Determine the ratio ##OA:AR## in the vector problem

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The discussion focuses on determining the ratio OA:AR in a vector problem. An alternative method using simultaneous equations is presented, leading to the conclusion that OA:OR is 2:3, which translates to OA:AR being 2:1. The similarity of triangles POR and SOA is also mentioned as a supporting argument for this ratio. Participants emphasize the value of exploring different problem-solving approaches to enhance understanding. The conversation highlights the importance of collaboration in mathematical problem-solving.
chwala
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Homework Statement
see attached
Relevant Equations
Vectors
My question is on part (c) only.

1645417467661.png
Find the markscheme solution below;
1645417555056.png


Mythoughts on this; (Alternative Method)
i used the simultaneous equation
##λ####\left[\dfrac {1}{2}a -\dfrac {1}{4}b\right]##=##\left[ -\dfrac {3}{4}b+ ka\right]## where ##OR=k OA##
##- \dfrac {1}{4}bλ##=## -\dfrac {3}{4}b## ⇒##λ=3## and also
## \dfrac {3}{2}a = ka## ⇒##k=1.5## therefore ##OA:OR=2:3## ⇒##OA:AR=2:1##

your thoughts guys...
 
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Say middle point of OB is S, we know SA and PR are parallel. Similarity of triangles POR and SOA gives OA:AR=2:1.
 
anuttarasammyak said:
Say middle point of OB is S, we know SA and PR are parallel. Similarity of triangles POR and SOA gives OA:AR=2:1.
Thanks Anutta...its always good to explore different ways in solving Math problems...by doing so i realize that i become better...cheers:cool:
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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