TrojanPoem
  • TrojanPoem
Vectors
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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TrojanPoem
  • 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 = .....
UsukiDoll
  • UsukiDoll
O_O .. um... m belongs to point AB ... n belongs to point AC. WHAT IS THIS? T__T!
dan815
  • dan815
is it really Cm/Am or Cn/An?

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TrojanPoem
  • TrojanPoem
|dw:1435133234099:dw|
dan815
  • dan815
okay and is that x cross or multiplication
TrojanPoem
  • TrojanPoem
It's cn / an Sorry dan.
TrojanPoem
  • TrojanPoem
I rechecked it from the book
dan815
  • dan815
okay
dan815
  • dan815
now is that multiplication or cross product?
TrojanPoem
  • TrojanPoem
cross product
TrojanPoem
  • TrojanPoem
But it's Am/Cm = Cn/nA so BC isn't parrell to BC
dan815
  • dan815
oh na okay
dan815
  • dan815
|dw:1435133638451:dw|
dan815
  • dan815
|dw:1435133683594:dw|
dan815
  • dan815
dan815
  • dan815
|dw:1435134020736:dw|
dan815
  • dan815
if they are equal that means their normal is in the same direction atleast
dan815
  • dan815
well thats a given xD
dan815
  • dan815
oh wait that angle theta 1 is wrong
dan815
  • dan815
|dw:1435134304911:dw|
dan815
  • dan815
|dw:1435134470362:dw|
Michele_Laino
  • 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: \[k = \frac{6}{{25}}\]
TrojanPoem
  • TrojanPoem
Thanks, right k = 6/25 ( book final answer)
TrojanPoem
  • TrojanPoem
Thanks all of you for the help.
dan815
  • dan815
nice
Michele_Laino
  • Michele_Laino
:)
dan815
  • dan815
how come he got AC = CN + AN
dan815
  • dan815
is that just a typo
dan815
  • dan815
ya probably nvm
dan815
  • dan815
@Michele_Laino hey i never know when its okay to distribute and stuff with cross products, how do you think of it?
TrojanPoem
  • TrojanPoem
I am trying to understand it too.
Michele_Laino
  • Michele_Laino
\[\begin{gathered} AB + BC + CA = 0 \hfill \\ AB + BC = - CA = AC \hfill \\ \end{gathered} \]
dan815
  • dan815
this rule A x (B+C) = A x B + A x B is there an intuition or some easy way to see this
dan815
  • dan815
er A x B + A x C
TrojanPoem
  • TrojanPoem
michele_Laino how did you get 5/3 (AN - MB)
Michele_Laino
  • Michele_Laino
it is a property of product vector
Michele_Laino
  • Michele_Laino
here are my steps: \[AC = NC + AN = \frac{2}{3}AN + AN = \frac{5}{3}AN\] \[\begin{gathered} AM + MB + BC = \frac{5}{3}AN \hfill \\ \hfill \\ \frac{2}{3}MB + MB + BC = \frac{5}{3}AN \hfill \\ \hfill \\ \frac{5}{3}MB + BC = \frac{5}{3}AN \hfill \\ \end{gathered} \]
Michele_Laino
  • 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
TrojanPoem
  • 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
Michele_Laino
  • Michele_Laino
\[AB = \frac{5}{3}MB\] it is right!
TrojanPoem
  • TrojanPoem
But in your answer AM + MB + BC = 5/3 AN with this logic AB is not equal to 5/3 MB
TrojanPoem
  • TrojanPoem
Otherwise MB = 0
TrojanPoem
  • TrojanPoem
and there is no triagnel
Michele_Laino
  • Michele_Laino
yes! we have: \[BC = \frac{5}{3}\left( {AN - MB} \right)\]
Michele_Laino
  • Michele_Laino
|dw:1435136410679:dw|
Michele_Laino
  • Michele_Laino
or equivalently, it comes from affine geometry
TrojanPoem
  • TrojanPoem
AB + BC + CA = 0 5/3 AN + BC + 5/3 AN = 0 BC = -10/3 AN ? right
Michele_Laino
  • Michele_Laino
no, since: \[AB = \frac{5}{3}MB\]
Michele_Laino
  • Michele_Laino
and: \[AC = - CA = \frac{5}{3}AN\]
TrojanPoem
  • 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
Michele_Laino
  • Michele_Laino
I understand your question
TrojanPoem
  • TrojanPoem
hey , what am I doing wrong ?
Michele_Laino
  • Michele_Laino
I'm pondering...
Michele_Laino
  • 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
TrojanPoem
  • TrojanPoem
So properties of proportion can't be applied here ?
Michele_Laino
  • Michele_Laino
no, it can be applied, nevertheless you have to keep in mind that you are working with vectors
dan815
  • dan815
ya unless u are incorporating angles too, but vector additions do that
Michele_Laino
  • Michele_Laino
better is don't apply that property, and to try to solve your problem using vectors relationships only
TrojanPoem
  • 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.
Michele_Laino
  • Michele_Laino
thanks! :)

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