Timeline for Rotate a vector by a randomly oriented angle
Current License: CC BY-SA 3.0
7 events
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| Oct 2, 2017 at 10:37 | history | migrated | from physics.stackexchange.com (revisions) | ||
| Sep 30, 2017 at 16:49 | comment | added | Floris | I think $\cos(rand(-\pi/2,\pi/2))$ might get you part of the way; but you still need to decide on the sign of the other components or you will end up in just some quadrant - and that means generating more than one random number. I suppose that generating a random number between $-\pi$ and $\pi$ might suffice - you could then take the sine and the cosine to give you the two pieces you need. Maybe. | |
| Sep 30, 2017 at 16:44 | comment | added | occamsrazor | @Floris Yeah, you're right! This is so cool! Should it be sinusoidal? | |
| Sep 30, 2017 at 16:42 | comment | added | Floris | The length of an individual component of $w$ (which is the randomly generated vector) does not follow a uniform distribution. This method won't work. | |
| Sep 30, 2017 at 16:20 | comment | added | occamsrazor | Why not [0,1]? This is a bijection. It should give you what you want. | |
| Sep 30, 2017 at 16:10 | comment | added | Airidas Korolkovas | okay, but what is the distribution from which you would draw the new random component? It must be demonstrated that on average the cross product $\langle \mathbf{w} \times \mathbf{v} \rangle = 0$, i.e. the new vector is equally likely to fall anywhere on the cone at an angle $\beta$ to the old vector. | |
| Sep 30, 2017 at 15:56 | history | answered | occamsrazor | CC BY-SA 3.0 |