OCR MEI C4 — Question 4 8 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeRatio division of line segment
DifficultyModerate -0.8 This is a straightforward multi-part vectors question requiring only basic operations: midpoint formula, vector subtraction, vector addition, and verifying collinearity. All parts use standard techniques with no problem-solving insight needed, making it easier than average but not trivial due to the multiple steps involved.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement
4 The points \(\mathrm { A } , \mathrm { B }\) and C are given by the position vectors \(\mathbf { a } = \binom { - 2 } { 1 } , \mathbf { b } = \binom { 0 } { 5 }\) and \(\mathbf { c } = \binom { 4 } { 3 }\). M is the midpoint of AC .
  1. Find the position vector of M .
  2. Find the vector \(\overrightarrow { B C }\).
  3. Find the position vector of the point D such that \(\overrightarrow { \mathrm { BC } } = \overrightarrow { \mathrm { AD } }\).
  4. Show that D lies on BM .
Question 4(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(\underline{m} = \begin{pmatrix}1\\2\end{pmatrix}\)B1
Total: 1 mark
Question 4(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\overrightarrow{BC} = \begin{pmatrix}4\\-2\end{pmatrix}\)M1 A1
Total: 2 marks
Question 4(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\overrightarrow{BC} = \overrightarrow{AD} \Rightarrow \underline{c} - \underline{b} = \underline{d} - \underline{a}\)
\(\Rightarrow \underline{d} = \underline{a} + \underline{c} - \underline{b} = \begin{pmatrix}2\\-1\end{pmatrix}\)M1 A1
Total: 2 marks
Question 4(iv):
AnswerMarks Guidance
AnswerMarks Guidance
\(\overrightarrow{BD} = \underline{d} - \underline{b} = \begin{pmatrix}2\\-6\end{pmatrix}\)M1
\(\overrightarrow{BM} = \underline{m} - \underline{b} = \begin{pmatrix}1\\-3\end{pmatrix}\)A1
\(\overrightarrow{BD} = 2\overrightarrow{BM}\), hence B, D, M are collinearB1
Total: 3 marks 
## Question 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\underline{m} = \begin{pmatrix}1\\2\end{pmatrix}$ | B1 | |
| **Total: 1 mark** | | |

## Question 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{BC} = \begin{pmatrix}4\\-2\end{pmatrix}$ | M1 A1 | |
| **Total: 2 marks** | | |

## Question 4(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{BC} = \overrightarrow{AD} \Rightarrow \underline{c} - \underline{b} = \underline{d} - \underline{a}$ | | |
| $\Rightarrow \underline{d} = \underline{a} + \underline{c} - \underline{b} = \begin{pmatrix}2\\-1\end{pmatrix}$ | M1 A1 | |
| **Total: 2 marks** | | |

## Question 4(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{BD} = \underline{d} - \underline{b} = \begin{pmatrix}2\\-6\end{pmatrix}$ | M1 | |
| $\overrightarrow{BM} = \underline{m} - \underline{b} = \begin{pmatrix}1\\-3\end{pmatrix}$ | A1 | |
| $\overrightarrow{BD} = 2\overrightarrow{BM}$, hence B, D, M are collinear | B1 | |
| **Total: 3 marks** | | |

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4 The points $\mathrm { A } , \mathrm { B }$ and C are given by the position vectors $\mathbf { a } = \binom { - 2 } { 1 } , \mathbf { b } = \binom { 0 } { 5 }$ and $\mathbf { c } = \binom { 4 } { 3 }$. M is the midpoint of AC .\\
(i) Find the position vector of M .\\
(ii) Find the vector $\overrightarrow { B C }$.\\
(iii) Find the position vector of the point D such that $\overrightarrow { \mathrm { BC } } = \overrightarrow { \mathrm { AD } }$.\\
(iv) Show that D lies on BM .

\hfill \mbox{\textit{OCR MEI C4  Q4 [8]}}
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