If X divides YZ, T divides XY, write OT as linear comb. of OX & OY

  • Thread starter Thread starter Mtl
  • Start date Start date
  • Tags Tags
    Linear
Mtl
Messages
6
Reaction score
0
Ok so I have an assignment for my Calculas/Geometry class and I seem to have not paid the greatest attention in class as I'm pretty much completely lost. I was wondering if anyone could help me with one of the questions.

The first part asked for a diagram which satified the following X divides YZ in a ration of
-5:7 and also that T divides XY in the ratio 1:3. (I'm pretty sure I have correctly completed this part)

The next part states that if O is not on the line, write OT as a linear combination of OX and OY.

Also it says to write OY as a linear combination of OT and OZ.
 
Physics news on Phys.org
Is that really a ratio of negative 5 to 7? Since lengths are never negative, how did you show that? Assuming that is 5:7, you have a line segment, YZ, divided into 5+ 7= 12 parts and X is 5 of those parts from Y. To put T dividing XY in the ratio 1:3, you would need to divide the segment XY into 4 parts. To simplify things, divide the original YZ into 4(12)= 48 parts and mark X 4(5)= 20 parts from Y (and so 4(7)= 28 parts from Z: 20:28= 5:7). Now mark T 5 parts from X, so 20-5= 15 parts from Y: 5:15= 1:3.

Now assume you have some "origin", O, not on that line. Think of OT as the vector from O to T, OX as the vector from O to X, and OY as the vector from O to Y. Now, you can write the various vectors as sums and difference of the others- follow along on your diagram. OT= OY+ YT and OT= OX- XT. Adding those equations, 2OT= OX+ OY+ YT- XT. Since T divides XY in the ratio 1:3, OX= (1/4)XY and OY= (3/4)XY so YT- XT= (3/4- 1/4)XY= (1/2)XY. That is, 2OT= OX+ OY+ (1/2)XY. XY= OX- OY, so (1/2)XY= (1/2)OX- (1/2)OY and we have 2OT= OX+ OY+ (1/2)(OX - OY)= (3/2)OX+ (1/2)OY. Dividing through by 2,OT= (3/4)OX+ (1/4)OY.

Now, can you do "OY as a linear combination of of OT and OZ?
 
"Is that really a ratio of negative 5 to 7?"

Yes, the negative affects the direction, or so my textbook says.

Anyways the first answer would look something along the lines of
X__T____Y__Z

Although, through reading through what you have read I'm pretty sure i can finish the problem, so thank you for your help.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

Similar threads

Replies
10
Views
5K
Replies
1
Views
3K
Replies
1
Views
4K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
39
Views
6K
Replies
1
Views
2K
Replies
1
Views
3K
Back
Top