Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple linear algebra problem (points on circle for a given vector and angle)

  1. Jul 20, 2012 #1
    Hi all,

    I have a seemingly simple linear algebra problem which I have trouble with and I would like to ask for some advice how to solve it. Here is the problem:

    Here are my thoughts about this:

    It is clear that a solution is not always defined for the whole range of [itex]\nu[/itex] and that for a given [itex]\nu[/itex] there can be 2 solutions. So I expect a quadratic formula to pop up. Also since the solution is on the circle I expect a sinus of some angle (or [itex]1-cos^2[/itex]) to be there as well.

    The way I tried to approach it was to find an expression for [itex]x[/itex] and [itex]z[/itex]:

    [itex]x=\sqrt{1-\mu_s^2-z^2}[/itex]

    [itex]z = \frac{\nu-xv_x-\mu_s y}{v_z}[/itex]

    and substitute z within the expression of [itex]x[/itex] which after some massaging gives me:

    [itex]x^2 - \frac{2\nu \mu_s v_y v_x}{v_z^2 - v_x^2}x - \frac{(1-\mu_s^2)v_z^2 + \nu + \mu_s v_y}{v_z^2 - v_x^2}=0[/itex]

    I get [itex]x_1[/itex] and [itex]x_2[/itex] from solving the quadratic formula and insert it into the formula I got for [itex]z[/itex]. The problem here is that I need to divide by [itex]v_z[/itex] which means there is no solution if [itex]v[/itex] lies in the [itex]xy[/itex] plane. I don't understand why this is geometrically.

    However, the problem I now have is that if I check my solutions they don't lie on the given circle and the angle doesnt satisify the given constrain. So I must be doing something wrong and my hope is that somebody here can point me into the right direction. Does the overall approach make sense? Is there something I miss? Is there a different way to solve the problem?

    Any comment is much appreciated.

    Thanks,
    David
     
  2. jcsd
  3. Jul 23, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey skydave and welcome to the forums.

    The first thing to notice is that your first equation is not a circle if it's in R^3: it's a cylinder since the y variable is allowed to take on any real value. If you want a circle you need to fix y = a for some constant a. This will then effect the equations you have derived later.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple linear algebra problem (points on circle for a given vector and angle)
Loading...