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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]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.

  2. jcsd
  3. Jul 23, 2012 #2


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    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.
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