Solve for y: Find Ratio of E Dividing CA in Triangle AOB

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In the discussion about finding the ratio in which point E divides line segment CA in triangle AOB, the problem involves a median drawn from vertex A to side OB, which is divided in a 2:1 ratio at point C. The position vectors of points A and B are represented as vectors a and b, with point D being the midpoint of OA. The user seeks additional equations to solve for the ratio y, leading to a suggestion to separate the equations for the independent vectors a and b. The problem was ultimately resolved with the help of forum members.
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Homework Statement


There is a triangle(AOB). One of its median is drawn. BO is divided in the ratio 2:1. The resultant point(say C) is joined with the vertex A. Find the ratio (y) in which E divides CA .

http://img88.imageshack.us/img88/8496/vectorszx0.png

The Attempt at a Solution


Let O b the origin an the position vectors of A and b are \vec{a} and \vec{b}.
Let the median meet OA at D. Position vector of d=\vec{a/2}
Position vector of C=\vec{b/3}
Let E divide BD in ratio x:1
equating the coordinates of e
\frac{x\vec{a/2}+\vec{b}}{x+1}=\frac{y\vec{b/3}+\vec{a}}{y+1}

What I need is more equation to solve for y! Please Help me!
 
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ritwik06 said:
\frac{x\vec{a/2}+\vec{b}}{x+1}=\frac{y\vec{b/3}+\vec{a}}{y+1}

What I need is more equation to solve for y!

Hi ritwik06! :smile:

(on this forum, its easier if you use bold for vectors, than arrows :wink:)

No, all you need to do is to split it into two equations, one for a and one for b.

Remember, a and b are independent, so if pa + qb= 0, then p = q = 0. :wink:
 
tiny-tim said:
hi ritwik06! :smile:

(on this forum, its easier if you use bold for vectors, than arrows :wink:)

no, all you need to do is to split it into two equations, one for a and one for b.

Remember, a and b are independent, so if pa + qb= 0, then p = q = 0. :wink:
SOLVED
thanks a lot.
 
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