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 Exercise [11.03] 
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Joined: 25 Feb 2008, 13:32
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Post Exercise [11.03]
If
q^{-1}=\bar{q}\left(q\bar{q}\right)^{-1}

Where
q=t+u i+v j+w k
\bar{q}=t-u i-v j-w k

Then
q^{-1}=\bar{q}\left(q\bar{q}\right)^{-1}

\frac{1}{(t+u i+v j+w k)}=\frac{(t-u i-v j-w k)}{(t+u i+v j+w k)(t-u i-v j-w k)}

\frac{1}{(t+u i+v j+w k)(t-u i-v j-w k)}=\frac{1}{(t+u i+v j+w k)(t-u i-v j-w k)}
\frac{1}{t^2+u^2 +v^2 +w^2}=\frac{1}{t^2+u^2 +v^2 +w^2}

1=1


The inverse quaternion is equal to the quaternion conjugate multiplied by the inverse of the norm.

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Last edited by Shaun Culver on 28 May 2008, 17:57, edited 3 times in total.
Edited


25 May 2008, 01:33

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Exercise [11.3]
The conjugate is defined by,
\overline{q}=t-ui-vj-wk
The inverse satisfies
q^{-1}q=qq^{-1}=1
We define q^{-1}=\frac{\overline{q}}{(q\overline{q})}
This definition works because
qq^{-1}=q\frac{\overline{q}}{(q\overline{q})}=\frac{q\overline{q}}{(q\overline{q})}=\frac{t^2+u^2+v^2+w^2}{t^2+u^2+v^2+w^2}=1and
q^{-1}q=\frac{\overline{q}}{(q\overline{q})}q=\frac{\overline{q}q}{(q\overline{q})}=\frac{t^2+u^2+v^2+w^2}{t^2+u^2+v^2+w^2}=1
(because q\overline{q}=\overline{q}q;see[11.2])
P.S.Shaun,your argument should go like this.
q^{-1}=\frac{1}{(t+ui+vj+wk)}=\frac{(t-ui-vj-wk)}{(t+ui+vj+wk)(t-ui-vj-wk)}
=\frac{\overline{q}}{t^2+u^2+v^2+w^2}=\frac{\overline{q}}{(q\overline{q})}


26 May 2008, 10:32
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Joined: 25 Feb 2008, 13:32
Posts: 106
Location: Cape Town, South Africa
Post Re: Exercise [11.3]
I've modified the solution slightly.

Is there something wrong with my argument?

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26 May 2008, 17:33

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Exercise [11.3]
Initially the argument did have an error,but you seem to have spotted it and corrected it.


27 May 2008, 09:25
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