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Help please! How to find position vector OD? (Why does OD makes the line CB in the ratio 1:2 ?) 

 Apr 30, 2021

Best Answer 

 #1
avatar+26117 
+2

Help please!

\(\text{Let $\vec{AC}=\vec{a}$} \\ \text{Let $\vec{CB}=\vec{b}$} \\ \text{Let $\vec{CD}=(1-\lambda)\vec{b}$} \\ \text{Let $\vec{DB}=\lambda\vec{b}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ABC}: \\ \hline \dfrac{1}{2}\vec{OB} &=& \vec{a} + \vec{b} \\ \mathbf{ \vec{OB} } &=& \mathbf{2(\vec{a} + \vec{b})} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{OAE}: \\ \hline \vec{OE} &=& \dfrac{1}{2}\vec{OB} + \dfrac{1}{2}\vec{a} \quad | \quad \mathbf{ \vec{OB}=2(\vec{a} + \vec{b})} \\\\ \vec{OE} &=& \dfrac{1}{2}\times 2(\vec{a} + \vec{b}) + \dfrac{1}{2}\vec{a} \\\\ \vec{OE} &=& \vec{a} + \vec{b} + \dfrac{1}{2}\vec{a} \\\\ \mathbf{ \vec{OE} } &=& \mathbf{\dfrac{3}{2}\vec{a} + \vec{b}} \\ \hline \end{array} \)

 

\(\text{Let $\vec{OD}=\mu\vec{OE}$} \\ \text{Let $\vec{OD}=\vec{OB}-\vec{DB}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ODB}: \\ \hline \vec{OD} = \mu\vec{OE} &=& \vec{OB}-\vec{DB} \\ \mu\vec{OE} &=& \vec{OB}-\vec{DB} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b},~\vec{OB}=2(\vec{a} + \vec{b}),~ \vec{DB}=\lambda\vec{b} \\\\ \mu \left(\dfrac{3}{2}\vec{a} + \vec{b}\right) &=& 2(\vec{a} + \vec{b})-\lambda\vec{b} \\\\ \dfrac{3}{2}\vec{a}\mu + \mu \vec{b} &=& 2\vec{a} + 2\vec{b}-\lambda\vec{b} \\\\ \vec{a}\underbrace{ \left(\dfrac{3}{2}\mu-2\right)}_{=0} &=& \vec{b}\underbrace{\left( 2-\mu-\lambda\right)}_{=0} \\\\ \hline \dfrac{3}{2}\mu-2 &=& 0 \\ \dfrac{3}{2}\mu &=& 2 \\ \mathbf{\mu} &=& \mathbf{\dfrac{4}{3}} \\\\ \vec{OD} &=& \mu\vec{OE} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b}\\ \vec{OD} &=& \dfrac{4}{3} \left(\dfrac{3}{2}\vec{a} + \vec{b} \right)\\ \mathbf{\vec{OD}} &=& \mathbf{2\vec{a} + \dfrac{4}{3} \vec{b}} \\ \hline 2-\mu-\lambda &=& 0 \\ \lambda &=& 2-\mu \quad | \quad \mathbf{\mu=\dfrac{4}{3}} \\ \lambda &=& 2-\dfrac{4}{3} \\ \lambda &=& \dfrac{6-4}{3} \\ \mathbf{\lambda} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{CD} &=& (1-\lambda)\vec{b} \\ \vec{CD} &=& \left(1-\dfrac{2}{3}\right)\vec{b} \\ \mathbf{\vec{CD}} &=& \mathbf{\dfrac{1}{3}\vec{b}} \\\\ \vec{DB} &=& \lambda\vec{b} \\ \mathbf{\vec{DB}} &=& \mathbf{\dfrac{2}{3}\vec{b}} \\ \hline \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}\vec{b}}{\dfrac{2}{3}\vec{b}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{1}{3} \times \dfrac{3}{2} \\\\ \mathbf{\dfrac{\vec{CD}}{\vec{DB}}} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

 

laugh

 Apr 30, 2021
 #1
avatar+26117 
+2
Best Answer

Help please!

\(\text{Let $\vec{AC}=\vec{a}$} \\ \text{Let $\vec{CB}=\vec{b}$} \\ \text{Let $\vec{CD}=(1-\lambda)\vec{b}$} \\ \text{Let $\vec{DB}=\lambda\vec{b}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ABC}: \\ \hline \dfrac{1}{2}\vec{OB} &=& \vec{a} + \vec{b} \\ \mathbf{ \vec{OB} } &=& \mathbf{2(\vec{a} + \vec{b})} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{OAE}: \\ \hline \vec{OE} &=& \dfrac{1}{2}\vec{OB} + \dfrac{1}{2}\vec{a} \quad | \quad \mathbf{ \vec{OB}=2(\vec{a} + \vec{b})} \\\\ \vec{OE} &=& \dfrac{1}{2}\times 2(\vec{a} + \vec{b}) + \dfrac{1}{2}\vec{a} \\\\ \vec{OE} &=& \vec{a} + \vec{b} + \dfrac{1}{2}\vec{a} \\\\ \mathbf{ \vec{OE} } &=& \mathbf{\dfrac{3}{2}\vec{a} + \vec{b}} \\ \hline \end{array} \)

 

\(\text{Let $\vec{OD}=\mu\vec{OE}$} \\ \text{Let $\vec{OD}=\vec{OB}-\vec{DB}$} \\ \begin{array}{|rcll|} \hline \text{In}~ \triangle \text{ODB}: \\ \hline \vec{OD} = \mu\vec{OE} &=& \vec{OB}-\vec{DB} \\ \mu\vec{OE} &=& \vec{OB}-\vec{DB} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b},~\vec{OB}=2(\vec{a} + \vec{b}),~ \vec{DB}=\lambda\vec{b} \\\\ \mu \left(\dfrac{3}{2}\vec{a} + \vec{b}\right) &=& 2(\vec{a} + \vec{b})-\lambda\vec{b} \\\\ \dfrac{3}{2}\vec{a}\mu + \mu \vec{b} &=& 2\vec{a} + 2\vec{b}-\lambda\vec{b} \\\\ \vec{a}\underbrace{ \left(\dfrac{3}{2}\mu-2\right)}_{=0} &=& \vec{b}\underbrace{\left( 2-\mu-\lambda\right)}_{=0} \\\\ \hline \dfrac{3}{2}\mu-2 &=& 0 \\ \dfrac{3}{2}\mu &=& 2 \\ \mathbf{\mu} &=& \mathbf{\dfrac{4}{3}} \\\\ \vec{OD} &=& \mu\vec{OE} \quad | \quad \vec{OE}=\dfrac{3}{2}\vec{a} + \vec{b}\\ \vec{OD} &=& \dfrac{4}{3} \left(\dfrac{3}{2}\vec{a} + \vec{b} \right)\\ \mathbf{\vec{OD}} &=& \mathbf{2\vec{a} + \dfrac{4}{3} \vec{b}} \\ \hline 2-\mu-\lambda &=& 0 \\ \lambda &=& 2-\mu \quad | \quad \mathbf{\mu=\dfrac{4}{3}} \\ \lambda &=& 2-\dfrac{4}{3} \\ \lambda &=& \dfrac{6-4}{3} \\ \mathbf{\lambda} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{CD} &=& (1-\lambda)\vec{b} \\ \vec{CD} &=& \left(1-\dfrac{2}{3}\right)\vec{b} \\ \mathbf{\vec{CD}} &=& \mathbf{\dfrac{1}{3}\vec{b}} \\\\ \vec{DB} &=& \lambda\vec{b} \\ \mathbf{\vec{DB}} &=& \mathbf{\dfrac{2}{3}\vec{b}} \\ \hline \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}\vec{b}}{\dfrac{2}{3}\vec{b}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} \\\\ \dfrac{\vec{CD}}{\vec{DB}} &=& \dfrac{1}{3} \times \dfrac{3}{2} \\\\ \mathbf{\dfrac{\vec{CD}}{\vec{DB}}} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

 

laugh

heureka Apr 30, 2021
 #2
avatar+121000 
+1

Thanks, heureka......good to see you  back  !!!!!

 

 

 

cool cool cool

CPhill  Apr 30, 2021
 #3
avatar
+1

Thank you very much heureka!!!

Guest May 1, 2021
 #4
avatar+26117 
0

Thank you, CPhill !

 

laugh

heureka  May 1, 2021
edited by heureka  May 1, 2021
edited by heureka  May 1, 2021

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