MHB Does Property (2) Imply Property (1) for Unit Quaternions?

[Math Amateur]

I am reading an article by Vernon Chi on quaternions and rotations in 3-space. The title of the article is as follows:View attachment 3979

I am concerned that I do not follow the proof of one of the properties of unit quaternions in Section 3.1.3 of the article.

Section 3.1.3 reads as follows:
View attachment 3980
In the above text we read the following (regarding an important property of unit quaternions):
" ... ... A less obvious, but very useful one (property) is,$$Q_u = R_u cos \phi + P_u sin \phi = cos \phi + P_u sin \phi ... ... ... (1)
$$where $$R_u = (1,0,0,0)$$ is a real quaternion and $$P_u = (0, i p_2 , j p_3 , k p_4 )$$ is a vector unit quaternion ... ..."


Chi then presents what is meant to be a proof of the property (1) above ... indeed he shows that if (1) is true then $$| Q_u |^2 = 1 $$ ... ... ... (2)follows.... ... BUT ... does he not have to show that for the property (1) to be true then we also have to show that if (2) is true then (1) follows ... ... Can someone please clarify this for me ...Further, if it is indeed the case that we need to show (2) implies (1) ... then can someone indicate how we would prove this ...Hope someone can help ...

Peter

To give readers an idea of the notation, definition and theory preceding the above section of Chi's article I am providing Sections 2 to 3.1.2 as follows:View attachment 3981
View attachment 3982
View attachment 3983

 

[I like Serena]

Hi Peter,

$Q_u$ will always be a linear combination of $R_u$ and $P_u$.
That is, the normalized real part respectively the normalized vector part, where the sign of the vector part is as yet unspecified.

Let's say that $Q_u=R_u\cdot a + P_u\cdot b$.
That means that $a$ is the real part of $Q_u$.

The size of $b$ is fully determined by the fact that $|Q_u|=1$.
Similar to what is shown, this means that $b^2=1-a^2$.

If we pick $\phi$ such that $\cos\phi =a$, then it follows that $b=\pm \sin\phi$.
In other words, it seems that Chi was not quite careful with the sign.
Then Chi specifies an interpretation where $\phi$ would be a certain ratio. He's sloppy again, since the ratio he means is actually $\tan\phi$. More precise would be that $\phi$ is the angle between $Q_u$ and the real axis.

Anyway, this interpretation determines the sign. So then it follows that $b=\sin\phi$.

 

[Math Amateur]

Hi Peter,

$Q_u$ will always be a linear combination of $R_u$ and $P_u$.
That is, the normalized real part respectively the normalized vector part, where the sign of the vector part is as yet unspecified.

Let's say that $Q_u=R_u\cdot a + P_u\cdot b$.
That means that $a$ is the real part of $Q_u$.

The size of $b$ is fully determined by the fact that $|Q_u|=1$.
Similar to what is shown, this means that $b^2=1-a^2$.

If we pick $\phi$ such that $\cos\phi =a$, then it follows that $b=\pm \sin\phi$.
In other words, it seems that Chi was not quite careful with the sign.
Then Chi specifies an interpretation where $\phi$ would be a certain ratio. He's sloppy again, since the ratio he means is actually $\tan\phi$. More precise would be that $\phi$ is the angle between $Q_u$ and the real axis.

Anyway, this interpretation determines the sign. So then it follows that $b=\sin\phi$.

Thanks I Like Serena ... your post is really helpful ...

BTW do you have a good source (online or textbook) for a detailed analysis of quarternions?

Peter