## TrojanPoem one year ago Vectors

1. TrojanPoem

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

O_O .. um... m belongs to point AB ... n belongs to point AC. WHAT IS THIS? T__T!

3. dan815

is it really Cm/Am or Cn/An?

4. TrojanPoem

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5. dan815

okay and is that x cross or multiplication

6. TrojanPoem

It's cn / an Sorry dan.

7. TrojanPoem

I rechecked it from the book

8. dan815

okay

9. dan815

now is that multiplication or cross product?

10. TrojanPoem

cross product

11. TrojanPoem

But it's Am/Cm = Cn/nA so BC isn't parrell to BC

12. dan815

oh na okay

13. dan815

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14. dan815

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15. dan815

<ABC doesnt have to be 90 degrees that shud be noted

16. dan815

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17. dan815

if they are equal that means their normal is in the same direction atleast

18. dan815

well thats a given xD

19. dan815

oh wait that angle theta 1 is wrong

20. dan815

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21. dan815

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22. Michele_Laino

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

Thanks, right k = 6/25 ( book final answer)

24. TrojanPoem

Thanks all of you for the help.

25. dan815

nice

26. Michele_Laino

:)

27. dan815

how come he got AC = CN + AN

28. dan815

is that just a typo

29. dan815

ya probably nvm

30. dan815

@Michele_Laino hey i never know when its okay to distribute and stuff with cross products, how do you think of it?

31. TrojanPoem

I am trying to understand it too.

32. Michele_Laino

$\begin{gathered} AB + BC + CA = 0 \hfill \\ AB + BC = - CA = AC \hfill \\ \end{gathered}$

33. dan815

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

er A x B + A x C

35. TrojanPoem

michele_Laino how did you get 5/3 (AN - MB)

36. Michele_Laino

it is a property of product vector

37. Michele_Laino

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

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

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

$AB = \frac{5}{3}MB$ it is right!

41. TrojanPoem

But in your answer AM + MB + BC = 5/3 AN with this logic AB is not equal to 5/3 MB

42. TrojanPoem

Otherwise MB = 0

43. TrojanPoem

and there is no triagnel

44. Michele_Laino

yes! we have: $BC = \frac{5}{3}\left( {AN - MB} \right)$

45. Michele_Laino

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46. Michele_Laino

or equivalently, it comes from affine geometry

47. TrojanPoem

AB + BC + CA = 0 5/3 AN + BC + 5/3 AN = 0 BC = -10/3 AN ? right

48. Michele_Laino

no, since: $AB = \frac{5}{3}MB$

49. Michele_Laino

and: $AC = - CA = \frac{5}{3}AN$

50. TrojanPoem

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

52. TrojanPoem

hey , what am I doing wrong ?

53. Michele_Laino

I'm pondering...

54. Michele_Laino

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

So properties of proportion can't be applied here ?

56. Michele_Laino

no, it can be applied, nevertheless you have to keep in mind that you are working with vectors

57. dan815

ya unless u are incorporating angles too, but vector additions do that

58. Michele_Laino

better is don't apply that property, and to try to solve your problem using vectors relationships only

59. TrojanPoem

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

thanks! :)