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Question Number 65325 by Tawa1 last updated on 28/Jul/19
Commented by Tony Lin last updated on 28/Jul/19
(a)(i) (i)OE^→ =λOD^→ +(1−λ)OC^→ =λ(4b^→ )+(1−λ)((4/3)a^→ ) (i i) =μOB^→ +(1−μ)OA^→ =μ(2b^→ )+(1−μ)(2a^→ ) (b)OE^→ =λOD^→ +(1−λ)OC^→ =λ(2OB^→ )+(1−λ)((2/3)OA^→ ) ⇒(4/3)λ+(2/3)=1⇒λ=(1/4) ⇒OE^→ =(1/4)OD^→ +(3/4)OC^→ ⇒CE: ED=1: 3 2λ=μ ⇒μ=(1/2) ⇒OE^→ =(1/2)OB^→ +(1/2)OA^→ ⇒AE: EB=1: 1
$$\left({a}\right)\left({i}\right) \\ $$$$\left({i}\right){O}\overset{\rightarrow} {{E}}=\lambda{O}\overset{\rightarrow} {{D}}+\left(\mathrm{1}−\lambda\right){O}\overset{\rightarrow} {{C}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\lambda\left(\mathrm{4}\overset{\rightarrow} {{b}}\right)+\left(\mathrm{1}−\lambda\right)\left(\frac{\mathrm{4}}{\mathrm{3}}\overset{\rightarrow} {{a}}\right) \\ $$$$\:\left({i}\:{i}\right)\:\:\:\:=\mu{O}\overset{\rightarrow} {{B}}+\left(\mathrm{1}−\mu\right){O}\overset{\rightarrow} {{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mu\left(\mathrm{2}\overset{\rightarrow} {{b}}\right)+\left(\mathrm{1}−\mu\right)\left(\mathrm{2}\overset{\rightarrow} {{a}}\right) \\ $$$$\left({b}\right){O}\overset{\rightarrow} {{E}}=\lambda{O}\overset{\rightarrow} {{D}}+\left(\mathrm{1}−\lambda\right){O}\overset{\rightarrow} {{C}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\lambda\left(\mathrm{2}{O}\overset{\rightarrow} {{B}}\right)+\left(\mathrm{1}−\lambda\right)\left(\frac{\mathrm{2}}{\mathrm{3}}{O}\overset{\rightarrow} {{A}}\right) \\ $$$$\Rightarrow\frac{\mathrm{4}}{\mathrm{3}}\lambda+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{1}\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{O}\overset{\rightarrow} {{E}}=\frac{\mathrm{1}}{\mathrm{4}}{O}\overset{\rightarrow} {{D}}+\frac{\mathrm{3}}{\mathrm{4}}{O}\overset{\rightarrow} {{C}} \\ $$$$\Rightarrow{CE}:\:{ED}=\mathrm{1}:\:\mathrm{3} \\ $$$$\mathrm{2}\lambda=\mu \\ $$$$\Rightarrow\mu=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{O}\overset{\rightarrow} {{E}}=\frac{\mathrm{1}}{\mathrm{2}}{O}\overset{\rightarrow} {{B}}+\frac{\mathrm{1}}{\mathrm{2}}{O}\overset{\rightarrow} {{A}} \\ $$$$\Rightarrow{AE}:\:{EB}=\mathrm{1}:\:\mathrm{1} \\ $$
Commented by Tony Lin last updated on 28/Jul/19
or you can use Menelaus theorem ((AC)/(CO))×((OD)/(DB))×((BE)/(EA))=1 (1/2)×(2/1)×((BE)/(EA))=1 ⇒BE: EA=1: 1 ⇒OE^→ =(1/2)OA^→ +(1/2)OB^→ ⇒μ=(1/2) ((DB)/(BO))×((OA)/(AC))×((CE)/(ED))=1 (1/1)×(3/1)×((CE)/(ED))=1 ⇒CE: ED=1: 3 ⇒OE^→ =(1/4)OD^→ +(3/4)OC^→ ⇒λ=(1/4)
$${or}\:{you}\:{can}\:{use}\:{Menelaus}\:{theorem} \\ $$$$\frac{{AC}}{{CO}}×\frac{{OD}}{{DB}}×\frac{{BE}}{{EA}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{1}}×\frac{{BE}}{{EA}}=\mathrm{1} \\ $$$$\Rightarrow{BE}:\:{EA}=\mathrm{1}:\:\mathrm{1} \\ $$$$\Rightarrow{O}\overset{\rightarrow} {{E}}=\frac{\mathrm{1}}{\mathrm{2}}{O}\overset{\rightarrow} {{A}}+\frac{\mathrm{1}}{\mathrm{2}}{O}\overset{\rightarrow} {{B}} \\ $$$$\Rightarrow\mu=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{DB}}{{BO}}×\frac{{OA}}{{AC}}×\frac{{CE}}{{ED}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}}×\frac{\mathrm{3}}{\mathrm{1}}×\frac{{CE}}{{ED}}=\mathrm{1} \\ $$$$\Rightarrow{CE}:\:{ED}=\mathrm{1}:\:\mathrm{3} \\ $$$$\Rightarrow{O}\overset{\rightarrow} {{E}}=\frac{\mathrm{1}}{\mathrm{4}}{O}\overset{\rightarrow} {{D}}+\frac{\mathrm{3}}{\mathrm{4}}{O}\overset{\rightarrow} {{C}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Tawa1 last updated on 28/Jul/19
God bless you sir. I appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$
Answered by mr W last updated on 28/Jul/19
(a) (i) CD=−CO+OD=−(4/3)a+4b CE=λCD=−((4λ)/3)a+4λb ⇒OE=OC+CE=(4/3)a−((4λ)/3)a+4λb=((4(1−λ))/3)a+4λb (ii) AB=AO+OB=−2a+2b AE=μAB=−2μa+2μb OE=OA+AE=2a−2μa+2μb=2(1−μ)a+2μb (b) ((4(1−λ))/3)=2(1−μ) ⇒2(1−λ)=3(1−μ) ⇒3μ−2λ=1 ...(i) 4λ=2μ ⇒2λ−μ=0 ...(ii) from (i) and (ii): 6λ−2λ=1 ⇒λ=(1/4) ⇒μ=(1/2)
$$\left({a}\right) \\ $$$$\left({i}\right) \\ $$$${CD}=−{CO}+{OD}=−\frac{\mathrm{4}}{\mathrm{3}}{a}+\mathrm{4}{b} \\ $$$${CE}=\lambda{CD}=−\frac{\mathrm{4}\lambda}{\mathrm{3}}{a}+\mathrm{4}\lambda{b} \\ $$$$\Rightarrow{OE}={OC}+{CE}=\frac{\mathrm{4}}{\mathrm{3}}{a}−\frac{\mathrm{4}\lambda}{\mathrm{3}}{a}+\mathrm{4}\lambda{b}=\frac{\mathrm{4}\left(\mathrm{1}−\lambda\right)}{\mathrm{3}}{a}+\mathrm{4}\lambda{b} \\ $$$$\left({ii}\right) \\ $$$${AB}={AO}+{OB}=−\mathrm{2}{a}+\mathrm{2}{b} \\ $$$${AE}=\mu{AB}=−\mathrm{2}\mu{a}+\mathrm{2}\mu{b} \\ $$$${OE}={OA}+{AE}=\mathrm{2}{a}−\mathrm{2}\mu{a}+\mathrm{2}\mu{b}=\mathrm{2}\left(\mathrm{1}−\mu\right){a}+\mathrm{2}\mu{b} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\frac{\mathrm{4}\left(\mathrm{1}−\lambda\right)}{\mathrm{3}}=\mathrm{2}\left(\mathrm{1}−\mu\right) \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{1}−\lambda\right)=\mathrm{3}\left(\mathrm{1}−\mu\right) \\ $$$$\Rightarrow\mathrm{3}\mu−\mathrm{2}\lambda=\mathrm{1}\:\:\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{4}\lambda=\mathrm{2}\mu \\ $$$$\Rightarrow\mathrm{2}\lambda−\mu=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\mathrm{6}\lambda−\mathrm{2}\lambda=\mathrm{1} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mu=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 28/Jul/19
God bless you sir, i appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

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