Active transformation of a vector

[KTC]

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

Homework Statement

The vector $$\overrightarrow{OP}$$ is rotated through an angle $$\theta$$ about an axis through $$O$$ in the direction of the unit vector $$\underline{n}$$.

Homework Equations

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

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.

 

[Jhero]

I can understand that you are having difficulties understanding the proof. Let me try to explain it to you.

Firstly, the notation \widehat{\overrightarrow{SP}} and \widehat{\overrightarrow{TQ}} represent unit vectors in the direction of \overrightarrow{SP} and \overrightarrow{TQ} respectively. These unit vectors are obtained by dividing the original vectors by their magnitudes.

Secondly, the equality | \underline{n}\times\underline{x} | = SP is based on the definition of the cross product. The magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by those two vectors. In this case, \underline{n} and \underline{x} form two sides of a parallelogram with SP as the base. Hence, the magnitude of the cross product is equal to the length of SP.

Lastly, to understand how the various dot products cancel out, let's break down the equation into smaller steps.

Starting from the third line, we have:

(\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{SQ \cos\theta \widehat{\overrightarrow{S P}}}_{| \overrightarrow{S T} |} + \underbrace{SQ \sin\theta \widehat{\overrightarrow{T Q}}}_{| \overrightarrow{T Q} |}

= (\underline{x}\cdot\underline{n})\underline{n} + \overrightarrow{S T} + \overrightarrow{T Q}

= \underline{x} \cos\theta + (\underline{n}\cdot\underline{x})\underline{n} + (\underline{n}\times\underline{x})\sin\theta