Add random spread to directional vector

Question

Let's say I have a unit vector a = Vector(0,1,0) and I want to add a random spread of something between x = Vector(-0.2,0,-0.2) and y = Vector(0.2,0,0.2), how would I go about doing that?

If I were to simply generate a random vector between x and y, I'd get a value somewhere in the bounds of a square:

What I'd like instead is a value within the circle made up by x and y:

This seems like a simple problem but I can't figure out the solution right now. Any help would be appreciated.

(I didn't ask this on mathoverflow since this isn't really a 'research level mathematics question')

Answer 1

If I read your question correctly, you want a vector in a random direction that's within a particular length (the radius of your circle).

The formula for a circle is: x² + y² = r²

So, if you have a maximum radius, r, that constrains the vector length, perhaps proceed something like this:

1. Choose a random value for x, that lies between -r and +r
2. Calculate a limit for randomising y, based on your chosen x, so y_lim = sqrt(r² - x²)
3. Finally, choose a random value of y between -y_lim and +y_lim

That way, you get a random direction in x and a random direction in y, but the vector length will remain within 0 to r and so will be constrained within a circle of that radius.

In your example, it seems that r should be sqrt(0.2²) which is approximately 0.28284.

UPDATE

As 3D vector has length (or magnitude) sqrt(x²+y²+z²) you could extend the technique to 3D although I would probably favour a different approach (which would also work for 2D).

1. Choose a random direction by choosing any x, y and z
2. Calculate the magnitude m = sqrt(x²+y²+z²)
3. Normalise the direction vector (by dividing each element by its magnitude), so x = x/m, y = y/m, z=z/m
4. Now choose a random length, L between 0 and r
5. Scale the direction vector by the random length. So x = x * L, y = y * L, z = z * L