Question Number 30120 by ajfour last updated on 16/Feb/18
Commented by ajfour last updated on 16/Feb/18
$${Determine}\:\boldsymbol{\rho},\:\boldsymbol{\eta},\:\boldsymbol{\sigma},\:\boldsymbol{\epsilon}\:\:\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{\lambda}\:{and}\:\boldsymbol{\mu}\:. \\ $$
Answered by mrW2 last updated on 17/Feb/18
Commented by ajfour last updated on 17/Feb/18
$${Thank}\:{you}\:{Sir}.\:{I}\:{have}\:{solved} \\ $$$${the}\:{vector}\:{way}. \\ $$
Commented by mrW2 last updated on 17/Feb/18
$${maybe}\:{one}\:{can}\:{use}\:{vector}\:{method}\:{to} \\ $$$${get}\:{the}\:{result}\:{more}\:{easily}. \\ $$
Commented by mrW2 last updated on 17/Feb/18
![let AB=c, BC=a, ∠ABC=θ B(0,0) D(λa,0) C(a,0) F(μc cos θ,μc sin θ) A(c cos θ,c sin θ) Eqn. of CF: ((y−0)/(x−a))=((μc sin θ−0)/(μc cos θ−a)) y(μc cos θ−a)=(x−a)μc sin θ Eqn. of AD ((y−0)/(x−λa))=((c sin θ−0)/(c cos θ−λa)) y(c cos θ−λa)=(x−λa)c sin θ Point G: y_G (μc cos θ−a)=(x_G −a)μc sin θ y_G (c cos θ−λa)=(x_G −λa)c sin θ ⇒((μc cos θ−a)/(c cos θ−λa))=(((x_G −a)μ)/(x_G −λa)) ⇒(μc cos θ −a)x_G −λμac cos θ+λa^2 =(μc cos θ−λμa)x_G −μac cos θ+λμa^2 ⇒(1−λμ)x_G =(1−λ)μc cos θ+λ(1−μ)a ⇒x_G =(((1−λ)μc cos θ+λ(1−μ)a)/(1−λμ)) ⇒y_G =((c sin θ(x_G −λa))/(c cos θ−λa)) ⇒y_G =((c sin θ)/(c cos θ−λa))×[(((1−λ)μc cos θ+λ(1−μ)a−λa(1−λμ))/(1−λμ))] ⇒y_G =(((1−λ)μc sin θ)/(1−λμ)) Eqn. of BE: (y/x)=((((1−λ)μc sin θ)/(1−λμ))/(((1−λ)μc cos θ+λ(1−μ)a)/(1−λμ)))=(((1−λ)μc sin θ)/((1−λ)μc cos θ+λ(1−μ)a)) y[(1−λ)μc cos θ+λ(1−μ)a]=x(1−λ)μc sin θ Eqn. of AC: ((y−0)/(x−a))=((c sin θ−0)/(c cos θ−a)) y(c cos θ−a)=(x−a)c sin θ Point E: y_E [(1−λ)μc cos θ+λ(1−μ)a]=x_E (1−λ)μc sin θ y_E (c cos θ−a)=(x_E −a)c sin θ ⇒(((1−λ)μc cos θ+λ(1−μ)a)/(c cos θ−a))=((x_E (1−λ)μ)/(x_E −a)) [(1−λ)μc cos θ+λ(1−μ)a]x_E −a[(1−λ)μc cos θ+λ(1−μ)a]=(1−λ)μ(c cos θ−a)x_E [(1−λ)μa+λ(1−μ)a]x_E =a[(1−λ)μc cos θ+λ(1−μ)a] [λ+μ(1−2λ)]x_E =(1−λ)μc cos θ+λ(1−μ)a ⇒x_E =(((1−λ)μc cos θ+λ(1−μ)a)/(λ+μ(1−2λ))) ⇒y_E =((c sin θ(x_E −a))/(c cos θ−a))=((c sin θ)/(c cos θ−a))×[(((1−λ)μc cos θ+λ(1−μ)a−aλ−aμ(1−2λ))/(λ+μ(1−2λ)))] ⇒y_E =((c sin θ)/(c cos θ−a))×[(((1−λ)μ(c cos θ−a))/(λ+μ(1−2λ)))] ⇒y_E =(((1−λ)μc sin θ)/(λ+μ(1−2λ))) η=(x_G /x_E )=(((1−λ)μc cos θ+λ(1−μ)a)/(1−λμ))×((λ+μ(1−2λ))/((1−λ)μc cos θ+λ(1−μ)a)) ⇒η=((λ+μ(1−2λ))/(1−λμ))=((λ+μ−2λμ)/(1−λμ)) ...(I) ε=(y_E /y_A )=(((1−λ)μc sin θ)/(λ+μ(1−2λ)))×(1/(c sin θ)) ⇒ε=(((1−λ)μ)/(λ+μ(1−2λ)))=((μ−λμ)/(λ+μ−2λμ)) ...(II) using these two relations all other values can be determined. σ=((ε+(1−λ)(1−2ε))/(1−ε(1−λ))) σ=(((((1−λ)μ)/(λ+μ(1−2λ)))+(1−λ)(1−((2(1−λ)μ)/(λ+μ(1−2λ)))))/(1−(((1−λ)μ)/(λ+μ(1−2λ)))(1−λ))) σ=(((1−λ)μ+(1−λ)[λ+μ(1−2λ)−2(1−λ)μ])/(λ+μ(1−2λ)−(1−λ)^2 μ)) ⇒σ=((1−λ)/(1−λμ)) ρ=(((1−μ)+(1−ε)[1−2(1−μ)])/(1−(1−μ)(1−ε))) ρ=(((1−μ)+[1−(((1−λ)μ)/(λ+μ(1−2λ)))][1−2(1−μ)])/(1−(1−μ)[1−(((1−λ)μ)/(λ+μ(1−2λ)))])) ρ=(((1−μ)[λ+μ(1−2λ)]+[λ+μ(1−2λ)−(1−λ)μ][1−2(1−μ)])/(λ+μ(1−2λ)−(1−μ)[λ+μ(1−2λ)−(1−λ)μ])) ⇒ρ=((1−μ)/(1−λμ))](https://www.tinkutara.com/question/Q30144.png)
$${let}\:{AB}={c},\:{BC}={a},\:\angle{ABC}=\theta \\ $$$${B}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${D}\left(\lambda{a},\mathrm{0}\right) \\ $$$${C}\left({a},\mathrm{0}\right) \\ $$$${F}\left(\mu{c}\:\mathrm{cos}\:\theta,\mu{c}\:\mathrm{sin}\:\theta\right) \\ $$$${A}\left({c}\:\mathrm{cos}\:\theta,{c}\:\mathrm{sin}\:\theta\right) \\ $$$$ \\ $$$${Eqn}.\:{of}\:{CF}: \\ $$$$\frac{{y}−\mathrm{0}}{{x}−{a}}=\frac{\mu{c}\:\mathrm{sin}\:\theta−\mathrm{0}}{\mu{c}\:\mathrm{cos}\:\theta−{a}} \\ $$$${y}\left(\mu{c}\:\mathrm{cos}\:\theta−{a}\right)=\left({x}−{a}\right)\mu{c}\:\mathrm{sin}\:\theta \\ $$$${Eqn}.\:{of}\:{AD} \\ $$$$\frac{{y}−\mathrm{0}}{{x}−\lambda{a}}=\frac{{c}\:\mathrm{sin}\:\theta−\mathrm{0}}{{c}\:\mathrm{cos}\:\theta−\lambda{a}} \\ $$$${y}\left({c}\:\mathrm{cos}\:\theta−\lambda{a}\right)=\left({x}−\lambda{a}\right){c}\:\mathrm{sin}\:\theta \\ $$$${Point}\:{G}: \\ $$$${y}_{{G}} \left(\mu{c}\:\mathrm{cos}\:\theta−{a}\right)=\left({x}_{{G}} −{a}\right)\mu{c}\:\mathrm{sin}\:\theta \\ $$$${y}_{{G}} \left({c}\:\mathrm{cos}\:\theta−\lambda{a}\right)=\left({x}_{{G}} −\lambda{a}\right){c}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{\mu{c}\:\mathrm{cos}\:\theta−{a}}{{c}\:\mathrm{cos}\:\theta−\lambda{a}}=\frac{\left({x}_{{G}} −{a}\right)\mu}{{x}_{{G}} −\lambda{a}} \\ $$$$\Rightarrow\left(\mu{c}\:\mathrm{cos}\:\theta\:−{a}\right){x}_{{G}} −\lambda\mu{ac}\:\mathrm{cos}\:\theta+\lambda{a}^{\mathrm{2}} =\left(\mu{c}\:\mathrm{cos}\:\theta−\lambda\mu{a}\right){x}_{{G}} −\mu{ac}\:\mathrm{cos}\:\theta+\lambda\mu{a}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{1}−\lambda\mu\right){x}_{{G}} =\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a} \\ $$$$\Rightarrow{x}_{{G}} =\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}}{\mathrm{1}−\lambda\mu} \\ $$$$\Rightarrow{y}_{{G}} =\frac{{c}\:\mathrm{sin}\:\theta\left({x}_{{G}} −\lambda{a}\right)}{{c}\:\mathrm{cos}\:\theta−\lambda{a}} \\ $$$$\Rightarrow{y}_{{G}} =\frac{{c}\:\mathrm{sin}\:\theta}{{c}\:\mathrm{cos}\:\theta−\lambda{a}}×\left[\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}−\lambda{a}\left(\mathrm{1}−\lambda\mu\right)}{\mathrm{1}−\lambda\mu}\right] \\ $$$$\Rightarrow{y}_{{G}} =\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta}{\mathrm{1}−\lambda\mu} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{BE}: \\ $$$$\frac{{y}}{{x}}=\frac{\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta}{\mathrm{1}−\lambda\mu}}{\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}}{\mathrm{1}−\lambda\mu}}=\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta}{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}} \\ $$$${y}\left[\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}\right]={x}\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta \\ $$$${Eqn}.\:{of}\:{AC}: \\ $$$$\frac{{y}−\mathrm{0}}{{x}−{a}}=\frac{{c}\:\mathrm{sin}\:\theta−\mathrm{0}}{{c}\:\mathrm{cos}\:\theta−{a}} \\ $$$${y}\left({c}\:\mathrm{cos}\:\theta−{a}\right)=\left({x}−{a}\right){c}\:\mathrm{sin}\:\theta \\ $$$${Point}\:{E}: \\ $$$${y}_{{E}} \left[\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}\right]={x}_{{E}} \left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta \\ $$$${y}_{{E}} \left({c}\:\mathrm{cos}\:\theta−{a}\right)=\left({x}_{{E}} −{a}\right){c}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}}{{c}\:\mathrm{cos}\:\theta−{a}}=\frac{{x}_{{E}} \left(\mathrm{1}−\lambda\right)\mu}{{x}_{{E}} −{a}} \\ $$$$\left[\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}\right]{x}_{{E}} −{a}\left[\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}\right]=\left(\mathrm{1}−\lambda\right)\mu\left({c}\:\mathrm{cos}\:\theta−{a}\right){x}_{{E}} \\ $$$$\left[\left(\mathrm{1}−\lambda\right)\mu{a}+\lambda\left(\mathrm{1}−\mu\right){a}\right]{x}_{{E}} ={a}\left[\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}\right] \\ $$$$\left[\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)\right]{x}_{{E}} =\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a} \\ $$$$\Rightarrow{x}_{{E}} =\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)} \\ $$$$\Rightarrow{y}_{{E}} =\frac{{c}\:\mathrm{sin}\:\theta\left({x}_{{E}} −{a}\right)}{{c}\:\mathrm{cos}\:\theta−{a}}=\frac{{c}\:\mathrm{sin}\:\theta}{{c}\:\mathrm{cos}\:\theta−{a}}×\left[\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}−{a}\lambda−{a}\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\right] \\ $$$$\Rightarrow{y}_{{E}} =\frac{{c}\:\mathrm{sin}\:\theta}{{c}\:\mathrm{cos}\:\theta−{a}}×\left[\frac{\left(\mathrm{1}−\lambda\right)\mu\left({c}\:\mathrm{cos}\:\theta−{a}\right)}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\right] \\ $$$$\Rightarrow{y}_{{E}} =\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)} \\ $$$$ \\ $$$$\eta=\frac{{x}_{{G}} }{{x}_{{E}} }=\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}}{\mathrm{1}−\lambda\mu}×\frac{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{cos}\:\theta+\lambda\left(\mathrm{1}−\mu\right){a}} \\ $$$$\Rightarrow\eta=\frac{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}{\mathrm{1}−\lambda\mu}=\frac{\lambda+\mu−\mathrm{2}\lambda\mu}{\mathrm{1}−\lambda\mu}\:\:\:…\left({I}\right) \\ $$$$\varepsilon=\frac{{y}_{{E}} }{{y}_{{A}} }=\frac{\left(\mathrm{1}−\lambda\right)\mu{c}\:\mathrm{sin}\:\theta}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}×\frac{\mathrm{1}}{{c}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\varepsilon=\frac{\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}=\frac{\mu−\lambda\mu}{\lambda+\mu−\mathrm{2}\lambda\mu}\:\:\:\:…\left({II}\right) \\ $$$$ \\ $$$${using}\:{these}\:{two}\:{relations}\:{all}\:{other} \\ $$$${values}\:{can}\:{be}\:{determined}. \\ $$$$\sigma=\frac{\varepsilon+\left(\mathrm{1}−\lambda\right)\left(\mathrm{1}−\mathrm{2}\varepsilon\right)}{\mathrm{1}−\varepsilon\left(\mathrm{1}−\lambda\right)} \\ $$$$\sigma=\frac{\frac{\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}+\left(\mathrm{1}−\lambda\right)\left(\mathrm{1}−\frac{\mathrm{2}\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\right)}{\mathrm{1}−\frac{\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\left(\mathrm{1}−\lambda\right)} \\ $$$$\sigma=\frac{\left(\mathrm{1}−\lambda\right)\mu+\left(\mathrm{1}−\lambda\right)\left[\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)−\mathrm{2}\left(\mathrm{1}−\lambda\right)\mu\right]}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)−\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \mu} \\ $$$$\Rightarrow\sigma=\frac{\mathrm{1}−\lambda}{\mathrm{1}−\lambda\mu} \\ $$$$ \\ $$$$\rho=\frac{\left(\mathrm{1}−\mu\right)+\left(\mathrm{1}−\varepsilon\right)\left[\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\mu\right)\right]}{\mathrm{1}−\left(\mathrm{1}−\mu\right)\left(\mathrm{1}−\varepsilon\right)} \\ $$$$\rho=\frac{\left(\mathrm{1}−\mu\right)+\left[\mathrm{1}−\frac{\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\right]\left[\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\mu\right)\right]}{\mathrm{1}−\left(\mathrm{1}−\mu\right)\left[\mathrm{1}−\frac{\left(\mathrm{1}−\lambda\right)\mu}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)}\right]} \\ $$$$\rho=\frac{\left(\mathrm{1}−\mu\right)\left[\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)\right]+\left[\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)−\left(\mathrm{1}−\lambda\right)\mu\right]\left[\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\mu\right)\right]}{\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)−\left(\mathrm{1}−\mu\right)\left[\lambda+\mu\left(\mathrm{1}−\mathrm{2}\lambda\right)−\left(\mathrm{1}−\lambda\right)\mu\right]} \\ $$$$\Rightarrow\rho=\frac{\mathrm{1}−\mu}{\mathrm{1}−\lambda\mu} \\ $$
Answered by ajfour last updated on 17/Feb/18
![let B be origin, BC^(−) = a^ ,BA^(−) = c^ , CA^(−) = b^ = c^ −a^ BG^(−) =ρ(λa^ )+(1−ρ)c^ =(1−σ)a^ +σ(μc^ ) = η(a^ +εb^ ) =η[a^ +ε(c^ −a^ )] coeff. of a^ is λρ =1−σ = η−εη ...(i) coeff. of c^ is (1−ρ)=σμ = ηε ...(ii) ⇒ 1−ρ = (1−λρ)μ 𝛒 = ((1−𝛍)/(1−𝛌𝛍)) σ=1−λρ = 1−(((λ−λμ)/(1−λμ))) ⇒ 𝛔 = ((1−𝛌)/(1−𝛌𝛍)) adding (i) and (ii) η = λρ+(1−ρ) =1−(1−λ)ρ = 1−(((1−λ)(1−μ))/(1−λμ)) 𝛈 = ((𝛍(1−𝛌)+𝛌(1−𝛍))/(1−𝛌𝛍)) ε = ((1−ρ)/η) = [1−(((1−μ)/(1−λμ)))]×((1−λμ)/(μ(1−λ)+λ(1−μ))) 𝛆 = ((𝛍(1−𝛌))/(𝛍(1−𝛌)+𝛌(1−𝛍))) .](https://www.tinkutara.com/question/Q30150.png)
$${let}\:\:{B}\:{be}\:{origin},\:\:\overline {{BC}}\:=\:\bar {{a}}\:\:,\overline {{BA}}\:=\:\bar {{c}}\:, \\ $$$$\overline {{CA}}\:=\:\bar {{b}}\:=\:\bar {{c}}−\bar {{a}} \\ $$$$\overline {{BG}}\:=\rho\left(\lambda\bar {{a}}\right)+\left(\mathrm{1}−\rho\right)\bar {{c}} \\ $$$$\:\:\:\:\:\:\:\:=\left(\mathrm{1}−\sigma\right)\bar {{a}}+\sigma\left(\mu\bar {{c}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\eta\left(\bar {{a}}+\epsilon\bar {{b}}\right)\:=\eta\left[\bar {{a}}+\epsilon\left(\bar {{c}}−\bar {{a}}\right)\right] \\ $$$${coeff}.\:{of}\:\bar {{a}}\:{is} \\ $$$$\:\:\:\:\:\lambda\rho\:=\mathrm{1}−\sigma\:=\:\eta−\epsilon\eta\:\:\:\:…\left({i}\right) \\ $$$${coeff}.\:{of}\:\bar {{c}}\:{is} \\ $$$$\:\:\:\:\:\left(\mathrm{1}−\rho\right)=\sigma\mu\:=\:\eta\epsilon\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{1}−\rho\:=\:\left(\mathrm{1}−\lambda\rho\right)\mu \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\rho}\:=\:\frac{\mathrm{1}−\boldsymbol{\mu}}{\mathrm{1}−\boldsymbol{\lambda\mu}}\: \\ $$$$\:\:\:\:\:\sigma=\mathrm{1}−\lambda\rho\:=\:\mathrm{1}−\left(\frac{\lambda−\lambda\mu}{\mathrm{1}−\lambda\mu}\right) \\ $$$$\:\:\:\Rightarrow\:\:\:\boldsymbol{\sigma}\:=\:\frac{\mathrm{1}−\boldsymbol{\lambda}}{\mathrm{1}−\boldsymbol{\lambda\mu}} \\ $$$$\:\:{adding}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$$\:\:\:\eta\:=\:\lambda\rho+\left(\mathrm{1}−\rho\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{1}−\left(\mathrm{1}−\lambda\right)\rho \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}−\frac{\left(\mathrm{1}−\lambda\right)\left(\mathrm{1}−\mu\right)}{\mathrm{1}−\lambda\mu} \\ $$$$\:\:\:\:\:\boldsymbol{\eta}\:=\:\frac{\boldsymbol{\mu}\left(\mathrm{1}−\boldsymbol{\lambda}\right)+\boldsymbol{\lambda}\left(\mathrm{1}−\boldsymbol{\mu}\right)}{\mathrm{1}−\boldsymbol{\lambda\mu}} \\ $$$$\:\:\epsilon\:=\:\frac{\mathrm{1}−\rho}{\eta}\: \\ $$$$\:\:=\:\left[\mathrm{1}−\left(\frac{\mathrm{1}−\mu}{\mathrm{1}−\lambda\mu}\right)\right]×\frac{\mathrm{1}−\lambda\mu}{\mu\left(\mathrm{1}−\lambda\right)+\lambda\left(\mathrm{1}−\mu\right)} \\ $$$$\:\:\:\:\:\:\boldsymbol{\epsilon}\:=\:\frac{\boldsymbol{\mu}\left(\mathrm{1}−\boldsymbol{\lambda}\right)}{\boldsymbol{\mu}\left(\mathrm{1}−\boldsymbol{\lambda}\right)+\boldsymbol{\lambda}\left(\mathrm{1}−\boldsymbol{\mu}\right)}\:\:. \\ $$
Commented by mrW2 last updated on 17/Feb/18
$${That}'{s}\:{really}\:{clearly}\:{laid}\:{out}!\:{Thanks}\:{sir}! \\ $$