Finding s for Parallel Vectors in R3

  • Thread starter Thread starter 75 johnson
  • Start date Start date
  • Tags Tags
    Parallel Vectors
AI Thread Summary
To determine the value of s that makes the vectors a=(2,2,-1) and b=(2,-1,s) parallel, the ratios of their components must be equal. The user finds that the ratio of the i components is 1, while the ratio of the j components is -1/2, indicating inconsistency. The goal is to find a value for s that aligns these ratios. The question is clarified as needing to find s such that both vectors are parallel and originate from the same point. The discussion emphasizes the need for consistent ratios among all vector components to establish parallelism.
75 johnson
Messages
4
Reaction score
0

Homework Statement



Let a=(2,2,-1) and let b=(2,-1,s)

The Attempt at a Solution



dividing the i components gives you 1 but then dividing the j components gives you -1/2. I know that the ratio has to be the same for i, j, k in order for the two vectors to be paralle. i need to find s so they are parrallel. am i not seeing something?
 
Physics news on Phys.org
hi 75 johnson! :smile:

what exactly is the question? :confused:
 
i need to find "s" so that a and b are parallel
 
but a and b are vectors which meet at the origin (0,0,0) :confused:
 
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