| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Point on line satisfying distance or other condition |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring vector equations of lines, dot product for angles, cross product for area, and geometric reasoning about scaling triangles. Part (e) requires insight that doubling the triangle area means doubling the base length, leading to two positions. While systematic, the length and combination of techniques (especially the final geometric insight) place it moderately above average difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either at point \(A\): \(\lambda = 1\) or at point \(B\): \(\lambda = 3\) | M1 | Finds/implies correct \(\lambda\) for at least one point |
| Leading to \(a = -3\) or \(b = 5\) | A1 | At least one of \(a\) or \(b\) correct |
| Leading to \(a = -3\) and \(b = 5\) | A1 | Both \(a\) and \(b\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\pm[('5\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (\mathbf{i} - 3\mathbf{j} + 5\mathbf{k})]\) | M1 | Subtraction either way round |
| \(\overrightarrow{AB} = 4\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) | A1 | Subtraction correct way round |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}3\\0\\-3\end{pmatrix}\) or \(\overrightarrow{CA} = \begin{pmatrix}-3\\0\\3\end{pmatrix}\) | M1 | Subtracts position vector of \(A\) from \(C\) or vice versa; ft their \(a\) |
| \(\cos C\hat{A}B = \dfrac{\begin{pmatrix}4\\2\\-2\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-3\end{pmatrix}}{\sqrt{(4)^2+(2)^2+(-2)^2}\cdot\sqrt{(3)^2+(0)^2+(-3)^2}}\) | dM1 | Applies dot product formula between \((\overrightarrow{AB}\) or \(\overrightarrow{BA})\) and \((\overrightarrow{AC}\) or \(\overrightarrow{CA})\) |
| \(\cos C\hat{A}B = \dfrac{12+0+6}{\sqrt{24}\cdot\sqrt{18}} = \dfrac{\sqrt{3}}{2}\) (o.e.) \(\Rightarrow C\hat{A}B = 30°\) | A1* | Correctly proves \(C\hat{A}B = 30°\); must use \(\overrightarrow{AB}\) with \(\overrightarrow{AC}\) or \(\overrightarrow{BA}\) with \(\overrightarrow{CA}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB = 2\sqrt{6},\ AC = 3\sqrt{2},\ BC = \sqrt{6}\) | M1 | Finds lengths of \(AB\), \(AC\) and \(BC\) |
| \(\cos C\hat{A}B = \dfrac{24+18-6}{2\sqrt{24}\sqrt{18}}\) or right angled triangle giving \(\cos C\hat{A}B = \dfrac{\sqrt{3}}{2}\) | dM1 | Uses cosine rule or trig of right-angled triangle |
| \(C\hat{A}B = 30°\) | A1 | Correct proof that angle \(= 30°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(CAB = \frac{1}{2}\sqrt{24}\sqrt{18}\sin 30°\) | M1 | Applies \(\frac{1}{2} |
| \(= 3\sqrt{3}\) (or \(k=3\)) | A1 | cao – must be exact in this form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OD_1} = \begin{pmatrix}-1\\-4\\6\end{pmatrix} + 5\begin{pmatrix}2\\1\\-1\end{pmatrix}\) or \(= \begin{pmatrix}1\\a\\5\end{pmatrix} + 2\begin{pmatrix}4\\2\\-2\end{pmatrix}\) or \(= \begin{pmatrix}b\\-1\\3\end{pmatrix} + \begin{pmatrix}4\\2\\-2\end{pmatrix}\) | M1 | Realises \(AD\) is twice length of \(AB\); complete method for one point; ft \(a\) or \(b\) |
| \(= \begin{pmatrix}9\\1\\1\end{pmatrix}\) | A1 | cao |
| \(\overrightarrow{OD_2} = \begin{pmatrix}-1\\-4\\6\end{pmatrix} - 3\begin{pmatrix}2\\1\\-1\end{pmatrix}\) or \(= \begin{pmatrix}1\\a\\5\end{pmatrix} - 2\begin{pmatrix}4\\2\\-2\end{pmatrix}\) or \(= \begin{pmatrix}b\\-1\\3\end{pmatrix} - 3\begin{pmatrix}2\\1\\-1\end{pmatrix}\) | M1 | \(AD\) twice length of \(AB\) but opposite direction; ft \(a\) or \(b\) |
| \(= \begin{pmatrix}-7\\-7\\9\end{pmatrix}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2\lambda-2)^2 + (\lambda-1)^2 + (1-\lambda)^2 = 96\), giving \(\lambda^2 - 2\lambda - 15 = 0\) | M1 | Sets up and simplifies equation; may make algebraic slip |
| \(\begin{pmatrix}9\\1\\1\end{pmatrix}\) (from \(\lambda = 5\)) | A1 | |
| Substitutes other value of \(\lambda\) | M1 | May make slip in algebra |
| \(\begin{pmatrix}-7\\-7\\9\end{pmatrix}\) (from \(\lambda = -3\)) | A1 |
# Question 14:
**Part (a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either at point $A$: $\lambda = 1$ or at point $B$: $\lambda = 3$ | M1 | Finds/implies correct $\lambda$ for at least one point |
| Leading to $a = -3$ **or** $b = 5$ | A1 | At least one of $a$ or $b$ correct |
| Leading to $a = -3$ **and** $b = 5$ | A1 | Both $a$ and $b$ correct |
**Part (b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\pm[('5\mathbf{i} - \mathbf{j} + 3\mathbf{k}) - (\mathbf{i} - 3\mathbf{j} + 5\mathbf{k})]$ | M1 | Subtraction either way round |
| $\overrightarrow{AB} = 4\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$ | A1 | Subtraction correct way round |
**Part (c) – Way 1**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}3\\0\\-3\end{pmatrix}$ or $\overrightarrow{CA} = \begin{pmatrix}-3\\0\\3\end{pmatrix}$ | M1 | Subtracts position vector of $A$ from $C$ or vice versa; ft their $a$ |
| $\cos C\hat{A}B = \dfrac{\begin{pmatrix}4\\2\\-2\end{pmatrix}\cdot\begin{pmatrix}3\\0\\-3\end{pmatrix}}{\sqrt{(4)^2+(2)^2+(-2)^2}\cdot\sqrt{(3)^2+(0)^2+(-3)^2}}$ | dM1 | Applies dot product formula between $(\overrightarrow{AB}$ or $\overrightarrow{BA})$ and $(\overrightarrow{AC}$ or $\overrightarrow{CA})$ |
| $\cos C\hat{A}B = \dfrac{12+0+6}{\sqrt{24}\cdot\sqrt{18}} = \dfrac{\sqrt{3}}{2}$ (o.e.) $\Rightarrow C\hat{A}B = 30°$ | A1* | Correctly proves $C\hat{A}B = 30°$; must use $\overrightarrow{AB}$ with $\overrightarrow{AC}$ or $\overrightarrow{BA}$ with $\overrightarrow{CA}$ |
**Part (c) – Way 2**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = 2\sqrt{6},\ AC = 3\sqrt{2},\ BC = \sqrt{6}$ | M1 | Finds lengths of $AB$, $AC$ and $BC$ |
| $\cos C\hat{A}B = \dfrac{24+18-6}{2\sqrt{24}\sqrt{18}}$ or right angled triangle giving $\cos C\hat{A}B = \dfrac{\sqrt{3}}{2}$ | dM1 | Uses cosine rule or trig of right-angled triangle |
| $C\hat{A}B = 30°$ | A1 | Correct proof that angle $= 30°$ |
**Part (d)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $CAB = \frac{1}{2}\sqrt{24}\sqrt{18}\sin 30°$ | M1 | Applies $\frac{1}{2}|\overrightarrow{AB}||\overrightarrow{AC}|\sin 30°$; must use vectors $(b-a)$ and $(c-a)$ |
| $= 3\sqrt{3}$ (or $k=3$) | A1 | cao – must be exact in this form |
**Part (e)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OD_1} = \begin{pmatrix}-1\\-4\\6\end{pmatrix} + 5\begin{pmatrix}2\\1\\-1\end{pmatrix}$ or $= \begin{pmatrix}1\\a\\5\end{pmatrix} + 2\begin{pmatrix}4\\2\\-2\end{pmatrix}$ or $= \begin{pmatrix}b\\-1\\3\end{pmatrix} + \begin{pmatrix}4\\2\\-2\end{pmatrix}$ | M1 | Realises $AD$ is twice length of $AB$; complete method for one point; ft $a$ or $b$ |
| $= \begin{pmatrix}9\\1\\1\end{pmatrix}$ | A1 | cao |
| $\overrightarrow{OD_2} = \begin{pmatrix}-1\\-4\\6\end{pmatrix} - 3\begin{pmatrix}2\\1\\-1\end{pmatrix}$ or $= \begin{pmatrix}1\\a\\5\end{pmatrix} - 2\begin{pmatrix}4\\2\\-2\end{pmatrix}$ or $= \begin{pmatrix}b\\-1\\3\end{pmatrix} - 3\begin{pmatrix}2\\1\\-1\end{pmatrix}$ | M1 | $AD$ twice length of $AB$ but opposite direction; ft $a$ or $b$ |
| $= \begin{pmatrix}-7\\-7\\9\end{pmatrix}$ | A1 | cao |
**Part (e) – Length of AD approach**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2\lambda-2)^2 + (\lambda-1)^2 + (1-\lambda)^2 = 96$, giving $\lambda^2 - 2\lambda - 15 = 0$ | M1 | Sets up and simplifies equation; may make algebraic slip |
| $\begin{pmatrix}9\\1\\1\end{pmatrix}$ (from $\lambda = 5$) | A1 | |
| Substitutes other value of $\lambda$ | M1 | May make slip in algebra |
| $\begin{pmatrix}-7\\-7\\9\end{pmatrix}$ (from $\lambda = -3$) | A1 | |
The image appears to be a blank page with only the "PMT" header/watermark and a Pearson Education Limited footer. There is no mark scheme content visible on this page to extract.14. Relative to a fixed origin $O$, the line $l$ has vector equation
$$\mathbf { r } = \left( \begin{array} { r }
- 1 \\
- 4 \\
6
\end{array} \right) + \lambda \left( \begin{array} { r }
2 \\
1 \\
- 1
\end{array} \right)$$
where $\lambda$ is a scalar parameter.
Points $A$ and $B$ lie on the line $l$, where $A$ has coordinates ( $1 , a , 5$ ) and $B$ has coordinates ( $b , - 1,3$ ).
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $a$ and the value of the constant $b$.
\item Find the vector $\overrightarrow { A B }$.
The point $C$ has coordinates ( $4 , - 3,2$ )
\item Show that the size of the angle $C A B$ is $30 ^ { \circ }$
\item Find the exact area of the triangle $C A B$, giving your answer in the form $k \sqrt { 3 }$, where $k$ is a constant to be determined.
The point $D$ lies on the line $l$ so that the area of the triangle $C A D$ is twice the area of the triangle $C A B$.
\item Find the coordinates of the two possible positions of $D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q14 [14]}}