CAIE P1 2011 June — Question 5 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2011
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeVector geometry in 3D shapes
DifficultyStandard +0.3 This is a straightforward 3D vector question requiring position vector identification from a diagram, basic vector arithmetic (finding PQ and RQ), and a standard scalar product calculation to find an angle. While it involves multiple steps and 3D visualization, all techniques are routine A-level procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors

5 \includegraphics[max width=\textwidth, alt={}, center]{d68c82ec-8c85-40b9-8e81-bd53c7f8dafe-2_748_1155_1146_495} In the diagram, \(O A B C D E F G\) is a rectangular block in which \(O A = O D = 6 \mathrm {~cm}\) and \(A B = 12 \mathrm {~cm}\). The unit vectors \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\) are parallel to \(\overrightarrow { O A } , \overrightarrow { O C }\) and \(\overrightarrow { O D }\) respectively. The point \(P\) is the mid-point of \(D G , Q\) is the centre of the square face \(C B F G\) and \(R\) lies on \(A B\) such that \(A R = 4 \mathrm {~cm}\).
  1. Express each of the vectors \(\overrightarrow { P Q }\) and \(\overrightarrow { R Q }\) in terms of \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\).
  2. Use a scalar product to find angle \(R Q P\).

AnswerMarks Guidance
(i) \(\overrightarrow{PQ} = 3i + 6j - 3k\)B2,1 B1 Allow B2,1 for either one, B1 for the other.
\(\overrightarrow{RQ} = -3i + 8j + 3k\)
[3]
(ii) \(\overrightarrow{PQ} \cdot \overrightarrow{RQ} = -9 + 48 - 9 = 30\)M1 Use of \(x_1x_2 + y_1y_2 + z_1z_2\)
\(= \sqrt{54} \sqrt{82} \cos RQP\)M1 Correct use of modulus
All linked correctlyM1
\(\rightarrow RQP = 63.2°\)A1 co
[4]
(i) $\overrightarrow{PQ} = 3i + 6j - 3k$ | B2,1 B1 | Allow B2,1 for either one, B1 for the other.
$\overrightarrow{RQ} = -3i + 8j + 3k$ | | 
| [3] |

(ii) $\overrightarrow{PQ} \cdot \overrightarrow{RQ} = -9 + 48 - 9 = 30$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$
$= \sqrt{54} \sqrt{82} \cos RQP$ | M1 | Correct use of modulus
All linked correctly | M1 | 
$\rightarrow RQP = 63.2°$ | A1 | co
| [4] |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{d68c82ec-8c85-40b9-8e81-bd53c7f8dafe-2_748_1155_1146_495}

In the diagram, $O A B C D E F G$ is a rectangular block in which $O A = O D = 6 \mathrm {~cm}$ and $A B = 12 \mathrm {~cm}$. The unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $\overrightarrow { O A } , \overrightarrow { O C }$ and $\overrightarrow { O D }$ respectively. The point $P$ is the mid-point of $D G , Q$ is the centre of the square face $C B F G$ and $R$ lies on $A B$ such that $A R = 4 \mathrm {~cm}$.\\
(i) Express each of the vectors $\overrightarrow { P Q }$ and $\overrightarrow { R Q }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Use a scalar product to find angle $R Q P$.

\hfill \mbox{\textit{CAIE P1 2011 Q5 [7]}}