v-x-f-x-f-x-mx-b-If-I-wish-to-rotate-v-counter-clockwise-by-degrees-how-does-one-do-so-where-v-is-rotated-about-the-origin-v-is-rotated-about-the-point-x-1-f-x-1-What-a

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Question Number 7372 by FilupSmith last updated on 25/Aug/16

$$\mathit{v}=\left[\begin{array}{c}x\\ f\left(x\right)\end{array}\right],f\left(x\right)=mx+b$$$$\mathrm{If}\mathrm{I}\mathrm{wish}\mathrm{to}\mathrm{rotate}\mathit{v}\mathrm{counter}\mathrm{clockwise}$$$$\mathrm{by}\theta \mathrm{degrees},\mathrm{how}\mathrm{does}\mathrm{one}\mathrm{do}\mathrm{so}\mathrm{where}:$$$$\bullet \mathit{v}\mathrm{is}\mathrm{rotated}\mathrm{about}\mathrm{the}\mathrm{origin}$$$$\bullet \mathit{v}\mathrm{is}\mathrm{rotated}\mathrm{about}\mathrm{the}\mathrm{point}(x_1,f\left(x_1\right))$$$$\mathrm{What}\mathrm{are}\mathrm{the}\mathrm{new}\mathrm{vectors}?$$

Answered by sandy_suhendra last updated on 25/Aug/16

$$ifwerotateby\theta °abouttheorigin,thematrixis\left[\begin{array}{c}cos\theta -sin\theta \\ sin\theta cos\theta \end{array}\right]$$$$thenewvector{\mathit{v}}^{\prime }=\left[\begin{array}{c}cos\theta -sin\theta \\ sin\theta cos\theta \end{array}\right]\left[\begin{array}{c}x\\ f\left(x\right)\end{array}\right]=\left[\begin{array}{c}xcos\theta -f\left(x\right)sin\theta \\ xsin\theta +f\left(x\right)cos\theta \end{array}\right]$$$$ifwerotatedaboutthepoint(x_1,f\left(x_1\right))$$$$thenewvector{\mathit{v}}^{\prime }=\left[\begin{array}{c}cos\theta -sin\theta \\ sin\theta cos\theta \end{array}\right]\left[\begin{array}{c}x-x_1\\ f\left(x\right)-f\left(x_1\right)\end{array}\right]+\left[\begin{array}{c}{x}_1\\ f\left(x_1\right)\end{array}\right]$$$$=\left[\begin{array}{c}(x-x_1)cos\theta +[f\left(x_1\right)-f\left(x\right)]sin\theta +x_1\\ (x-x_1)sin\theta +[f\left(x\right)-f\left(x_1\right)]cos\theta +f\left(x_1\right)\end{array}\right]$$

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