| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Vector geometry in 3D shapes |
| Difficulty | Standard +0.3 This is a straightforward 3D vector question requiring coordinate setup in a cube, finding position vectors, calculating a scalar product for an angle, and computing distances. All steps are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{PR} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}\), \(\overrightarrow{PQ} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}\) | B1, B2,1 [3] | All elements of \(\overrightarrow{PR}\) – any notation ok. Loses one mark for each error in \(\overrightarrow{PQ}\) |
| (ii) \(\overrightarrow{PQ}·\overrightarrow{PR} = -4 + 4 + 8 = 8\); \( | \overrightarrow{PQ} | = \sqrt{24}\), \( |
| \(\overrightarrow{PQ}·\overrightarrow{PR} = \sqrt{12}\sqrt{24}\cos QPR\); Angle \(QPR = 61.9°\) or 1.08 rad | M1, A1 [4] | Everything linked (\(\overrightarrow{QP}·\overrightarrow{PR}\) used – still gains all M marks) Co |
| (iii) \(\overrightarrow{QR} = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix}\), \( | \overrightarrow{QR} | = \sqrt{20}\); Perimeter = \(\sqrt{12} + \sqrt{24} + \sqrt{20} = 12.8\) cm |
**(i)** $\overrightarrow{PR} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}$, $\overrightarrow{PQ} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}$ | B1, B2,1 [3] | All elements of $\overrightarrow{PR}$ – any notation ok. Loses one mark for each error in $\overrightarrow{PQ}$
**(ii)** $\overrightarrow{PQ}·\overrightarrow{PR} = -4 + 4 + 8 = 8$; $|\overrightarrow{PQ}| = \sqrt{24}$, $|\overrightarrow{PR}| = \sqrt{12}$ | M1, M1 [4] | Must be scalar; As long as this is used with dot product
$\overrightarrow{PQ}·\overrightarrow{PR} = \sqrt{12}\sqrt{24}\cos QPR$; Angle $QPR = 61.9°$ or 1.08 rad | M1, A1 [4] | Everything linked ($\overrightarrow{QP}·\overrightarrow{PR}$ used – still gains all M marks) Co
**(iii)** $\overrightarrow{QR} = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix}$, $|\overrightarrow{QR}| = \sqrt{20}$; Perimeter = $\sqrt{12} + \sqrt{24} + \sqrt{20} = 12.8$ cm | M1, M1 A1 [3] | For correct $\overrightarrow{QR}$ - cosine rule ok. Adds three roots. co - beware fortuitous answers from incorrect sign in vectors.10\\
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The diagram shows a cube $O A B C D E F G$ in which the length of each side is 4 units. 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 mid-points of $O A$ and $D G$ are $P$ and $Q$ respectively and $R$ is the centre of the square face $A B F E$.\\
(i) Express each of the vectors $\overrightarrow { P R }$ and $\overrightarrow { P Q }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(ii) Use a scalar product to find angle $Q P R$.\\
(iii) Find the perimeter of triangle $P Q R$, giving your answer correct to 1 decimal place.
\hfill \mbox{\textit{CAIE P1 2007 Q10 [10]}}