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

1. Jul 20, 2012

### skydave

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:

It is clear that a solution is not always defined for the whole range of $\nu$ and that for a given $\nu$ 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 $1-cos^2$) to be there as well.

The way I tried to approach it was to find an expression for $x$ and $z$:

$x=\sqrt{1-\mu_s^2-z^2}$

$z = \frac{\nu-xv_x-\mu_s y}{v_z}$

and substitute z within the expression of $x$ which after some massaging gives me:

$x^2 - \frac{2\nu \mu_s v_y v_x}{v_z^2 - v_x^2}x - \frac{(1-\mu_s^2)v_z^2 + \nu + \mu_s v_y}{v_z^2 - v_x^2}=0$

I get $x_1$ and $x_2$ from solving the quadratic formula and insert it into the formula I got for $z$. The problem here is that I need to divide by $v_z$ which means there is no solution if $v$ lies in the $xy$ 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. Jul 23, 2012

### chiro

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.