| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Standard +0.3 This is a straightforward two-part vector question requiring standard techniques: (i) finding angle between vectors using dot product formula, and (ii) applying the area formula ½|AB||AC|sin(θ). The calculation is routine with no conceptual challenges beyond A-level basics, making it slightly easier than average. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Angle \(BAC\) needs sides \(AB, AC\) or \(BA, CA\) | Ignore their labels: | |
| \(AB \cdot AC = (b-a) \cdot (c-a)\) | ||
| \(= \begin{vmatrix} -4 \\ -2 \\ 4 \end{vmatrix} \begin{vmatrix} 0 \\ 3 \\ 4 \end{vmatrix} = 10\) | B1 M1 | One of AB, BA, AC, CA correct. Use of \(v_1x v_2 + y_1y_2\), etc. |
| \(= \sqrt{36} \times \sqrt{25} \cos BAC\) | M1M1 | M1 prod of moduli. M1 all linked |
| \(\rightarrow BAC = \cos^{-1}\frac{1}{3}\) AG | A1 | If e.g. BA.OC max B1M1M1. If both vectors wrong 0/5. If e.g. BA.AC used \(\rightarrow \cos^{-1}\left[-\frac{1}{3}\right]\) final mark A0 |
| [5] | ||
| (ii) \(\sin BAC = \sqrt{1 - \frac{1}{9}}\) | B1 | Use of \(s^2 + c^2 = 1\) – not decimals |
| Area \(= \frac{1}{2} \times 6 \times 5 \times \sqrt{\frac{8}{9}} = 5\sqrt{8}\) oe | M1 A1 | Correct formula for area. Decimals seen A0 |
| [3] |
(i) Angle $BAC$ needs sides $AB, AC$ or $BA, CA$ | | Ignore their labels: |
$AB \cdot AC = (b-a) \cdot (c-a)$ | | |
$= \begin{vmatrix} -4 \\ -2 \\ 4 \end{vmatrix} \begin{vmatrix} 0 \\ 3 \\ 4 \end{vmatrix} = 10$ | B1 M1 | One of AB, BA, AC, CA correct. Use of $v_1x v_2 + y_1y_2$, etc. |
$= \sqrt{36} \times \sqrt{25} \cos BAC$ | M1M1 | M1 prod of moduli. M1 all linked |
$\rightarrow BAC = \cos^{-1}\frac{1}{3}$ AG | A1 | If e.g. **BA.OC** max B1M1M1. If both vectors wrong 0/5. If e.g. **BA.AC** used $\rightarrow \cos^{-1}\left[-\frac{1}{3}\right]$ final mark A0 |
| | [5] |
(ii) $\sin BAC = \sqrt{1 - \frac{1}{9}}$ | B1 | Use of $s^2 + c^2 = 1$ – not decimals |
Area $= \frac{1}{2} \times 6 \times 5 \times \sqrt{\frac{8}{9}} = 5\sqrt{8}$ oe | M1 A1 | Correct formula for area. Decimals seen A0 |
| | [3] |7 The position vectors of points $A , B$ and $C$ relative to an origin $O$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
2 \\
1 \\
3
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
6 \\
- 1 \\
7
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
2 \\
4 \\
7
\end{array} \right)$$
(i) Show that angle $B A C = \cos ^ { - 1 } \left( \frac { 1 } { 3 } \right)$.\\
(ii) Use the result in part (i) to find the exact value of the area of triangle $A B C$.
\hfill \mbox{\textit{CAIE P1 2014 Q7 [8]}}