Solve the given vector problem

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The forum discussion centers on solving a vector problem from an international past paper, specifically focusing on part iii of the question. The solution involves establishing the relationships between vectors OD and DC using scalar multipliers λ and κ, leading to simultaneous equations that yield λ = 2/3. The mark scheme indicates that part iii is worth two marks, awarded for stating the condition for parallel vectors and correctly calculating vector OD. The discussion highlights the need for clarity in mark allocation, suggesting that the question should have been awarded more marks for its complexity.

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chwala
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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.
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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.
 
Last edited:
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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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