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TrojanPoem

  • one year ago

Vectors

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  1. TrojanPoem
    • one year ago
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    ABC is a triangle : in which \[m \in AB\] , \[n \in AC\] as \[\frac{ Am }{ mB } = \frac{ Cm }{ Am } = \frac{ 2 }{ 3 }\] if Am x mn = AC x (kBC) so K = .....

  2. UsukiDoll
    • one year ago
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    O_O .. um... m belongs to point AB ... n belongs to point AC. WHAT IS THIS? T__T!

  3. dan815
    • one year ago
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    is it really Cm/Am or Cn/An?

  4. TrojanPoem
    • one year ago
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    |dw:1435133234099:dw|

  5. dan815
    • one year ago
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    okay and is that x cross or multiplication

  6. TrojanPoem
    • one year ago
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    It's cn / an Sorry dan.

  7. TrojanPoem
    • one year ago
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    I rechecked it from the book

  8. dan815
    • one year ago
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    okay

  9. dan815
    • one year ago
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    now is that multiplication or cross product?

  10. TrojanPoem
    • one year ago
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    cross product

  11. TrojanPoem
    • one year ago
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    But it's Am/Cm = Cn/nA so BC isn't parrell to BC

  12. dan815
    • one year ago
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    oh na okay

  13. dan815
    • one year ago
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    |dw:1435133638451:dw|

  14. dan815
    • one year ago
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    |dw:1435133683594:dw|

  15. dan815
    • one year ago
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    <ABC doesnt have to be 90 degrees that shud be noted

  16. dan815
    • one year ago
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    |dw:1435134020736:dw|

  17. dan815
    • one year ago
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    if they are equal that means their normal is in the same direction atleast

  18. dan815
    • one year ago
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    well thats a given xD

  19. dan815
    • one year ago
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    oh wait that angle theta 1 is wrong

  20. dan815
    • one year ago
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    |dw:1435134304911:dw|

  21. dan815
    • one year ago
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    |dw:1435134470362:dw|

  22. Michele_Laino
    • one year ago
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    here is my reasoning: we start from this condition: \[AM \times MN = AC \times kBC\] we can note this: \[AC = CN + AN = \frac{2}{3}AN + AN = \frac{5}{3}AN\] furthermore we can write: \[\begin{gathered} AB + BC + CA = 0 \hfill \\ AB + BC = AC \hfill \\ \end{gathered} \] being \[AM = \frac{2}{3}MB\] by substitution, we can write: \[BC = \frac{5}{3}\left( {AN - MB} \right)\] Now I substitute all that into the condition: \[AM \times MN = AC \times kBC\] and I get: \[\begin{gathered} \frac{2}{3}MB \times MN = \frac{5}{3}AN \times \frac{5}{3}k\left( {AN - MB} \right) \hfill \\ \hfill \\ \frac{2}{3}MB \times MN = - \frac{{25k}}{9}AN \times MB \hfill \\ \hfill \\ \frac{2}{3}MB \times MN = \frac{{25k}}{9}MB \times AN \hfill \\ \end{gathered} \] then by comparison, we got: \[k = \frac{6}{{25}}\]

  23. TrojanPoem
    • one year ago
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    Thanks, right k = 6/25 ( book final answer)

  24. TrojanPoem
    • one year ago
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    Thanks all of you for the help.

  25. dan815
    • one year ago
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    nice

  26. Michele_Laino
    • one year ago
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    :)

  27. dan815
    • one year ago
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    how come he got AC = CN + AN

  28. dan815
    • one year ago
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    is that just a typo

  29. dan815
    • one year ago
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    ya probably nvm

  30. dan815
    • one year ago
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    @Michele_Laino hey i never know when its okay to distribute and stuff with cross products, how do you think of it?

  31. TrojanPoem
    • one year ago
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    I am trying to understand it too.

  32. Michele_Laino
    • one year ago
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    \[\begin{gathered} AB + BC + CA = 0 \hfill \\ AB + BC = - CA = AC \hfill \\ \end{gathered} \]

  33. dan815
    • one year ago
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    this rule A x (B+C) = A x B + A x B is there an intuition or some easy way to see this

  34. dan815
    • one year ago
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    er A x B + A x C

  35. TrojanPoem
    • one year ago
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    michele_Laino how did you get 5/3 (AN - MB)

  36. Michele_Laino
    • one year ago
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    it is a property of product vector

  37. Michele_Laino
    • one year ago
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    here are my steps: \[AC = NC + AN = \frac{2}{3}AN + AN = \frac{5}{3}AN\] \[\begin{gathered} AM + MB + BC = \frac{5}{3}AN \hfill \\ \hfill \\ \frac{2}{3}MB + MB + BC = \frac{5}{3}AN \hfill \\ \hfill \\ \frac{5}{3}MB + BC = \frac{5}{3}AN \hfill \\ \end{gathered} \]

  38. Michele_Laino
    • one year ago
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    these ratios: \[\frac{{AM}}{{MB}} = \frac{2}{3} = \frac{{CN}}{{AN}}\] are ratios between lengths not between vectors, since a ratio between vectors is not defined

  39. TrojanPoem
    • one year ago
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    AM + MB = AB right ? we have AM/MB = 2/3 we can say AM +MB / MB = 5/3 ( right ?) so we have AB / MB = 5/3 by substituting AB/AN = 5/3 SO AB = 5/3 AN spot the mistake

  40. Michele_Laino
    • one year ago
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    \[AB = \frac{5}{3}MB\] it is right!

  41. TrojanPoem
    • one year ago
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    But in your answer AM + MB + BC = 5/3 AN with this logic AB is not equal to 5/3 MB

  42. TrojanPoem
    • one year ago
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    Otherwise MB = 0

  43. TrojanPoem
    • one year ago
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    and there is no triagnel

  44. Michele_Laino
    • one year ago
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    yes! we have: \[BC = \frac{5}{3}\left( {AN - MB} \right)\]

  45. Michele_Laino
    • one year ago
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    |dw:1435136410679:dw|

  46. Michele_Laino
    • one year ago
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    or equivalently, it comes from affine geometry

  47. TrojanPoem
    • one year ago
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    AB + BC + CA = 0 5/3 AN + BC + 5/3 AN = 0 BC = -10/3 AN ? right

  48. Michele_Laino
    • one year ago
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    no, since: \[AB = \frac{5}{3}MB\]

  49. Michele_Laino
    • one year ago
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    and: \[AC = - CA = \frac{5}{3}AN\]

  50. TrojanPoem
    • one year ago
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    yeah that's my problem am/mb = cn/na = 2/3 we can do am+mb/mb ( with the proportion ) so we have am+mb/mb = 2+3/3 ab/mb = 5/3 cn + na / na = 5/3 ca/na = 5/3 we can do exchange so ca/mb = ab/na = 5/3 so ab = 5/3 na

  51. Michele_Laino
    • one year ago
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    I understand your question

  52. TrojanPoem
    • one year ago
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    hey , what am I doing wrong ?

  53. Michele_Laino
    • one year ago
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    I'm pondering...

  54. Michele_Laino
    • one year ago
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    I think that your procedure involve only vector lengths, nevertheless, we have to keep care when we substitute into the starting equation: \[AM \times MN = AC \times kBC\] since that is a vector equation, which involves vectors and not lengths or real numbers

  55. TrojanPoem
    • one year ago
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    So properties of proportion can't be applied here ?

  56. Michele_Laino
    • one year ago
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    no, it can be applied, nevertheless you have to keep in mind that you are working with vectors

  57. dan815
    • one year ago
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    ya unless u are incorporating angles too, but vector additions do that

  58. Michele_Laino
    • one year ago
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    better is don't apply that property, and to try to solve your problem using vectors relationships only

  59. TrojanPoem
    • one year ago
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    O.k, so I concluded that I don't understand vectors properly. I will try to solve it with vectors relationships although using the properties eases everything ! Thanks for your time , Good Luck.

  60. Michele_Laino
    • one year ago
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    thanks! :)

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