Question Number 7372 by FilupSmith last updated on 25/Aug/16
![v= [(x),((f(x))) ], f(x)=mx+b If I wish to rotate v counter clockwise by θ 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 are the new vectors?](Q7372.png)
$$\boldsymbol{{v}}=\begin{bmatrix}{{x}}\\{{f}\left({x}\right)}\end{bmatrix},\:\:{f}\left({x}\right)={mx}+{b} \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{rotate}\:\boldsymbol{{v}}\:\mathrm{counter}\:\mathrm{clockwise} \\ $$$$\mathrm{by}\:\theta\:\mathrm{degrees},\:\mathrm{how}\:\mathrm{does}\:\mathrm{one}\:\mathrm{do}\:\mathrm{so}\:\mathrm{where}: \\ $$$$\bullet\:\:\:\boldsymbol{{v}}\:\mathrm{is}\:\mathrm{rotated}\:\mathrm{about}\:\mathrm{the}\:\mathrm{origin} \\ $$$$\bullet\:\:\:\boldsymbol{{v}}\:\mathrm{is}\:\mathrm{rotated}\:\mathrm{about}\:\mathrm{the}\:\mathrm{point}\:\left({x}_{\mathrm{1}} ,\:{f}\left({x}_{\mathrm{1}} \right)\right) \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{new}\:\mathrm{vectors}? \\ $$
Answered by sandy_suhendra last updated on 25/Aug/16
![if we rotate by θ° about the origin, the matrix is [((cos θ −sin θ)),((sin θ cos θ)) ] the new vector v′ = [((cos θ −sin θ)),((sin θ cos θ)) ] [(x),((f(x))) ]= [((x cos θ−f(x) sin θ)),((x sin θ+f(x) cos θ)) ] if we rotated about the point (x_1 , f(x_1 )) the new vector v′ = [((cos θ −sin θ)),((sin θ cos θ)) ] [((x−x_1 )),((f(x)−f(x_1 ))) ]+ [(x_1 ),((f(x_1 ))) ] = [(((x−x_1 )cos θ+[f(x_1 )−f(x)]sin θ+x_1 )),(((x−x_1 )sin θ+[f(x)−f(x_1 )]cos θ+f(x_1 ))) ]](Q7378.png)
$${if}\:{we}\:{rotate}\:{by}\:\theta°\:{about}\:{the}\:{origin},\:{the}\:{matrix}\:{is}\:\begin{bmatrix}{{cos}\:\theta\:\:\:\:\:−{sin}\:\theta}\\{{sin}\:\theta\:\:\:\:\:\:\:\:\:{cos}\:\theta}\end{bmatrix} \\ $$$${the}\:{new}\:{vector}\:\boldsymbol{{v}}'\:=\begin{bmatrix}{{cos}\:\theta\:\:\:−{sin}\:\theta}\\{{sin}\:\theta\:\:\:\:\:\:{cos}\:\theta}\end{bmatrix}\begin{bmatrix}{{x}}\\{{f}\left({x}\right)}\end{bmatrix}=\:\begin{bmatrix}{{x}\:{cos}\:\theta−{f}\left({x}\right)\:{sin}\:\theta}\\{{x}\:{sin}\:\theta+{f}\left({x}\right)\:{cos}\:\theta}\end{bmatrix} \\ $$$${if}\:{we}\:{rotated}\:{about}\:{the}\:{point}\:\left({x}_{\mathrm{1}} \:,\:{f}\left({x}_{\mathrm{1}} \right)\right) \\ $$$${the}\:{new}\:{vector}\:\boldsymbol{{v}}'\:=\:\begin{bmatrix}{{cos}\:\theta\:\:\:\:−{sin}\:\theta}\\{{sin}\:\theta\:\:\:\:\:\:\:\:{cos}\:\theta}\end{bmatrix}\begin{bmatrix}{{x}−{x}_{\mathrm{1}} }\\{{f}\left({x}\right)−{f}\left({x}_{\mathrm{1}} \right)}\end{bmatrix}+\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{f}\left({x}_{\mathrm{1}} \right)}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\begin{bmatrix}{\left({x}−{x}_{\mathrm{1}} \right){cos}\:\theta+\left[{f}\left({x}_{\mathrm{1}} \right)−{f}\left({x}\right)\right]{sin}\:\theta+{x}_{\mathrm{1}} }\\{\left({x}−{x}_{\mathrm{1}} \right){sin}\:\theta+\left[{f}\left({x}\right)−{f}\left({x}_{\mathrm{1}} \right)\right]{cos}\:\theta+{f}\left({x}_{\mathrm{1}} \right)}\end{bmatrix} \\ $$