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KTC
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I'm revising for exam, and trying to understand the following proof, and am having difficulties.
The vector [tex]\overrightarrow{OP}[/tex] is rotated through an angle [tex]\theta[/tex] about an axis through [tex]O[/tex] in the direction of the unit vector [tex]\underline{n}[/tex].
http://img172.imageshack.us/img172/4739/diagsx5.png
[tex]\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*}[/tex]
3. Questions
How was [tex]\widehat{\overrightarrow{SP}}[/tex] and [tex]\widehat{\overrightarrow{TQ}}[/tex] arrived at?
How does [tex]| \underline{n}\times\underline{x} | = SP[/tex]?
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.
Homework Statement
The vector [tex]\overrightarrow{OP}[/tex] is rotated through an angle [tex]\theta[/tex] about an axis through [tex]O[/tex] in the direction of the unit vector [tex]\underline{n}[/tex].
Homework Equations
http://img172.imageshack.us/img172/4739/diagsx5.png
[tex]\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*}[/tex]
3. Questions
How was [tex]\widehat{\overrightarrow{SP}}[/tex] and [tex]\widehat{\overrightarrow{TQ}}[/tex] arrived at?
How does [tex]| \underline{n}\times\underline{x} | = SP[/tex]?
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.
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