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+1
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smileysmileysmiley

qualitystreet  Sep 16, 2018
edited by qualitystreet  Sep 16, 2018
 #1
avatar+20591 
+8

Vectors:

 

a)
Find CB:

\(\begin{array}{|rcll|} \hline -DC+DA+AB&=& CB \\ -p+2q-p+5p &=& CB \\ 2q + 3p &=& CB \\ \mathbf{\vec{CB}} & \mathbf{=} & \mathbf{2q + 3p} \\ \hline \end{array}\)

 

b)

\(\begin{array}{|rcll|} \hline AQ &=& \dfrac25 AB \quad & | \quad AB =5p \\ &=& \dfrac25 \cdot 5p \\ \mathbf{AQ} & \mathbf{=} & \mathbf{2p} \\\\ QB &=& \dfrac35 AB \quad & | \quad AB =5p \\ &=& \dfrac35 \cdot 5p \\ \mathbf{QB} & \mathbf{=} & \mathbf{3p} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{DA}{2}+PQ+QB-CB-DC &=& 0 \quad | \quad DA =2q-p \quad QB = 3p \quad DC = p \\ \dfrac{2q-p}{2}+PQ+3p-CB-p &=& 0 \\ \dfrac{2q-p}{2}+PQ+2p-CB &=& 0 \\ \dfrac{2q-p}{2}+PQ+2p &=& CB \\ && \mathbf{\vec{CB}=2q + 3p} \\ && 2q = CB - 3p \\ && q = \dfrac{CB - 3p}{2} \\ \dfrac{2\left(\dfrac{CB - 3p}{2}\right)-p}{2}+PQ+2p &=& CB \\ \dfrac{CB - 3p-p}{2}+PQ+2p &=& CB \\ \dfrac{CB - 4p}{2}+PQ+2p &=& CB \\ \dfrac{CB}{2} - 2p+PQ+2p &=& CB \\ \dfrac{CB}{2}+PQ&=& CB \\ PQ&=& CB -\dfrac{CB}{2} \\ \mathbf{PQ} & \mathbf{=} & \mathbf{\dfrac12\cdot CB} \\ \hline \end{array}\)

 

laugh

heureka  Sep 17, 2018
edited by heureka  Sep 18, 2018

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