| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Angle between two vectors |
| Difficulty | Moderate -0.8 This is a straightforward application of standard vector formulas (dot product for angle, magnitude of difference for length, area formula) with simple arithmetic. All three parts are direct recall of basic vector techniques with no problem-solving insight required, making it easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{a \cdot b} = | \mathbf{a} |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(c^2 = a^2 + b^2 - 2ab\cos C \Rightarrow c^2 = 5^2 + 6^2 - 2\times5\times6\times\frac{2}{3}\) | M1 | Or using \(c^2 = a^2+b^2 - 2\mathbf{a\cdot b} \Rightarrow c^2 = 5^2+6^2-2\times20\) |
| \(\Rightarrow | AB | = \sqrt{21}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta = \frac{2}{3} \Rightarrow \sin\theta = \frac{\sqrt{5}}{3}\) or exact height \(= \frac{5\sqrt{5}}{3}\) | M1 | Use exact \(\cos\theta\) to find exact \(\sin\theta\); use of non-exact angle is M0 |
| Area \(= \frac{1}{2}\times5\times6\times\sin(AOB)\) | M1 | Correct area formula method |
| \(\frac{1}{2}\times5\times6\times\sin(AOB) = 5\sqrt{5}\) | A1* | No evidence of calculator; clear working with surds required |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{a \cdot b} = |\mathbf{a}||\mathbf{b}|\cos\theta \Rightarrow 20 = 5 \times 6\cos\theta \Rightarrow \cos\theta = \frac{20}{30} = \frac{2}{3}$ | M1A1 | M1 for using dot product formula; A1 for $\cos\theta = \frac{2}{3}$ or $\frac{20}{30}$ |
**(2 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $c^2 = a^2 + b^2 - 2ab\cos C \Rightarrow c^2 = 5^2 + 6^2 - 2\times5\times6\times\frac{2}{3}$ | M1 | Or using $c^2 = a^2+b^2 - 2\mathbf{a\cdot b} \Rightarrow c^2 = 5^2+6^2-2\times20$ |
| $\Rightarrow |AB| = \sqrt{21}$ | A1 | Exact answer only |
**(2 marks)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta = \frac{2}{3} \Rightarrow \sin\theta = \frac{\sqrt{5}}{3}$ or exact height $= \frac{5\sqrt{5}}{3}$ | M1 | Use exact $\cos\theta$ to find exact $\sin\theta$; use of non-exact angle is M0 |
| Area $= \frac{1}{2}\times5\times6\times\sin(AOB)$ | M1 | Correct area formula method |
| $\frac{1}{2}\times5\times6\times\sin(AOB) = 5\sqrt{5}$ | A1* | No evidence of calculator; clear working with surds required |
**(3 marks)**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c08fbab-283e-4c92-89a4-10f68f37e133-07_330_494_210_724}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the points $A$ and $B$ with position vectors $\mathbf { a }$ and $\mathbf { b }$ respectively, relative to a fixed origin $O$.
Given that $| \mathbf { a } | = 5 , | \mathbf { b } | = 6$ and a.b $= 20$
\begin{enumerate}[label=(\alph*)]
\item find the cosine of angle $A O B$,
\item find the exact length of $A B$.
\item Show that the area of triangle $O A B$ is $5 \sqrt { 5 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q4 [7]}}