We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+3
234
2
avatar+429 

Please help. Thank you!

smileysmileysmiley

 Sep 18, 2018
edited by qualitystreet  Sep 18, 2018

Best Answer 

 #1
avatar+21977 
+11

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

 Sep 19, 2018
edited by heureka  Sep 19, 2018
 #1
avatar+21977 
+11
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+99351 
+1

Very nice, heureka!!!!

 

cool cool cool

CPhill  Sep 19, 2018

17 Online Users