| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| 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 standard vectors question requiring routine application of dot product for angles, cross product for area, and a simple geometric relationship for perpendicular distance. All three parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(\begin{pmatrix}2\\3\\-2\end{pmatrix}\cdot\begin{pmatrix}5\\-6\\1\end{pmatrix} = \sqrt{(2^2+3^2+(-2)^2)}\sqrt{(5^2+(-6)^2+1^2)}\cos(\angle CAB)\) | M1 | Correct use of Pythagoras to find and multiply lengths, multiplies and adds components for dot product |
| \(\angle CAB = \arccos\left(-\frac{10}{\sqrt{17}\sqrt{62}}\right) = 107.94°\) | dM1 | Dependent on M1; continuing to find \(\angle CAB\) using invcos. Allow \(\arccos\left(\frac{\pm10}{\sqrt{17}\sqrt{62}}\right)\) |
| \(107.94°\) | A1 | awrt \(107.94°\) only; \(108°\) or \(72.1°\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \frac{1}{2}\sqrt{17}\sqrt{62}\sin(107.94°) = 15.44\) | M1A1 | Uses Area \(= \frac{1}{2} |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | BC | = \sqrt{(5-2)^2+(-6-3)^2+(1--2)^2}\) |
| Uses Area \(= \frac{1}{2} | BC | \times |
# Question 9:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $\begin{pmatrix}2\\3\\-2\end{pmatrix}\cdot\begin{pmatrix}5\\-6\\1\end{pmatrix} = \sqrt{(2^2+3^2+(-2)^2)}\sqrt{(5^2+(-6)^2+1^2)}\cos(\angle CAB)$ | M1 | Correct use of Pythagoras to find and multiply lengths, multiplies and adds components for dot product |
| $\angle CAB = \arccos\left(-\frac{10}{\sqrt{17}\sqrt{62}}\right) = 107.94°$ | dM1 | Dependent on M1; continuing to find $\angle CAB$ using invcos. Allow $\arccos\left(\frac{\pm10}{\sqrt{17}\sqrt{62}}\right)$ |
| $107.94°$ | A1 | awrt $107.94°$ only; $108°$ or $72.1°$ scores A0 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\sqrt{17}\sqrt{62}\sin(107.94°) = 15.44$ | M1A1 | Uses Area $= \frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin C$ with correct vectors and their angle; Area = awrt 15.44 (allow if $72.06°$ used) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|BC| = \sqrt{(5-2)^2+(-6-3)^2+(1--2)^2}$ | M1 | Attempts length of $BC$; must be used in (c) to score this mark |
| Uses Area $= \frac{1}{2}|BC|\times|AD| \Rightarrow 15.44 = \frac{1}{2}\times\sqrt{99}\times|AD| \Rightarrow |AD| = 3.10$ | M1A1 | Attempts Area formula with their area from (b) and $|BC|$; $|AD|$ = awrt 3.10 (not 3.1) |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-28_615_328_210_808}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of a triangle $A B C$.\\
Given $\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k }$ and $\overrightarrow { A C } = 5 \mathbf { i } - 6 \mathbf { j } + \mathbf { k }$,
\begin{enumerate}[label=(\alph*)]
\item find the size of angle $C A B$, giving your answer in degrees to 2 decimal places,
\item find the area of triangle $A B C$, giving your answer to 2 decimal places.
Using your answer to part (b), or otherwise,
\item find the shortest distance from $A$ to $B C$, giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q9 [8]}}