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Please help. Thank you!

smileysmileysmiley

qualitystreet  Sep 18, 2018
edited by qualitystreet  Sep 18, 2018

Best Answer 

 #1
avatar+20744 
+10

Vectors:

\(\text{Let $\vec{AN}=2\vec{OA}=2a$} \\ \text{Let $\vec{ON}=\vec{OA}+\vec{AN}=a+2a=3a$} \\ \text{Let $\vec{OM}=\dfrac12\vec{OB}=\dfrac{b}{2}$} \)

 

\(\mathbf{\vec{AB}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OA} + \vec{AB} &=& \vec{OB} \quad & | \quad \vec{OA}=a \qquad \vec{OB}=b \\ a + \vec{AB} &=& b \\ \mathbf{\vec{AB}} &\mathbf{=}& \mathbf{b-a} \\ \hline \end{array}\)

 

\(\mathbf{\vec{MN}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{MN} &=& \vec{ON} - \vec{OM} \quad & | \quad \vec{ON}=3a \qquad \vec{OM}=\dfrac{b}{2} \\ \mathbf{\vec{MN}} &\mathbf{=}& \mathbf{3a - \dfrac{b}{2}} \\ \hline \end{array}\)


\(\mathbf{\vec{MP}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\vec{MP} &=& a + k(b-a) \\ \vec{MP} &=& a + k(b-a)-\dfrac{b}{2} \\ \vec{MP} &=& a + kb -ka-\dfrac{b}{2} \\ \vec{MP} &=& a(1-k) + b(k-\dfrac12) \\ \mathbf{\vec{MP}} &\mathbf{=}& \mathbf{a(1-k) + b\left(k-\dfrac12 \right)} \\ \hline \end{array}\)

 

\(\mathbf{k} =\ ? \\ \text{Let $\vec{MP}=\lambda\vec{MN}$, where $\lambda$ is a scalar quantity.} \\ \text{Let $\vec{MN}=\vec{ON}-\vec{OM}=3a-\dfrac{b}{2}$} \)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{MP}= \lambda\vec{MN} \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\lambda\vec{MN} &=& a + k(b-a) \quad & | \quad \vec{MN}= 3a-\dfrac{b}{2} \\ \dfrac{b}{2}+\lambda \left(3a-\dfrac{b}{2} \right) &=& a + k(b-a) \\ \dfrac{b}{2}+ 3\lambda a-\dfrac{\lambda}{2}b &=& a + kb-ka \\ 3\lambda a -a+ka &=& kb + \dfrac{\lambda}{2}b -\dfrac{b}{2} \\ a\underbrace{\left( 3\lambda -1+k \right)}_{=0} &=& b\underbrace{\left( k + \dfrac{\lambda}{2} -\dfrac{1}{2} \right)}_{=0} \quad & | \quad \text{linearly independent vectors }~a \text{ and } ~ b \\ \hline k + \dfrac{\lambda}{2} -\dfrac{1}{2} &=& 0 \quad & | \quad \cdot 2 \\ 2k + \lambda -1 &=& 0 \\ \mathbf{ \lambda } & \mathbf{=} & \mathbf{1-2k} \\ \hline 3\lambda -1+k &=& 0 \quad & | \quad \lambda=1-2k \\ 3(1-2k) -1+k &=& 0 \\ 3-6k -1+k &=& 0 \\ 2-5k &=& 0 \\ 5k &=& 2\\ \mathbf{k} &\mathbf{=}& \mathbf{\dfrac25} \\ \hline \end{array}\)

 

 

 

laugh

heureka  Sep 19, 2018
edited by heureka  Sep 19, 2018
 #1
avatar+20744 
+10
Best Answer

Vectors:

\(\text{Let $\vec{AN}=2\vec{OA}=2a$} \\ \text{Let $\vec{ON}=\vec{OA}+\vec{AN}=a+2a=3a$} \\ \text{Let $\vec{OM}=\dfrac12\vec{OB}=\dfrac{b}{2}$} \)

 

\(\mathbf{\vec{AB}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OA} + \vec{AB} &=& \vec{OB} \quad & | \quad \vec{OA}=a \qquad \vec{OB}=b \\ a + \vec{AB} &=& b \\ \mathbf{\vec{AB}} &\mathbf{=}& \mathbf{b-a} \\ \hline \end{array}\)

 

\(\mathbf{\vec{MN}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{MN} &=& \vec{ON} - \vec{OM} \quad & | \quad \vec{ON}=3a \qquad \vec{OM}=\dfrac{b}{2} \\ \mathbf{\vec{MN}} &\mathbf{=}& \mathbf{3a - \dfrac{b}{2}} \\ \hline \end{array}\)


\(\mathbf{\vec{MP}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\vec{MP} &=& a + k(b-a) \\ \vec{MP} &=& a + k(b-a)-\dfrac{b}{2} \\ \vec{MP} &=& a + kb -ka-\dfrac{b}{2} \\ \vec{MP} &=& a(1-k) + b(k-\dfrac12) \\ \mathbf{\vec{MP}} &\mathbf{=}& \mathbf{a(1-k) + b\left(k-\dfrac12 \right)} \\ \hline \end{array}\)

 

\(\mathbf{k} =\ ? \\ \text{Let $\vec{MP}=\lambda\vec{MN}$, where $\lambda$ is a scalar quantity.} \\ \text{Let $\vec{MN}=\vec{ON}-\vec{OM}=3a-\dfrac{b}{2}$} \)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{MP}= \lambda\vec{MN} \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\lambda\vec{MN} &=& a + k(b-a) \quad & | \quad \vec{MN}= 3a-\dfrac{b}{2} \\ \dfrac{b}{2}+\lambda \left(3a-\dfrac{b}{2} \right) &=& a + k(b-a) \\ \dfrac{b}{2}+ 3\lambda a-\dfrac{\lambda}{2}b &=& a + kb-ka \\ 3\lambda a -a+ka &=& kb + \dfrac{\lambda}{2}b -\dfrac{b}{2} \\ a\underbrace{\left( 3\lambda -1+k \right)}_{=0} &=& b\underbrace{\left( k + \dfrac{\lambda}{2} -\dfrac{1}{2} \right)}_{=0} \quad & | \quad \text{linearly independent vectors }~a \text{ and } ~ b \\ \hline k + \dfrac{\lambda}{2} -\dfrac{1}{2} &=& 0 \quad & | \quad \cdot 2 \\ 2k + \lambda -1 &=& 0 \\ \mathbf{ \lambda } & \mathbf{=} & \mathbf{1-2k} \\ \hline 3\lambda -1+k &=& 0 \quad & | \quad \lambda=1-2k \\ 3(1-2k) -1+k &=& 0 \\ 3-6k -1+k &=& 0 \\ 2-5k &=& 0 \\ 5k &=& 2\\ \mathbf{k} &\mathbf{=}& \mathbf{\dfrac25} \\ \hline \end{array}\)

 

 

 

laugh

heureka  Sep 19, 2018
edited by heureka  Sep 19, 2018
 #2
avatar+93038 
+1

Very nice, heureka!!!!

 

cool cool cool

CPhill  Sep 19, 2018

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