July 14th, 2018, 02:42 AM  #1 
Newbie Joined: Jul 2018 From: india Posts: 2 Thanks: 0  vector geometry
ABC is a triangle, P divides BC in the ratio 2:1, and Q divides AC in the same ratio. AP and BQ meet at X. Find in what ratio AP and BQ are divided by X? (Prove by vector method.)
Last edited by skipjack; July 15th, 2018 at 12:39 AM. 
July 14th, 2018, 06:58 AM  #2  
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135  Quote:
$\displaystyle \vec{p} = \dfrac{\vec{b}+2\vec{c}}{3}$ and $\displaystyle \vec{q} = \dfrac{\vec{a}+2\vec{c}}{3}$. So, $\displaystyle 3\vec{p} = \vec{b}+2\vec{c}$ and $\displaystyle 3\vec{q} = \vec{a}+2\vec{c}$. Subtracting we get $\displaystyle 3\vec{p}3\vec{q}=\vec{b}\vec{a}$ or $\displaystyle 3\vec{p}+\vec{a}=3\vec{q}+\vec{b}$ and $\displaystyle \dfrac{3\vec{p}+\vec{a}}{4}=\dfrac{3\vec{q}+\vec{b }}{4}=\vec{x}$. So $\displaystyle X$ divides each cevian into a ratio of $\displaystyle 3:1$. Last edited by skipjack; July 15th, 2018 at 12:40 AM.  
July 15th, 2018, 03:27 AM  #3 
Newbie Joined: Jul 2018 From: india Posts: 2 Thanks: 0 
Thank you


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geometry, vector 
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