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Need help on this question please

 May 14, 2018
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Need help on this question please

 

\(\text{Let $\vec{OA} = \vec{a}$ } \\ \text{Let $\vec{OB} = \vec{b}$ } \\ \text{Let $\vec{OM} = \frac12\vec{OB} = \frac12 \vec{b}$ } \\ \text{Let $\vec{AP} = k\vec{AB} $ } \)

 

\(\begin{array}{|rcll|} \hline AN = 2\cdot OA \quad & \Rightarrow & \quad \vec{AN} = 2\vec{OA} \\ & & \quad \vec{AN} = 2\vec{a} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \vec{ON} &=& \vec{OA} + \vec{AN} \\ \vec{ON} &=& \vec{a} + 2\vec{a} \\ \vec{ON} &=& 3\vec{a} \\ \hline \end{array} \)

 

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

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

\(\begin{array}{|rcll|} \hline \vec{AP} &=& k\vec{AB} \\ \mathbf{\vec{AP}} & \mathbf{=} & \mathbf{k(\vec{b}-\vec{a})} \\ \hline \end{array}\)


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

\(\begin{array}{|rcll|} \hline \vec{MN} &=& \vec{ON}-\vec{OM} \\ \mathbf{\vec{MN}} & \mathbf{=} & \mathbf{3\vec{a}-\frac12\vec{b}} \\ \hline \end{array}\)


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

\(\begin{array}{|rcll|} \hline \vec{MP} &=& \vec{OA} + \vec{AP} - \vec{OM} \\ \vec{MP} &=& \vec{a} + k(\vec{b}-\vec{a}) - \frac12 \vec{b} \\ \vec{MP} &=& \vec{a}-k\vec{a} + k\vec{b} - \frac12 \vec{b} \\ \mathbf{\vec{MP}} & \mathbf{=} & \mathbf{ \vec{a}(1-k) +\vec{b}(k-\frac12) } \\ \hline \end{array}\)

 

\(k=\ ?\)

\(\begin{array}{|rclrcl|} \hline \vec{OA} + \vec{AP} &=& \vec{OM} + \vec{MP} \\ \vec{a} + k(\vec{b}-\vec{a}) &=& \frac12 \vec{b} + \lambda\vec{MN} \quad & | \quad \vec{MN} = 3\vec{a}-\frac12\vec{b} \\ \vec{a} + k(\vec{b}-\vec{a}) &=& \frac12 \vec{b} + \lambda(3\vec{a}-\frac12\vec{b}) \\ \vec{a} + k\vec{b}-k\vec{a} &=& \frac12 \vec{b} + 3\lambda\vec{a}-\frac{\lambda}{2}\vec{b} \\ \vec{a} -k\vec{a} - 3\lambda\vec{a} &=& \frac12 \vec{b} -\frac{\lambda}{2}\vec{b} - k\vec{b} \\ \vec{a}(\underbrace{1 -k - 3\lambda}_{=0}) &=& \vec{b}(\underbrace{\frac12 -\frac{\lambda}{2} - k}_{=0}) \\ \frac12 -\frac{\lambda}{2} - k &=& 0 \quad | \cdot 6 & 1 -k - 3\lambda &=& 0 \\ 3 -3\lambda - 6k &=& 0 & 3\lambda &=& 1 -k \\ 3 -(1 -k ) - 6k &=& 0 \\ 3 - 1 +k - 6k &=& 0 \\ 2 - 5k &=& 0 \\ 5k &=& 2 \\ \mathbf{k} & \mathbf{=} & \mathbf{\dfrac25 } \\ \hline \end{array}\)

 

 

laugh

 May 15, 2018

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