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caozeyuan

  • one year ago

True or false (give a reason if true or a counterexample if false): (a) If u is perpendicular (in three dimensions) to v and w, those vectors v and w are parallel. " (b) If u is perpendicular to v and w, then u is perpendicular to v + 2 w, (c) If u and v are perpendicular unit vectors then II u - v" = ,.,fi, g

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  1. caozeyuan
    • one year ago
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    (c) if u and v are perpendicular unit vectors then \[\left| u-v \right|\]=\[\sqrt{2}\]

  2. caozeyuan
    • one year ago
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    @terenzreignz , hey the 99 guy there! do you have any idea?

  3. terenzreignz
    • one year ago
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    Which one are you currently working on?

  4. caozeyuan
    • one year ago
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    I'm working on b. (a) solved

  5. terenzreignz
    • one year ago
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    My gut feel says no...

  6. caozeyuan
    • one year ago
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    counterexample needed then

  7. terenzreignz
    • one year ago
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    What if u is not perpendicular to w?

  8. caozeyuan
    • one year ago
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    but the question said that!

  9. terenzreignz
    • one year ago
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    Wait nvm... misread it... recalculating..

  10. terenzreignz
    • one year ago
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    okay, it's true, then :P

  11. terenzreignz
    • one year ago
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    What's a convenient way to find if two vectors are perpendicular?

  12. caozeyuan
    • one year ago
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    u.v=0! ofcuz!

  13. terenzreignz
    • one year ago
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    Here's the <ehem> convenient truth... if u is perpendicular to v, then... \[\Large \vec u \cdot \vec v = 0\] Yup

  14. terenzreignz
    • one year ago
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    Same goes for w. What then bodes for... \[\Large \vec u \cdot(\vec v + 2\vec w)\] :P

  15. terenzreignz
    • one year ago
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    Well, you get the idea :P

  16. terenzreignz
    • one year ago
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    If you don't (you probably do, but anyway...) remember that dot product distributes over addition...

  17. caozeyuan
    • one year ago
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    so how about (c) . I think it should be True, but I cannot give a reason

  18. terenzreignz
    • one year ago
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    (c) doesn't make sense, I think you didn't copy it correctly :)

  19. terenzreignz
    • one year ago
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    Oh, I saw it just now... derp :3

  20. terenzreignz
    • one year ago
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    Tough one :D

  21. terenzreignz
    • one year ago
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    Okay, got it :P

  22. terenzreignz
    • one year ago
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    Sorry it took so long, I didn't feel like getting a piece of paper XD

  23. terenzreignz
    • one year ago
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    Let's let \[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_u \vec k \]

  24. terenzreignz
    • one year ago
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    crud, I meant \[\Large \vec v= x_v\vec i + y_v\vec j + z_{\color{red}v}\vec k\]

  25. caozeyuan
    • one year ago
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    I got an idea! It is given that u and v are unit vectors so I can say that they are ( 1,0,0) and (0,1,0) respectively. the difference gives the vector(1,-1,0) and the length of that vector is clearly sqrt(2)

  26. terenzreignz
    • one year ago
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    Yes, but that's a specific case... you're supposed to prove it for *any* unit vector...

  27. terenzreignz
    • one year ago
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    Things are not that simple, but they are... rather elegant... Let's work out the magic, shall we? :)

  28. terenzreignz
    • one year ago
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    \[\Large \vec u = x_u\vec i + y_u\vec j + z_u\vec k\]\[\Large \vec v= x_v\vec i + y_v\vec j + z_v \vec k\] Just tell me when you're ready .. ;)

  29. caozeyuan
    • one year ago
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    Yep, I understand that. move on

  30. terenzreignz
    • one year ago
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    Someone's bossy :/

  31. terenzreignz
    • one year ago
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    Okay, we know u and v are unit vectors, which can only mean... \[\Large x_u^2 + y_u^2 + z_u^2 = 1 = x_v^2 + y_v^2+z_v^2\]

  32. terenzreignz
    • one year ago
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    Right? Their magnitudes are equal to 1.

  33. caozeyuan
    • one year ago
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    so that gives sqrt(1^2+1^2)=sqrt(2), right?

  34. terenzreignz
    • one year ago
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    How did you figure that? -.-

  35. terenzreignz
    • one year ago
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    I like your initiative, but patience... is a virtue :)

  36. terenzreignz
    • one year ago
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    And no, your reasoning is faulty ;)

  37. terenzreignz
    • one year ago
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    So, why don't we acquire u - v... \[\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k\]

  38. terenzreignz
    • one year ago
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    Catch me so far?

  39. caozeyuan
    • one year ago
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    My brain explodes when things get too complicated

  40. terenzreignz
    • one year ago
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    Well, there is no avoiding this, so brace your brain :) Don't worry, it's not going to be THAT MUCH complicated... I just need to know that you understand everything up to here \[\Large \vec u - \vec v =(x_u - x_v)\vec i + (y_u-y_v)\vec j + (z_u - z_v )\vec k\]so far... do you?

  41. caozeyuan
    • one year ago
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    Yes, I'm following

  42. terenzreignz
    • one year ago
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    Let's get its magnitude... \[\Large ||\vec u - \vec v ||\]

  43. terenzreignz
    • one year ago
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    By definition, this is \[\Large =\sqrt{(x_u-x_v)^2+(y_u-y_v)^2+(z_u-z_v)^2}\]

  44. caozeyuan
    • one year ago
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    so we have to show it is indeed\[\sqrt{2}\]

  45. terenzreignz
    • one year ago
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    Yup :)

  46. terenzreignz
    • one year ago
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    Ready? The tricky part begins now...

  47. caozeyuan
    • one year ago
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    oh NO! Don't torture me!

  48. terenzreignz
    • one year ago
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    This is for your own good <force feeds you lima beans> Now, let's simplify them squares... \[\Large = \sqrt{x_u^2-2x_ux_v+x_v^2+y_u^2-2y_uy_v+y_v^2+z_u^2-2z_uz_v+z_v^2}\]

  49. terenzreignz
    • one year ago
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    Everything good so far? I only simplified the squares... you know... \[\large (a-b)^2 = a^2-2ab+b^2\]

  50. caozeyuan
    • one year ago
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    Yes, I understand so far

  51. terenzreignz
    • one year ago
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    I think, I shall rearrange the terms... \[\Large = \sqrt{x_u^2+y_u^2+z_u^2+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\]

  52. terenzreignz
    • one year ago
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    Are you okay with the rearrangement? Something you didn't get? :)

  53. caozeyuan
    • one year ago
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    Get it w/o prob.

  54. terenzreignz
    • one year ago
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    Okay, by hypothesis, since u is a unit vector, this part \[\Large = \sqrt{\color{red}{x_u^2+y_u^2+z_u^2}+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\] Is just \[\Large = \sqrt{\color{red}1+x_v^2+y_v^2+z_v^2-2x_ux_v-2y_uy_v-2z_uz_v}\] Are you following me so far? :)

  55. caozeyuan
    • one year ago
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    and the next three stuff becomes 1 as well?

  56. terenzreignz
    • one year ago
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    That is correct, since v is also a unit vector... \[\Large = \sqrt{\color{red}1+\color{green}1-2x_ux_v-2y_uy_v-2z_uz_v}\] \[\Large = \sqrt{\color{blue}2-2x_ux_v-2y_uy_v-2z_uz_v}\] Are we done here? :P

  57. terenzreignz
    • one year ago
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    Nah, of course not :) Allow me to factor out the -2 here... \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\]

  58. terenzreignz
    • one year ago
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    See anything familiar...?

  59. caozeyuan
    • one year ago
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    no, I don't see any familiar

  60. terenzreignz
    • one year ago
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    You're lacking creativity :P \[\Large = \sqrt{2-2\color{blue}{(x_ux_v+y_uy_v+z_uz_v)}}\] \[\Large = \sqrt{2-2\color{blue}{(\vec u \cdot \vec v)}}\]

  61. terenzreignz
    • one year ago
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    "aha!" moment?

  62. caozeyuan
    • one year ago
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    Oh my goodness! why am I so STUPID?!

  63. terenzreignz
    • one year ago
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    You're not... Just a little exhausted, maybe :) We haven't used the fact that they are perpendicular yet, and since they are, their dot product is zero :P \[\Large = \sqrt{2-2\color{blue}{(0)}}= \sqrt2\] tadaa

  64. caozeyuan
    • one year ago
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    ok, QED.

  65. terenzreignz
    • one year ago
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    Now say it... Say Terence is awesome :3

  66. caozeyuan
    • one year ago
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    NO! you are arrogant! LOL!

  67. terenzreignz
    • one year ago
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    Arrogant is not the opposite of awesome. Now say it >:)

  68. terenzreignz
    • one year ago
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    LOL JK have a nice day :3

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