Vectors

- TrojanPoem

Vectors

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- 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

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

- dan815

is it really Cm/Am or Cn/An?

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## More answers

- TrojanPoem

|dw:1435133234099:dw|

- dan815

okay and is that x cross or multiplication

- TrojanPoem

It's cn / an Sorry dan.

- TrojanPoem

I rechecked it from the book

- dan815

okay

- dan815

now is that multiplication or cross product?

- TrojanPoem

cross product

- TrojanPoem

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

- dan815

oh na okay

- dan815

|dw:1435133638451:dw|

- dan815

|dw:1435133683594:dw|

- dan815

- dan815

|dw:1435134020736:dw|

- dan815

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

- dan815

well thats a given xD

- dan815

oh wait that angle theta 1 is wrong

- dan815

|dw:1435134304911:dw|

- dan815

|dw:1435134470362:dw|

- 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

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

- TrojanPoem

Thanks all of you for the help.

- dan815

nice

- Michele_Laino

:)

- dan815

how come he got AC = CN + AN

- dan815

is that just a typo

- dan815

ya probably nvm

- dan815

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

- TrojanPoem

I am trying to understand it too.

- Michele_Laino

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

- 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

er A x B + A x C

- TrojanPoem

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

- Michele_Laino

it is a property of product vector

- 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

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

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

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

- TrojanPoem

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

- TrojanPoem

Otherwise MB = 0

- TrojanPoem

and there is no triagnel

- Michele_Laino

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

- Michele_Laino

|dw:1435136410679:dw|

- Michele_Laino

or equivalently, it comes from affine geometry

- TrojanPoem

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

- Michele_Laino

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

- Michele_Laino

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

- 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

I understand your question

- TrojanPoem

hey , what am I doing wrong ?

- Michele_Laino

I'm pondering...

- 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

So properties of proportion can't be applied here ?

- Michele_Laino

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

- dan815

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

- Michele_Laino

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

- 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

thanks! :)

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