Vector geometry, some problems.

  • Thread starter Thread starter Gramsci
  • Start date Start date
  • Tags Tags
    Geometry Vector
AI Thread Summary
The discussion revolves around solving two vector geometry problems related to a parallelogram ABCD. The first problem involves expressing the vector AG as a linear combination of vectors u (AB) and v (AD). The second problem requires demonstrating that the area of triangle AFG is a consistent fractional part of the area of parallelogram ABCD and determining that fractional part. Participants suggest a step-by-step approach to find the necessary vectors and points. The conversation emphasizes breaking down the problems into manageable parts for clarity and understanding.
Gramsci
Messages
64
Reaction score
0
I have two problems I need help with.

Homework Statement


ABCD is a parallellogram. E is the midpoint on BC. F is on AD so that |DF| = 8|AF|. G is the intersection between AE and BF.
a) Express the vector AG as a linear combination of u = AB and v = AD.
b) Show that the area of the triangle AFG always is the same fractional part of ABCD. Find that fractional part!

2.


Homework Equations


-


The Attempt at a Solution


I have no idae. Any help would be nice.
 
Physics news on Phys.org
Hi Gramsci! :wink:

Do it step by step …

i] what is AE (in terms of u and v)?

ii] what is AF?

iii] so what is a general point on AE, and on BF? :smile:
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top