Solve the given vector problem

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The discussion centers on solving a vector problem from an international past paper, specifically focusing on part iii. The mark scheme awards marks for correctly identifying methods and obtaining the right answers, with part (iii) worth two marks for stating conditions for parallel vectors and finding vector OD. The participant's approach involves setting up simultaneous equations based on vector relationships, ultimately determining that OD equals (4/3)i - 2j. There is a suggestion that the question should have been allocated more marks due to its complexity. Overall, the mark scheme appears consistent in its evaluation criteria for the problem.
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Homework Statement
This is an international past paper question- I have attached the question and the markscheme... the ms was a bit confusing for 2 marks hence my post.
Relevant Equations
vectors.
This is an international past paper question- I have attached the question and the markscheme... the ms was a bit confusing for 2 marks hence my post.

Question; interest is on part iii. only

1673515205191.png


Mark scheme solution;

1673515315922.png
My thinking;

Let

##OD=λOA## Where ##λ## is a scalar.

##OD=λ
\begin{pmatrix}
2 & \\
-3 & \\
\end{pmatrix}##

Let

##DC=κOB## Where ##κ## is a scalar.

##DC=κ
\begin{pmatrix}
11 & \\
42 & \\
\end{pmatrix}##



##OD+DC=OC##

##λ
\begin{pmatrix}
2 & \\
-3 & \\
\end{pmatrix}

\begin{pmatrix}
11 & \\
42 & \\
\end{pmatrix}
=
\begin{pmatrix}
5 & \\
12 & \\
\end{pmatrix}
##

We end up with the simultaneous equation;

##2λ+11κ=5##
##-3λ+42κ=12##

##39λ=26##

##λ=\dfrac{2}{3}##

therefore,

##OD=\dfrac{2}{3}
\begin{pmatrix}
2 & \\
-3 & \\
\end{pmatrix}=\dfrac{4}{3} i -2j
##

Unless, there is something i have overlooked on the ms... the question ought to have been given more marks...cheers

Your insight highly appreciated.
 
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The mark scheme seems pretty consistent in awarding one mark for choosing a correct method and one mark for getting the correct answer using that method. So part (i) is worth two marks: one for observing that \vec{AB} = \vec{OB} - \vec{OA} and one for doing the subtraction. Part (ii) is worth 4 marks: 2 for finding \vec{OC} and 2 for calculating its length. Part (iii) is worth 2 marks: one for stating a condition which ensures that \vec{DC} and \vec{OB} are parallel, and one for using that condition to find \vec{OD}.

For part (iii), the mark scheme gives two methods to find \vec{OD}, each of which get the same number of marks:
  • If \vec{DC} and \vec{OB} are parallel, then OAB and DAC are similar triangles. We know \vec{AC} = \frac13 \vec{AB}, so \vec{AD} = \frac13\vec{AO} and hence \vec{OD} = \frac23 \vec{OA}. (This is the most obvious method if you've drawn a diagram.)
  • If vectors are parallel then the ratios of the \mathbf{i} and \mathbf{j} components are equal; we know that \vec{DC} = (5 - 2\lambda)\mathbf{i} + (12 + 3\lambda)\mathbf{j} for some \lambda and we know \vec{OB}.
So whatever method you use to determine \vec{OD}, and yours involves more steps than both of these, you get one method mark and one answer mark.
 
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