## Quaternion parallel or perpendicular to all basis vectors [closed]

Given an object's quaternion q, and basis vectors vx, vy, vz forming a 3D space, how can I check whether the quaternion is parallel or perpendicular to all of the basis vectors?

For example, I have basis vectors: vx = (0.447410, 0, -0.894329) vy = (0, 1, 0) vz = (0.894329, 0, 0.447410)

and quaternion q(w,x,y,z) = (-0.973224, 0, -0.229860, 0)

I know the quaternion is perpendicular or parallel (or anti-parallel) to all of the basis vectors but how can I actually calculate it?

Another example, q(w,x,y,z) = (0.823991, 0, 0.566602, 0)

This is NOT perpendicular or parallel (or anti-parallel) to all of the basis vectors.

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What do you mean by being parallel to "all of the basis vectors"? Do you mean that it belongs to their linear span? Take the following as constructive criticism, please. This question suffers from the following problems. Firstly, it is poorly written and not using proper mathematical language. Secondly, it is "too localised": you have not made the effort to ask a general question, instead writing some unsightly rational numbers which obscure more than illuminate. Finally, and this is the crucial one, it's probably not of interest to research mathematicians. I'm voting to close. – JosÃ© Figueroa-O'Farrill Aug 20 2010 at 8:22
Parallel OR perpendicular to all of the basis vectors. Of course it cannot be parallel at the same time to all of the basis vectors, but it can be parallel OR perpendicular to all of the basis vectors. – Quaternional Aug 20 2010 at 8:35
@Quaternional: in what way do you mean for a quaternion to be parallel or perpendicular to a 3-vector? The relationships of 'parallelism' and 'perpendicularity' hold for two elements of the same space. So your question currently does not make any sense. If you mean to embed R^3 into the quaternions, with the standard basis e1, e2, e3 identified with i,j,k, then you should explicitly say so. – Niel de Beaudrap Aug 20 2010 at 9:42
Yes, exactly so. – Quaternional Aug 20 2010 at 9:55
...wait, are you just using quaternions to mean R^4? I don't see the algebraic structure used anywhere. If you have an embedding of R^3 into R^4 with coordinate basis chosen so that the image of R^3 is just the linear span of three of the basis vectors AND you have expression of the vectors in coordinates, then you can just do the trivial thing and take dot products and be done with it. – Willie Wong Aug 20 2010 at 11:04
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