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Question Number 30120 by ajfour last updated on 16/Feb/18
Commented by ajfour last updated on 16/Feb/18
Determine 𝛒, 𝛈, 𝛔, 𝛆 in terms of 𝛌 and 𝛍 .
Determine\boldsymbolρ,\boldsymbolη,\boldsymbolσ,\boldsymbolϵintermsof\boldsymbolλand\boldsymbolμ.
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.
ThankyouSir.Ihavesolvedthevectorway.
Commented by mrW2 last updated on 17/Feb/18
maybe one can use vector method to get the result more easily.
maybeonecanusevectormethodtogettheresultmoreeasily.
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))
letAB=c,BC=a,∠ABC=θB(0,0)D(λa,0)C(a,0)F(μccosθ,μcsinθ)A(ccosθ,csinθ)Eqn.ofCF:y−0x−a=μcsinθ−0μccosθ−ay(μccosθ−a)=(x−a)μcsinθEqn.ofADy−0x−λa=csinθ−0ccosθ−λay(ccosθ−λa)=(x−λa)csinθPointG:yG(μccosθ−a)=(xG−a)μcsinθyG(ccosθ−λa)=(xG−λa)csinθ⇒μccosθ−accosθ−λa=(xG−a)μxG−λa⇒(μccosθ−a)xG−λμaccosθ+λa2=(μccosθ−λμa)xG−μaccosθ+λμa2⇒(1−λμ)xG=(1−λ)μccosθ+λ(1−μ)a⇒xG=(1−λ)μccosθ+λ(1−μ)a1−λμ⇒yG=csinθ(xG−λa)ccosθ−λa⇒yG=csinθccosθ−λa×[(1−λ)μccosθ+λ(1−μ)a−λa(1−λμ)1−λμ]⇒yG=(1−λ)μcsinθ1−λμEqn.ofBE:yx=(1−λ)μcsinθ1−λμ(1−λ)μccosθ+λ(1−μ)a1−λμ=(1−λ)μcsinθ(1−λ)μccosθ+λ(1−μ)ay[(1−λ)μccosθ+λ(1−μ)a]=x(1−λ)μcsinθEqn.ofAC:y−0x−a=csinθ−0ccosθ−ay(ccosθ−a)=(x−a)csinθPointE:yE[(1−λ)μccosθ+λ(1−μ)a]=xE(1−λ)μcsinθyE(ccosθ−a)=(xE−a)csinθ⇒(1−λ)μccosθ+λ(1−μ)accosθ−a=xE(1−λ)μxE−a[(1−λ)μccosθ+λ(1−μ)a]xE−a[(1−λ)μccosθ+λ(1−μ)a]=(1−λ)μ(ccosθ−a)xE[(1−λ)μa+λ(1−μ)a]xE=a[(1−λ)μccosθ+λ(1−μ)a][λ+μ(1−2λ)]xE=(1−λ)μccosθ+λ(1−μ)a⇒xE=(1−λ)μccosθ+λ(1−μ)aλ+μ(1−2λ)⇒yE=csinθ(xE−a)ccosθ−a=csinθccosθ−a×[(1−λ)μccosθ+λ(1−μ)a−aλ−aμ(1−2λ)λ+μ(1−2λ)]⇒yE=csinθccosθ−a×[(1−λ)μ(ccosθ−a)λ+μ(1−2λ)]⇒yE=(1−λ)μcsinθλ+μ(1−2λ)η=xGxE=(1−λ)μccosθ+λ(1−μ)a1−λμ×λ+μ(1−2λ)(1−λ)μccosθ+λ(1−μ)a⇒η=λ+μ(1−2λ)1−λμ=λ+μ−2λμ1−λμ…(I)ε=yEyA=(1−λ)μcsinθλ+μ(1−2λ)×1csinθ⇒ε=(1−λ)μλ+μ(1−2λ)=μ−λμλ+μ−2λμ…(II)usingthesetworelationsallothervaluescanbedetermined.σ=ε+(1−λ)(1−2ε)1−ε(1−λ)σ=(1−λ)μλ+μ(1−2λ)+(1−λ)(1−2(1−λ)μλ+μ(1−2λ))1−(1−λ)μλ+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)2μ⇒σ=1−λ1−λμρ=(1−μ)+(1−ε)[1−2(1−μ)]1−(1−μ)(1−ε)ρ=(1−μ)+[1−(1−λ)μλ+μ(1−2λ)][1−2(1−μ)]1−(1−μ)[1−(1−λ)μλ+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−μ1−λμ
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))) .
letBbeorigin,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.ofa¯isλρ=1−σ=η−ϵη…(i)coeff.ofc¯is(1−ρ)=σμ=ηϵ…(ii)⇒1−ρ=(1−λρ)μ\boldsymbolρ=1−\boldsymbolμ1−\boldsymbolλμσ=1−λρ=1−(λ−λμ1−λμ)⇒\boldsymbolσ=1−\boldsymbolλ1−\boldsymbolλμadding(i)and(ii)η=λρ+(1−ρ)=1−(1−λ)ρ=1−(1−λ)(1−μ)1−λμ\boldsymbolη=\boldsymbolμ(1−\boldsymbolλ)+\boldsymbolλ(1−\boldsymbolμ)1−\boldsymbolλμϵ=1−ρη=[1−(1−μ1−λμ)]×1−λμμ(1−λ)+λ(1−μ)\boldsymbolϵ=\boldsymbolμ(1−\boldsymbolλ)\boldsymbolμ(1−\boldsymbolλ)+\boldsymbolλ(1−\boldsymbolμ).
Commented by mrW2 last updated on 17/Feb/18
That′s really clearly laid out! Thanks sir!
That′sreallyclearlylaidout!Thankssir!That′sreallyclearlylaidout!Thankssir!

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