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 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
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Quote:
 Originally Posted by chandan gowda 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.)
Assuming you mean $\displaystyle \dfrac{BP}{PC}=\dfrac{2}{1}$ and $\displaystyle \dfrac{AQ}{QC}=\dfrac{2}{1}$ and assuming that $\displaystyle \vec{a}$ is the vector from an arbitrary origin to the point $\displaystyle A$, we can say the following:

$\displaystyle \vec{p} = \dfrac{\vec{b}+2\vec{c}}{3}$ 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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