tuned. I'm sure I'll be back soon! ;)

  27. amistre64
    • 5 years ago
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    and all vectors can be expressed as a sum of their unit vectors and scalars right?

  28. anonymous
    • 5 years ago
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    Sheeeeeeeeesh. They wanted it as a unit vector parallel to the original....

  29. anonymous
    • 5 years ago
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    Yeah, as a linear combination. Gottttt it!

  30. anonymous
    • 5 years ago
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    Thanks!

  31. amistre64
    • 5 years ago
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    yay!!

  32. anonymous
    • 5 years ago
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    How do I become your fan? Lol

  33. amistre64
    • 5 years ago
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    you might have to hit f5 to refresh your browser; but if you want there is a "become a fan" to the side of my name.

  34. anonymous
    • 5 years ago
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    :D

  35. amistre64
    • 5 years ago
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    lol...thanx ;)

  36. anonymous
    • 5 years ago
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    Are you savvy with dot products?

  37. amistre64
    • 5 years ago
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    sorta savvy; <xa,ya>.<xb,yb> = xa xb + ya yb right?

  38. anonymous
    • 5 years ago
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    Yeah

  39. anonymous
    • 5 years ago
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    For this question, I have (xv) . v And the answer is 2x.

  40. anonymous
    • 5 years ago
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    They give me the length of v to be √2

  41. amistre64
    • 5 years ago
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    I recall multiplication not being associative with vectors for some odd reason.

  42. anonymous
    • 5 years ago
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    Is that just since SQRT(2)*SQRT(2) = 2?

  43. anonymous
    • 5 years ago
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    Maybe..

  44. amistre64
    • 5 years ago
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    The associative property is meaningless for the dot product because is not defined since is a scalar and therefore cannot itself be dotted. However, it does satisfy the property (rX).Y = r(X.Y) for a scalar.

  45. amistre64
    • 5 years ago
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    The associative property is meaningless for the dot product because (a.b).c is not defined since a.b is a scalar and therefore cannot itself be dotted. However, it does satisfy the property

  46. amistre64
    • 5 years ago
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    i know this means something lol

  47. anonymous
    • 5 years ago
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    Okay... so. That gives me X(V.V)=2X, when ||V||=√2

  48. anonymous
    • 5 years ago
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    What's the significance of a vector dotted with itself?

  49. anonymous
    • 5 years ago
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    WAIT

  50. amistre64
    • 5 years ago
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    lets see :) <1,2>.<1,2> 1(1) + 2(2) = 2+4 = 2(1+2)

  51. amistre64
    • 5 years ago
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    1+2 = 1i + 2j right?

  52. anonymous
    • 5 years ago
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    If ||V|| = √2, and ||V||= √(V.V), then (V.V)= (||V||)^2!

  53. anonymous
    • 5 years ago
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    Where ||V|| is the length of V.

  54. amistre64
    • 5 years ago
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    making sense for you :)

  55. anonymous
    • 5 years ago
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    But not for you? :P

  56. amistre64
    • 5 years ago
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    little bit, but it aint my question lol.

  57. anonymous
    • 5 years ago
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    hahahaha, fair nuff!

  58. amistre64
    • 5 years ago
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    if the length is sqrt(2) then yes, twice the length = 2

  59. anonymous
    • 5 years ago
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    Nooooo. Twice the length is 2√2. Lol. √2 squared is 2.

  60. amistre64
    • 5 years ago
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    thats better :) been staring at math too long

  61. anonymous
    • 5 years ago
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    Lol. Bedtime!

  62. anonymous
    • 5 years ago
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    I think you need to be my fan now..

  63. anonymous
    • 5 years ago
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    We all just want to be loved, don't we? Hahahahaha

  64. amistre64
    • 5 years ago
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    what? these aint free ya know ;)

  65. anonymous
    • 5 years ago
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    Pfffffft

  66. anonymous
    • 5 years ago
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    Just trying to get some street cred, yo.

  67. amistre64
    • 5 years ago
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    yeah, pupil and 0 is soooo unbecoming these days :)

  68. anonymous
    • 5 years ago
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    ;)

  69. amistre64
    • 5 years ago
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    I do actually have to head out, library is closing up soon... Ciao :)

  70. anonymous
    • 5 years ago
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    Adios amigo!

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