Question-10612

Question Number 10612 by FilupS last updated on 20/Feb/17

Commented by FilupS last updated on 20/Feb/17

Solve for A′ and B′ via: 1. Vectors 2. Triganometry

$$\mathrm{Solve}\:\mathrm{for}\:\boldsymbol{{A}}'\:\mathrm{and}\:\boldsymbol{{B}}'\:\mathrm{via}: \\ $$$$\mathrm{1}.\:\:\mathrm{Vectors} \\ $$$$\mathrm{2}.\:\:\mathrm{Triganometry} \\ $$

Answered by sandy_suhendra last updated on 20/Feb/17

by trigonometry QA=QA′=(√(1^2 +0^2 ))=1 x_(A′) = 1 cos θ = cos θ y_(A′) = 1 sin θ = sin θ so A′ = [((cos θ)),((sin θ)) ] QB=QB′ = (√(0^2 +2^2 )) = 2 x_(B′) = 2 cos (90°+θ) = −2 sin θ y_(B′) = 2 sin (90°+θ) = 2 cos θ so B′ = [((−2 sin θ)),(( 2 cos θ)) ]

$$\mathrm{by}\:\mathrm{trigonometry} \\ $$$$\mathrm{QA}=\mathrm{QA}'=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{A}'} \:=\:\mathrm{1}\:\mathrm{cos}\:\theta\:=\:\mathrm{cos}\:\theta \\ $$$$\mathrm{y}_{\mathrm{A}'} \:=\:\mathrm{1}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\theta \\ $$$$\mathrm{so}\:\mathrm{A}'\:=\:\begin{bmatrix}{\mathrm{cos}\:\theta}\\{\mathrm{sin}\:\theta}\end{bmatrix} \\ $$$$\mathrm{QB}=\mathrm{QB}'\:=\:\sqrt{\mathrm{0}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\mathrm{x}_{\mathrm{B}'} \:=\:\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{90}°+\theta\right)\:=\:−\mathrm{2}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{y}_{\mathrm{B}'} \:=\:\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{90}°+\theta\right)\:=\:\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{so}\:\mathrm{B}'\:=\:\begin{bmatrix}{−\mathrm{2}\:\mathrm{sin}\:\theta}\\{\:\:\:\mathrm{2}\:\mathrm{cos}\:\theta}\end{bmatrix} \\ $$

Answered by mrW1 last updated on 21/Feb/17

by vectors: QA^(→) =1i+0j AA′^(→) =−(1sin θtan (θ/2))i+1sin θj QA′^(→) =QA^(→) +AA′^(→) =(1−1sin θtan (θ/2))i+(0+1sin θ)j=1cos θi+1sin θj ⇒x_(A′) =cos θ ⇒y_(A′) =sin θ QB^(→) =0i+2j BB′^(→) =−2sin θi−(2sin θtan (θ/2))j QB′^(→) =QB^(→) +BB′^(→) =(0−2sin θ)i+(2−2sin θtan (θ/2))j=−2sin θi+2cos θj ⇒x_(B′) =−2sin θ ⇒y_(B′) =2cos θ note: 1−sin θtan (θ/2)=1−2sin (θ/2)cos (θ/2)tan (θ/2)=1−2sin^2 (θ/2)=cos θ

$${by}\:{vectors}: \\ $$$$ \\ $$$$\overset{\rightarrow} {{QA}}=\mathrm{1}{i}+\mathrm{0}{j} \\ $$$$\overset{\rightarrow} {{AA}'}=−\left(\mathrm{1sin}\:\theta\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){i}+\mathrm{1sin}\:\theta{j} \\ $$$$\overset{\rightarrow} {{QA}'}=\overset{\rightarrow} {{QA}}+\overset{\rightarrow} {{AA}'}=\left(\mathrm{1}−\mathrm{1sin}\:\theta\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){i}+\left(\mathrm{0}+\mathrm{1sin}\:\theta\right){j}=\mathrm{1cos}\:\theta{i}+\mathrm{1sin}\:\theta{j} \\ $$$$ \\ $$$$\Rightarrow{x}_{{A}'} =\mathrm{cos}\:\theta \\ $$$$\Rightarrow{y}_{{A}'} =\mathrm{sin}\:\theta \\ $$$$ \\ $$$$\overset{\rightarrow} {{QB}}=\mathrm{0}{i}+\mathrm{2}{j} \\ $$$$\overset{\rightarrow} {{BB}'}=−\mathrm{2sin}\:\theta{i}−\left(\mathrm{2sin}\:\:\theta\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){j} \\ $$$$\overset{\rightarrow} {{QB}'}=\overset{\rightarrow} {{QB}}+\overset{\rightarrow} {{BB}'}=\left(\mathrm{0}−\mathrm{2sin}\:\theta\right){i}+\left(\mathrm{2}−\mathrm{2sin}\:\:\theta\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right){j}=−\mathrm{2sin}\:\theta{i}+\mathrm{2cos}\:\theta{j} \\ $$$$ \\ $$$$\Rightarrow{x}_{{B}'} =−\mathrm{2sin}\:\theta \\ $$$$\Rightarrow{y}_{{B}'} =\mathrm{2cos}\:\theta \\ $$$$ \\ $$$${note}: \\ $$$$\mathrm{1}−\mathrm{sin}\:\theta\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\mathrm{1}−\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}=\mathrm{cos}\:\theta \\ $$

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