Question 9 - A-Level Maths

CAIE P1 2017 November — Question 9 9 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2017
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Triangle and parallelogram problems
Difficulty	Standard +0.3 This is a straightforward 3D vectors question requiring standard techniques: dot product to verify perpendicularity, using parallel vectors with given magnitude ratio, and trapezium area formula. All steps are routine applications of basic vector operations with no novel insight required, making it slightly easier than average.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10d Vector operations: addition and scalar multiplication 1.10f Distance between points: using position vectors 1.10g Problem solving with vectors: in geometry

9 \includegraphics[max width=\textwidth, alt={}, center]{518bb805-5b14-4b41-94fd-38a31a90c218-16_533_601_258_772} The diagram shows a trapezium \(O A B C\) in which \(O A\) is parallel to \(C B\). The position vectors of \(A\) and \(B\) relative to the origin \(O\) are given by \(\overrightarrow { O A } = \left( \begin{array} { r } 2 \\ - 2 \\ - 1 \end{array} \right)\) and \(\overrightarrow { O B } = \left( \begin{array} { l } 6 \\ 1 \\ 1 \end{array} \right)\).

Show that angle \(O A B\) is \(90 ^ { \circ }\).
The magnitude of \(\overrightarrow { C B }\) is three times the magnitude of \(\overrightarrow { O A }\).
Find the position vector of \(C\).
Find the exact area of the trapezium \(O A B C\), giving your answer in the form \(a \sqrt { } b\), where \(a\) and \(b\) are integers.

Show mark scheme Show mark scheme source

Question 9(i):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix}\) or \(\overrightarrow{BA} = \begin{pmatrix} -4 \\ -3 \\ -2 \end{pmatrix}\)	M1	Use of \(\mathbf{b} - \mathbf{a}\) or \(\mathbf{a} - \mathbf{b}\)
e.g. \(\overrightarrow{AO} \cdot \overrightarrow{AB} = -8 + 6 + 2 = 0 \rightarrow O\hat{A}B = 90°\) AG	M1 A1	Use of dot product with either \(\overrightarrow{AO}\) or \(\overrightarrow{OA}\) & either \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\). Must see 3 component products
OR \(	\overrightarrow{OA}	= 3,
3

Question 9(ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(\overrightarrow{CB} = \begin{pmatrix} 6 \\ -6 \\ -3 \end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix}\)	B1	Must correctly identify the vector
\(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}\) (or \(-\overrightarrow{CB}\)) \(= \begin{pmatrix} 0 \\ 7 \\ 4 \end{pmatrix}\)	M1 A1	Correct link leading to \(\overrightarrow{OC}\)
3

Question 9(iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(	\overrightarrow{OA}	= 3,
Area \(= \frac{1}{2}(3 + 9)\sqrt{29}\) or \(3\sqrt{29} + 3\sqrt{29}\)	M1	Correct formula(e) used for trapezium or (rectangle + triangle) or two triangles using their lengths
\(= 6\sqrt{29}\) \((1\sqrt{1044}, 2\sqrt{261}\) or \(3\sqrt{116})\)	A1	Exact answer in correct form
3

## Question 9(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \\ 2 \end{pmatrix}$ or $\overrightarrow{BA} = \begin{pmatrix} -4 \\ -3 \\ -2 \end{pmatrix}$ | M1 | Use of $\mathbf{b} - \mathbf{a}$ or $\mathbf{a} - \mathbf{b}$ |
| e.g. $\overrightarrow{AO} \cdot \overrightarrow{AB} = -8 + 6 + 2 = 0 \rightarrow O\hat{A}B = 90°$ AG | M1 A1 | Use of dot product with either $\overrightarrow{AO}$ or $\overrightarrow{OA}$ & either $\overrightarrow{AB}$ or $\overrightarrow{BA}$. Must see 3 component products |
| **OR** $|\overrightarrow{OA}| = 3, |\overrightarrow{OB}| = \sqrt{38}, |\overrightarrow{AB}| = \sqrt{29}$; $OA^2 + AB^2 = OB^2 \rightarrow O\hat{A}B = 90°$ AG | | OR Correct use of Pythagoras. In both methods must state angle or $\Theta = 90°$ or similar for **A1** |
| | **3** | |

## Question 9(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{CB} = \begin{pmatrix} 6 \\ -6 \\ -3 \end{pmatrix}$ or $\overrightarrow{BC} = \begin{pmatrix} -6 \\ 6 \\ 3 \end{pmatrix}$ | B1 | Must correctly identify the vector |
| $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC}$ (or $-\overrightarrow{CB}$) $= \begin{pmatrix} 0 \\ 7 \\ 4 \end{pmatrix}$ | M1 A1 | Correct link leading to $\overrightarrow{OC}$ |
| | **3** | |

## Question 9(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $|\overrightarrow{OA}| = 3, |\overrightarrow{BC}| = 9, |\overrightarrow{AB}| = \sqrt{29}$ (5.39) | B1 | For any one of these |
| Area $= \frac{1}{2}(3 + 9)\sqrt{29}$ or $3\sqrt{29} + 3\sqrt{29}$ | M1 | Correct formula(e) used for trapezium or (rectangle + triangle) or two triangles using their lengths |
| $= 6\sqrt{29}$ $(1\sqrt{1044}, 2\sqrt{261}$ or $3\sqrt{116})$ | A1 | Exact answer in correct form |
| | **3** | |

Show LaTeX source

9\\
\includegraphics[max width=\textwidth, alt={}, center]{518bb805-5b14-4b41-94fd-38a31a90c218-16_533_601_258_772}

The diagram shows a trapezium $O A B C$ in which $O A$ is parallel to $C B$. The position vectors of $A$ and $B$ relative to the origin $O$ are given by $\overrightarrow { O A } = \left( \begin{array} { r } 2 \\ - 2 \\ - 1 \end{array} \right)$ and $\overrightarrow { O B } = \left( \begin{array} { l } 6 \\ 1 \\ 1 \end{array} \right)$.\\
(i) Show that angle $O A B$ is $90 ^ { \circ }$.\\

The magnitude of $\overrightarrow { C B }$ is three times the magnitude of $\overrightarrow { O A }$.\\
(ii) Find the position vector of $C$.\\

(iii) Find the exact area of the trapezium $O A B C$, giving your answer in the form $a \sqrt { } b$, where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{CAIE P1 2017 Q9 [9]}}

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