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Question Number 30120 by ajfour last updated on 16/Feb/18

Commented by ajfour last updated on 16/Feb/18

$$Determine\mathit{\rho },\mathit{\eta },\mathit{\sigma },\mathit{ϵ}interms$$$$of\mathit{\lambda }and\mathit{\mu }.$$

Answered by mrW2 last updated on 17/Feb/18

Commented by ajfour last updated on 17/Feb/18

$$ThankyouSir.Ihavesolved$$$$thevectorway.$$

Commented by mrW2 last updated on 17/Feb/18

$$maybeonecanusevectormethodto$$$$gettheresultmoreeasily.$$

Commented by mrW2 last updated on 17/Feb/18

$$letAB=c,BC=a,\mathrm{\angle }ABC=\theta$$$$B(0,0)$$$$D(\lambda a,0)$$$$C(a,0)$$$$F(\mu c\cos \theta ,\mu c\sin \theta )$$$$A(c\cos \theta ,c\sin \theta )$$$$$$$$Eqn.ofCF:$$$$\frac{y-0}{x-a}=\frac{\mu c\sin \theta -0}{\mu c\cos \theta -a}$$$$y(\mu c\cos \theta -a)=(x-a)\mu c\sin \theta$$$$Eqn.ofAD$$$$\frac{y-0}{x-\lambda a}=\frac{c\sin \theta -0}{c\cos \theta -\lambda a}$$$$y(c\cos \theta -\lambda a)=(x-\lambda a)c\sin \theta$$$$PointG:$$$$y_G(\mu c\cos \theta -a)=(x_G-a)\mu c\sin \theta$$$$y_G(c\cos \theta -\lambda a)=(x_G-\lambda a)c\sin \theta$$$$\Rightarrow \frac{\mu c\cos \theta -a}{c\cos \theta -\lambda a}=\frac{(x_G-a)\mu }{x_G-\lambda a}$$$$\Rightarrow (\mu c\cos \theta -a)x_G-\lambda \mu ac\cos \theta +\lambda a^2=(\mu c\cos \theta -\lambda \mu a)x_G-\mu ac\cos \theta +\lambda \mu a^2$$$$\Rightarrow (1-\lambda \mu )x_G=(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a$$$$\Rightarrow x_G=\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}{1-\lambda \mu }$$$$\Rightarrow y_G=\frac{c\sin \theta (x_G-\lambda a)}{c\cos \theta -\lambda a}$$$$\Rightarrow y_G=\frac{c\sin \theta }{c\cos \theta -\lambda a}\times \left[\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a-\lambda a(1-\lambda \mu )}{1-\lambda \mu }\right]$$$$\Rightarrow y_G=\frac{(1-\lambda )\mu c\sin \theta }{1-\lambda \mu }$$$$$$$$Eqn.ofBE:$$$$\frac{y}{x}=\frac{\frac{(1-\lambda )\mu c\sin \theta }{1-\lambda \mu }}{\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}{1-\lambda \mu }}=\frac{(1-\lambda )\mu c\sin \theta }{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}$$$$y[(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a]=x(1-\lambda )\mu c\sin \theta$$$$Eqn.ofAC:$$$$\frac{y-0}{x-a}=\frac{c\sin \theta -0}{c\cos \theta -a}$$$$y(c\cos \theta -a)=(x-a)c\sin \theta$$$$PointE:$$$$y_E[(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a]=x_E(1-\lambda )\mu c\sin \theta$$$$y_E(c\cos \theta -a)=(x_E-a)c\sin \theta$$$$\Rightarrow \frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}{c\cos \theta -a}=\frac{x_E(1-\lambda )\mu }{x_E-a}$$$$[(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a]x_E-a[(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a]=(1-\lambda )\mu (c\cos \theta -a)x_E$$$$[(1-\lambda )\mu a+\lambda (1-\mu )a]x_E=a[(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a]$$$$[\lambda +\mu (1-2\lambda )]x_E=(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a$$$$\Rightarrow x_E=\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}{\lambda +\mu (1-2\lambda )}$$$$\Rightarrow y_E=\frac{c\sin \theta (x_E-a)}{c\cos \theta -a}=\frac{c\sin \theta }{c\cos \theta -a}\times \left[\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a-a\lambda -a\mu (1-2\lambda )}{\lambda +\mu (1-2\lambda )}\right]$$$$\Rightarrow y_E=\frac{c\sin \theta }{c\cos \theta -a}\times \left[\frac{(1-\lambda )\mu (c\cos \theta -a)}{\lambda +\mu (1-2\lambda )}\right]$$$$\Rightarrow y_E=\frac{(1-\lambda )\mu c\sin \theta }{\lambda +\mu (1-2\lambda )}$$$$$$$$\eta =\frac{x_G}{x_E}=\frac{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}{1-\lambda \mu }\times \frac{\lambda +\mu (1-2\lambda )}{(1-\lambda )\mu c\cos \theta +\lambda (1-\mu )a}$$$$\Rightarrow \eta =\frac{\lambda +\mu (1-2\lambda )}{1-\lambda \mu }=\frac{\lambda +\mu -2\lambda \mu }{1-\lambda \mu }...\left(I\right)$$$$\epsilon =\frac{y_E}{y_A}=\frac{(1-\lambda )\mu c\sin \theta }{\lambda +\mu (1-2\lambda )}\times \frac{1}{c\sin \theta }$$$$\Rightarrow \epsilon =\frac{(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )}=\frac{\mu -\lambda \mu }{\lambda +\mu -2\lambda \mu }...\left(II\right)$$$$$$$$usingthesetworelationsallother$$$$valuescanbedetermined.$$$$\sigma =\frac{\epsilon +(1-\lambda )(1-2\epsilon )}{1-\epsilon (1-\lambda )}$$$$\sigma =\frac{\frac{(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )}+(1-\lambda )(1-\frac{2(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )})}{1-\frac{(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )}(1-\lambda )}$$$$\sigma =\frac{(1-\lambda )\mu +(1-\lambda )[\lambda +\mu (1-2\lambda )-2(1-\lambda )\mu ]}{\lambda +\mu (1-2\lambda )-{(1-\lambda )}^2\mu }$$$$\Rightarrow \sigma =\frac{1-\lambda }{1-\lambda \mu }$$$$$$$$\rho =\frac{(1-\mu )+(1-\epsilon )[1-2(1-\mu )]}{1-(1-\mu )(1-\epsilon )}$$$$\rho =\frac{(1-\mu )+[1-\frac{(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )}][1-2(1-\mu )]}{1-(1-\mu )[1-\frac{(1-\lambda )\mu }{\lambda +\mu (1-2\lambda )}]}$$$$\rho =\frac{(1-\mu )[\lambda +\mu (1-2\lambda )]+[\lambda +\mu (1-2\lambda )-(1-\lambda )\mu ][1-2(1-\mu )]}{\lambda +\mu (1-2\lambda )-(1-\mu )[\lambda +\mu (1-2\lambda )-(1-\lambda )\mu ]}$$$$\Rightarrow \rho =\frac{1-\mu }{1-\lambda \mu }$$

Answered by ajfour last updated on 17/Feb/18

$$letBbeorigin,\overline{BC}=\overline{a},\overline{BA}=\overline{c},$$$$\overline{CA}=\overline{b}=\overline{c}-\overline{a}$$$$\overline{BG}=\rho \left(\lambda \overline{a}\right)+(1-\rho )\overline{c}$$$$=(1-\sigma )\overline{a}+\sigma \left(\mu \overline{c}\right)$$$$=\eta (\overline{a}+ϵ\overline{b})=\eta [\overline{a}+ϵ(\overline{c}-\overline{a})]$$$$coeff.of\overline{a}is$$$$\lambda \rho =1-\sigma =\eta -ϵ\eta ...\left(i\right)$$$$coeff.of\overline{c}is$$$$(1-\rho )=\sigma \mu =\eta ϵ...\left(ii\right)$$$$\Rightarrow 1-\rho =(1-\lambda \rho )\mu$$$$\mathit{\rho }=\frac{1-\mathit{\mu }}{1-\mathit{\lambda }\mathit{\mu }}$$$$\sigma =1-\lambda \rho =1-\left(\frac{\lambda -\lambda \mu }{1-\lambda \mu }\right)$$$$\Rightarrow \mathit{\sigma }=\frac{1-\mathit{\lambda }}{1-\mathit{\lambda }\mathit{\mu }}$$$$adding\left(i\right)and\left(ii\right)$$$$\eta =\lambda \rho +(1-\rho )$$$$=1-(1-\lambda )\rho$$$$=1-\frac{(1-\lambda )(1-\mu )}{1-\lambda \mu }$$$$\mathit{\eta }=\frac{\mathit{\mu }(1-\mathit{\lambda })+\mathit{\lambda }(1-\mathit{\mu })}{1-\mathit{\lambda }\mathit{\mu }}$$$$ϵ=\frac{1-\rho }{\eta }$$$$=[1-\left(\frac{1-\mu }{1-\lambda \mu }\right)]\times \frac{1-\lambda \mu }{\mu (1-\lambda )+\lambda (1-\mu )}$$$$\mathit{ϵ}=\frac{\mathit{\mu }(1-\mathit{\lambda })}{\mathit{\mu }(1-\mathit{\lambda })+\mathit{\lambda }(1-\mathit{\mu })}.$$

Commented by mrW2 last updated on 17/Feb/18

$${That}^{\prime }sreallyclearlylaidout!Thankssir!$$

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