CAIE P1 2009 November — Question 9 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2009
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicularity conditions
DifficultyStandard +0.3 This is a straightforward multi-part vectors question testing standard techniques: angle between vectors using dot product (routine calculation), finding a unit vector and scaling it (direct application), and using perpendicularity condition (dot product = 0, solve linear equation). All parts are textbook exercises requiring only recall and basic algebraic manipulation, making it slightly easier than average.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence

9 Relative to an origin \(O\), the position vectors of the points \(A , B\) and \(C\) are given by $$\overrightarrow { O A } = \left( \begin{array} { r } 2 \\ 3 \\ - 6 \end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r } 0 \\ - 6 \\ 8 \end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r } - 2 \\ 5 \\ - 2 \end{array} \right)$$
  1. Find angle \(A O B\).
  2. Find the vector which is in the same direction as \(\overrightarrow { A C }\) and has magnitude 30 .
  3. Find the value of the constant \(p\) for which \(\overrightarrow { O A } + p \overrightarrow { O B }\) is perpendicular to \(\overrightarrow { O C }\).

M1: \(\vec{OA}= \begin{pmatrix}2\\3\\-6\end{pmatrix}\), \(\vec{OB}= \begin{pmatrix}0\\-6\\8\end{pmatrix}\), \(\vec{OC} = \begin{pmatrix}-2\\5\\-2\end{pmatrix}\) — Use of \(x_1 x_2 + y_1 y_2 + z_1 z_2\)
(i) M1: Scalar product = \(−18 − 48\) — Linking everything correctly
M1: \(−66 = ab\cos\theta\) — Linking everything correctly
M1: \(a= \sqrt{7}\) and \(b= \sqrt{10}\) — Correct modulus of either \(a\) or \(b\)
A1: \(\to\) Angle \(AOB = 160.5°\) — co; allow \(\pm\)
[4]
(ii) B1: \(\vec{AC} = \vec{c}-\vec{a}= \begin{pmatrix}-4\\2\\4\end{pmatrix}\) — For modulus and multiplying by "5"
M1: Modulus = \(6\)
A1: Vector = \(\begin{pmatrix}-20\\10\\20\end{pmatrix}\) (or \(\begin{pmatrix}-4\\2\\4\end{pmatrix}\) scaled)
[3]
(iii) B1: $\begin{pmatrix}2\\3\\-6\end{pmatrix} + 6p\begin
M1: $\vec{OA}= \begin{pmatrix}2\\3\\-6\end{pmatrix}$, $\vec{OB}= \begin{pmatrix}0\\-6\\8\end{pmatrix}$, $\vec{OC} = \begin{pmatrix}-2\\5\\-2\end{pmatrix}$ — Use of $x_1 x_2 + y_1 y_2 + z_1 z_2$

(i) M1: Scalar product = $−18 − 48$ — Linking everything correctly

M1: $−66 = ab\cos\theta$ — Linking everything correctly

M1: $a= \sqrt{7}$ and $b= \sqrt{10}$ — Correct modulus of either $a$ or $b$

A1: $\to$ Angle $AOB = 160.5°$ — co; allow $\pm$

[4]

(ii) B1: $\vec{AC} = \vec{c}-\vec{a}= \begin{pmatrix}-4\\2\\4\end{pmatrix}$ — For modulus and multiplying by "5"

M1: Modulus = $6$

A1: Vector = $\begin{pmatrix}-20\\10\\20\end{pmatrix}$ (or $\begin{pmatrix}-4\\2\\4\end{pmatrix}$ scaled)

[3]

(iii) B1: $\begin{pmatrix}2\\3\\-6\end{pmatrix} + 6p\begin
9 Relative to an origin $O$, the position vectors of the points $A , B$ and $C$ are given by

$$\overrightarrow { O A } = \left( \begin{array} { r } 
2 \\
3 \\
- 6
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r } 
0 \\
- 6 \\
8
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r } 
- 2 \\
5 \\
- 2
\end{array} \right)$$

(i) Find angle $A O B$.\\
(ii) Find the vector which is in the same direction as $\overrightarrow { A C }$ and has magnitude 30 .\\
(iii) Find the value of the constant $p$ for which $\overrightarrow { O A } + p \overrightarrow { O B }$ is perpendicular to $\overrightarrow { O C }$.

\hfill \mbox{\textit{CAIE P1 2009 Q9 [10]}}