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

[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:

Given is a circle on the unit sphere:

<br /> x^2 + \mu_s^2 + z^2 = 1<br />

where \mu_s is a known constant. So the circle lies in the x z plane at a given height.

Also given is an arbitrary vector v.

The problem is to find the vector function s(\nu) with \nu \in [-1,1] such that the dot product between the solution and vector v equals \nu:

<br /> xv_x + yv_y + zv_z = \nu<br />

Here are my thoughts about this:

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 doesn't 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

 

[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.