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Usually, there are two representations of a quaternion, say $\mathbf q = q_0+q_1 \mathbf i+q_2 \mathbf j+q_3 \mathbf k$ ,alternatively $\mathbf q = q_x \mathbf i+q_y \mathbf j+q_z \mathbf k + q_z$. Now, provide a quaternion of the form: $$[-0.5739 \quad 0.5831 \quad-0.4229 \quad0.3895].$$ what is the simplest way to distinguish its real part from the imaginary part?

Update
According to transformation from Axis Angle to Quaternion, given a quad of $[x\; y \; z \;w]$, normalize it to $[x'\; y'\; z'\; w']$, then evaluate the expression $$(x'^2 +y'^2+z'^2)/(sin^2(arccos(w')))$$ $w'$ is the real part if above expression equals 1, but it seemed the same result arose in any other numbers in the quad. Can anyone tell me what was going on?

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    $\begingroup$ I believe the first way you wrote it is standard $\endgroup$ Sep 9, 2019 at 12:50

2 Answers 2

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If all you know are the four real coefficients and you are not told what convention identifies which is which then there is no way to distinguish the real part.

I agree with @G.Chiusole that the first way is the most likely.

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As has already been mentioned, with no other information about this lone coordinate tuple or how it’s used, there’s no way to tell. That’s not really any different from ordinary vectors: given the coordinate tuple $(1,2,3)$ and nothing else, there’s no way to tell what vector it’s meant to represent other than it’s the linear combination $v_1+2v_2+3v_3$ of some ordered set of basis vectors.

As to your follow-up question, after normalization you’ve got a unit quaternion, so that $x'^2+y'^2+z'^2+w'^2=1$. The problem is that the four variables in this expression are entirely interchangeable. In particular, no matter which one you isolate as representing the real part or the quaternion, the above expression is equivalent to one of the form $a^2+b^2=1$, and there’s some angle $\phi$ for which $a=\cos\phi$ and $b=\sin\phi$. So, for instance, if you decided that $w'$ represents the real part, $w'=\cos\phi$ and $x^2+y^2+z^2=sin^2\phi$, which satisfies your identity, but when choosing $x'$ instead we’d have $x'=\cos\phi'$ and $y^2+z^2+w^2=\sin^2\phi'$, which yields the same identity.

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