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TrojanPoem
 one year ago
Vectors
TrojanPoem
 one year ago
Vectors

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0ABC 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
 one year ago
Best ResponseYou've already chosen the best response.0O_O .. um... m belongs to point AB ... n belongs to point AC. WHAT IS THIS? T__T!

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is it really Cm/Am or Cn/An?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435133234099:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay and is that x cross or multiplication

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0It's cn / an Sorry dan.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I rechecked it from the book

dan815
 one year ago
Best ResponseYou've already chosen the best response.0now is that multiplication or cross product?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But it's Am/Cm = Cn/nA so BC isn't parrell to BC

dan815
 one year ago
Best ResponseYou've already chosen the best response.0<ABC doesnt have to be 90 degrees that shud be noted

dan815
 one year ago
Best ResponseYou've already chosen the best response.0if they are equal that means their normal is in the same direction atleast

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh wait that angle theta 1 is wrong

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3here 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, right k = 6/25 ( book final answer)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Thanks all of you for the help.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0how come he got AC = CN + AN

dan815
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino hey i never know when its okay to distribute and stuff with cross products, how do you think of it?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to understand it too.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\begin{gathered} AB + BC + CA = 0 \hfill \\ AB + BC =  CA = AC \hfill \\ \end{gathered} \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0this rule A x (B+C) = A x B + A x B is there an intuition or some easy way to see this

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0michele_Laino how did you get 5/3 (AN  MB)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is a property of product vector

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3here 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
 one year ago
Best ResponseYou've already chosen the best response.3these 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
 one year ago
Best ResponseYou've already chosen the best response.0AM + 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
 one year ago
Best ResponseYou've already chosen the best response.3\[AB = \frac{5}{3}MB\] it is right!

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But in your answer AM + MB + BC = 5/3 AN with this logic AB is not equal to 5/3 MB

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0and there is no triagnel

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes! we have: \[BC = \frac{5}{3}\left( {AN  MB} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435136410679:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3or equivalently, it comes from affine geometry

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0AB + BC + CA = 0 5/3 AN + BC + 5/3 AN = 0 BC = 10/3 AN ? right

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3no, since: \[AB = \frac{5}{3}MB\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3and: \[AC =  CA = \frac{5}{3}AN\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0yeah 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
 one year ago
Best ResponseYou've already chosen the best response.3I understand your question

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0hey , what am I doing wrong ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.0So properties of proportion can't be applied here ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3no, it can be applied, nevertheless you have to keep in mind that you are working with vectors

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya unless u are incorporating angles too, but vector additions do that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3better is don't apply that property, and to try to solve your problem using vectors relationships only

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0O.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.
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