| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two vectors/lines (direct) |
| Difficulty | Moderate -0.3 This is a straightforward application of standard vector techniques: part (i) uses the scalar product formula for angles (routine calculation), and part (ii) requires vector addition using a scalar multiple. Both are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a \cdot b = -3 + 12 + 12 = 27\) \(a \cdot b = \sqrt{54} \times \sqrt{21} \cos\theta\) \(\rightarrow \theta = 36.7°\) or 0.641 radians | M1 M1 A1 [3] | Ok to work throughout with column vectors or with i,j,k. Use of \(x_1x_2 + y_1y_2 + z_1z_2\). Use of \(\sqrt{\mathbf{v}} \cdot \mathbf{v} \cos\theta\). In either degrees or in radians. |
| (ii) Vector \(AB = b - a = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}\) | M1 | For use of \(b - a\). |
| Vector \(OC = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}\) | M1 | For \(a + 3(b - a)\) or equivalent |
| Unit vector = Vector \(OC \div 9\), \(= \frac{1}{3}i - \frac{2}{3}j + \frac{2}{3}k\) | M1 A1 [4] | For division by Modulus of \(OC\). Co. |
$a = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix}$ $b = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}$
(i) $a \cdot b = -3 + 12 + 12 = 27$ $a \cdot b = \sqrt{54} \times \sqrt{21} \cos\theta$ $\rightarrow \theta = 36.7°$ or 0.641 radians | M1 M1 A1 [3] | Ok to work throughout with column vectors or with i,j,k. Use of $x_1x_2 + y_1y_2 + z_1z_2$. Use of $\sqrt{\mathbf{v}} \cdot \mathbf{v} \cos\theta$. In either degrees or in radians.
(ii) Vector $AB = b - a = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$ | M1 | For use of $b - a$.
Vector $OC = \begin{pmatrix} -3 \\ 6 \\ 3 \end{pmatrix} + 3 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}$ | M1 | For $a + 3(b - a)$ or equivalent
Unit vector = Vector $OC \div 9$, $= \frac{1}{3}i - \frac{2}{3}j + \frac{2}{3}k$ | M1 A1 [4] | For division by Modulus of $OC$. Co.
---4 The position vectors of points $A$ and $B$ are $\left( \begin{array} { r } - 3 \\ 6 \\ 3 \end{array} \right)$ and $\left( \begin{array} { r } - 1 \\ 2 \\ 4 \end{array} \right)$ respectively, relative to an origin $O$.\\
(i) Calculate angle $A O B$.\\
(ii) The point $C$ is such that $\overrightarrow { A C } = 3 \overrightarrow { A B }$. Find the unit vector in the direction of $\overrightarrow { O C }$.
\hfill \mbox{\textit{CAIE P1 2006 Q4 [7]}}