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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

$$\left(a\right)\left(i\right)$$$$\left(i\right)O\overrightarrow{E}=\lambda O\overrightarrow{D}+(1-\lambda )O\overrightarrow{C}$$$$=\lambda \left(4\overrightarrow{b}\right)+(1-\lambda )\left(\frac{4}{3}\overrightarrow{a}\right)$$$$\left(ii\right)=\mu O\overrightarrow{B}+(1-\mu )O\overrightarrow{A}$$$$=\mu \left(2\overrightarrow{b}\right)+(1-\mu )\left(2\overrightarrow{a}\right)$$$$\left(b\right)O\overrightarrow{E}=\lambda O\overrightarrow{D}+(1-\lambda )O\overrightarrow{C}$$$$=\lambda \left(2O\overrightarrow{B}\right)+(1-\lambda )\left(\frac{2}{3}O\overrightarrow{A}\right)$$$$\Rightarrow \frac{4}{3}\lambda +\frac{2}{3}=1\Rightarrow \lambda =\frac{1}{4}$$$$\Rightarrow O\overrightarrow{E}=\frac{1}{4}O\overrightarrow{D}+\frac{3}{4}O\overrightarrow{C}$$$$\Rightarrow CE:ED=1:3$$$$2\lambda =\mu$$$$\Rightarrow \mu =\frac{1}{2}$$$$\Rightarrow O\overrightarrow{E}=\frac{1}{2}O\overrightarrow{B}+\frac{1}{2}O\overrightarrow{A}$$$$\Rightarrow AE:EB=1:1$$

Commented by Tony Lin last updated on 28/Jul/19

$$oryoucanuseMenelaustheorem$$$$\frac{AC}{CO}\times \frac{OD}{DB}\times \frac{BE}{EA}=1$$$$\frac{1}{2}\times \frac{2}{1}\times \frac{BE}{EA}=1$$$$\Rightarrow BE:EA=1:1$$$$\Rightarrow O\overrightarrow{E}=\frac{1}{2}O\overrightarrow{A}+\frac{1}{2}O\overrightarrow{B}$$$$\Rightarrow \mu =\frac{1}{2}$$$$\frac{DB}{BO}\times \frac{OA}{AC}\times \frac{CE}{ED}=1$$$$\frac{1}{1}\times \frac{3}{1}\times \frac{CE}{ED}=1$$$$\Rightarrow CE:ED=1:3$$$$\Rightarrow O\overrightarrow{E}=\frac{1}{4}O\overrightarrow{D}+\frac{3}{4}O\overrightarrow{C}$$$$\Rightarrow \lambda =\frac{1}{4}$$

Commented by Tawa1 last updated on 28/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

Answered by mr W last updated on 28/Jul/19

$$\left(a\right)$$$$\left(i\right)$$$$CD=-CO+OD=-\frac{4}{3}a+4b$$$$CE=\lambda CD=-\frac{4\lambda }{3}a+4\lambda b$$$$\Rightarrow OE=OC+CE=\frac{4}{3}a-\frac{4\lambda }{3}a+4\lambda b=\frac{4(1-\lambda )}{3}a+4\lambda b$$$$\left(ii\right)$$$$AB=AO+OB=-2a+2b$$$$AE=\mu AB=-2\mu a+2\mu b$$$$OE=OA+AE=2a-2\mu a+2\mu b=2(1-\mu )a+2\mu b$$$$$$$$\left(b\right)$$$$\frac{4(1-\lambda )}{3}=2(1-\mu )$$$$\Rightarrow 2(1-\lambda )=3(1-\mu )$$$$\Rightarrow 3\mu -2\lambda =1...\left(i\right)$$$$4\lambda =2\mu$$$$\Rightarrow 2\lambda -\mu =0...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$6\lambda -2\lambda =1$$$$\Rightarrow \lambda =\frac{1}{4}$$$$\Rightarrow \mu =\frac{1}{2}$$

Commented by Tawa1 last updated on 28/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$

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