# Active transformation of a vector

1. Apr 19, 2007

### KTC

I'm revising for exam, and trying to understand the following proof, and am having difficulties.

1. The problem statement, all variables and given/known data
The vector $$\overrightarrow{OP}$$ is rotated through an angle $$\theta$$ about an axis through $$O$$ in the direction of the unit vector $$\underline{n}$$.

2. Relevant equations
http://img172.imageshack.us/img172/4739/diagsx5.png [Broken]
\begin{align*} \underline{y} &= \overrightarrow{O S} + \overrightarrow{S T} + \overrightarrow{T Q} \\ &= (\underline{x}\cdot\underline{n})\underline{n} + \underbrace{SQ \cos\theta}_{| \overrightarrow{S T} |} \underbrace{ \left ( \frac{\underline{x} - (\underline{x}\cdot\underline{n})\underline{n}}{SP} \right )}_{\widehat{\overrightarrow{S P}}} + \underbrace{SQ \sin\theta}_{| \overrightarrow{T Q} |} \underbrace{ \frac{\underline{n} \times \underline{x}}{ |\underline{n} \times \underline{x} |}}_{\widehat{\overrightarrow{T Q}}} \\ &= (\underline{x}\cdot\underline{n})\underline{n} + \underbrace{\cos\theta (\underline{x}-(\underline{x}\cdot\underline{n})\underline{n})}_{(\mbox{as } SP = SQ)} + \underbrace{\sin\theta ~\underline{n} \times \underline{x}}_{(\mbox{as } | \underline{n}\times\underline{x} | = SP = SQ )} \\ \mbox{So} \\ \underline{y} &= \underline{x} \cos\theta + (1 - \cos\theta)(\underline{n}\cdot\underline{x})\underline{n} + (\underline{n}\times\underline{x})\sin\theta \end{align*}

3. Questions
How was $$\widehat{\overrightarrow{SP}}$$ and $$\widehat{\overrightarrow{TQ}}$$ arrived at?
How does $$| \underline{n}\times\underline{x} | = SP$$?
I'm being stupid and don't understand how the various dot product etc. cancel out from the third to the last line.

Thanks for any help.

Last edited by a moderator: May 2, 2017