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Active transformation of a vector

  1. Apr 19, 2007 #1


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    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 [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].

    2. Relevant equations
    \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

    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.
  2. jcsd
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