Originally Posted by

**Showcase_22** (It was funny when I got your post back: "Failure has replied to your post" (Crying) )

I followed it through and it was much easier than using coordinates.

$\displaystyle \vec{s}=(\text{Rot}(\alpha)-\text{Rot}(- \beta))^{-1}(\vec{q}-\vec{p}-\text{Rot}(-\beta) \vec{q}+\text{Rot}(\alpha) \vec{p})$

I could write it out in terms of coordinates, but I don't think it would get any simpler.

I'm also sorry for springing this on you, but I hoped this would help me solve part ii) of the question. Unfortunately, i'm still lost:

So I know that:

$\displaystyle \mathrm{Rot}(P,\alpha)\vec{s}=\mathrm{Rot}(\alpha) (\vec{s}-\vec{p})+\vec{p}$

and:

$\displaystyle \mathrm{Rot}(Q,\beta)\vec{s}=\mathrm{Rot}(\beta)(\ vec{s}-\vec{q})+\vec{q}$

Therefore:

$\displaystyle \text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\mathrm{Rot}(\beta)(\mathrm{Rot}(\alpha)(\ vec{x}-\vec{p})+\vec{p}-\vec{q})+\vec{q}$

(putting the first one into the second one, for some vector $\displaystyle \vec{x}$).

Expanding this out gives:

$\displaystyle \text{Rot}(Q, \beta) \circ \text{Rot}(P, \alpha) \vec{x}=\text{Rot}(\beta). \text{Rot}(\alpha)(\vec{x}-\vec{p})+\text{Rot}(\beta) \vec{p}-\text{Rot}(\beta) \vec{q}+\vec{q}$

If it didn't say "hence calculate...", I probably would have left it there.

At this point I thought I could group the terms together to get $\displaystyle \vec{s}$ appearing somewhere (or at least $\displaystyle \text{Rot}(\alpha)-\text{Rot}(-\beta)) \vec{s}$). A big obstacle is that $\displaystyle \vec{s}$ doesn't really pop up. I tried seeing if there was a link between $\displaystyle \text{Rot}(Q, \beta)$ and $\displaystyle \text{Rot}(Q, -\beta)$, but writing them out doesn't produce anything.