| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: unknown constant then intersect |
| Difficulty | Standard +0.2 This is a straightforward vector line problem requiring students to verify a point lies on a line and find a constant by substitution. It involves basic manipulation of vector equations with no geometric insight or multi-step problem-solving—slightly easier than average for C4. |
| 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 multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B\) lies on \(l_2 \Rightarrow \mu = -1 \Rightarrow p = 5\) | B1 | \(p = 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Setting \(l_1 = l_2\), writes equation involving one parameter, e.g. i: \(7 + 3\mu = 1\) | M1 | Writes equation involving only one parameter |
| \(\mu = -2\) | A1 | |
| Point of intersection \(\overrightarrow{OC} = \mathbf{i} + 10\mathbf{j} - \mathbf{k}\) | B1 | |
| Checks \(\lambda = 4\) and \(\mu = -2\) in third component, AND substitutes into both lines to verify | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix}-3\\5\\-4\end{pmatrix}\) | M1 | Attempt to find both vectors \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) and \(\overrightarrow{BC}\) or \(\overrightarrow{CB}\) |
| \(\cos ACB = \dfrac{\overrightarrow{AC}\cdot\overrightarrow{BC}}{ | \overrightarrow{AC} | |
| \(\cos ACB = \dfrac{56}{\sqrt{4000}} \Rightarrow ACB = 27.7°\ (3\text{ sf})\) | A1 | Anything that rounds to \(27.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(ACB = \dfrac{1}{2}(\sqrt{80})(\sqrt{50})\sin 27.69...° = 14.696...\) | M1 | |
| \(\approx 14.7\) | A1 | Anything that rounds to \(14.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{d}_1 = \begin{pmatrix}0\\2\\-1\end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix}3\\-5\\4\end{pmatrix}\) | ||
| \(\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{ | \mathbf{d}_1 | |
| \(\cos\theta = \frac{0-10-4}{\sqrt{5}\cdot\sqrt{50}} = \frac{-7\sqrt{10}}{25} \Rightarrow \theta = 152.3054385...\) | ||
| Angle \(ACB = 180 - 152.3054385... = 27.69446145... = 27.7°\) (3 sf) | A1 | Anything that rounds to 27.7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}1\\10\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}1\\10\\-1\end{pmatrix}-\begin{pmatrix}4\\5\\3\end{pmatrix}=\begin{pmatrix}-3\\5\\-4\end{pmatrix}\) | M1 | An attempt to find both vectors \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) and \(\overrightarrow{BC}\) or \(\overrightarrow{CB}\) |
| Also \(\overrightarrow{AB}=\begin{pmatrix}4\\5\\3\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}3\\3\\0\end{pmatrix}\) | ||
| \( | \overrightarrow{AC} | =\sqrt{80}\), \( |
| \((\sqrt{18})^2 = (\sqrt{80})^2 + (\sqrt{50})^2 - 2(\sqrt{80})(\sqrt{50})\cos\theta\) | M1 oe | Applies cosine rule the correct way round |
| \(\cos\theta = \frac{7\sqrt{10}}{25} \Rightarrow \theta = 27.69446145... = 27.7°\) (3 sf) | A1 | Anything that rounds to 27.7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC}=\begin{pmatrix}0\\8\\-4\end{pmatrix}\) and \(\overrightarrow{BC}=\begin{pmatrix}-3\\5\\-4\end{pmatrix}\) | M1 | An attempt to find both vectors \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) and \(\overrightarrow{BC}\) or \(\overrightarrow{CB}\) |
| \(\overrightarrow{AC}\times\overrightarrow{BC}=\begin{pmatrix}0\\8\\-4\end{pmatrix}\times\begin{pmatrix}-3\\5\\-4\end{pmatrix}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&8&-4\\-3&5&-4\end{vmatrix}=24\mathbf{i}+12\mathbf{j}+24\mathbf{k}\) | M1 | Full method for applying vector cross product formula between \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\) and their \(\overrightarrow{BC}\) or \(\overrightarrow{CB}\) |
| \(\sin ACB = \frac{\sqrt{(24)^2+(12)^2+(12)^2}}{\sqrt{(0)^2+(8)^2+(-4)^2}\cdot\sqrt{(-3)^2+(5)^2+(-4)^2}}\) | ||
| \(\sin ACB = \frac{\sqrt{864}}{\sqrt{80}\cdot\sqrt{50}}=\frac{3\sqrt{15}}{25} \Rightarrow \theta = 27.69446145... = 27.7°\) (3 sf) | A1 | Anything that rounds to 27.7 |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B$ lies on $l_2 \Rightarrow \mu = -1 \Rightarrow p = 5$ | B1 | $p = 5$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Setting $l_1 = l_2$, writes equation involving one parameter, e.g. **i**: $7 + 3\mu = 1$ | M1 | Writes equation involving only one parameter |
| $\mu = -2$ | A1 | |
| Point of intersection $\overrightarrow{OC} = \mathbf{i} + 10\mathbf{j} - \mathbf{k}$ | B1 | |
| Checks $\lambda = 4$ and $\mu = -2$ in third component, AND substitutes into both lines to verify | B1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}0\\8\\-4\end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix}-3\\5\\-4\end{pmatrix}$ | M1 | Attempt to find both vectors $\overrightarrow{AC}$ or $\overrightarrow{CA}$ **and** $\overrightarrow{BC}$ or $\overrightarrow{CB}$ |
| $\cos ACB = \dfrac{\overrightarrow{AC}\cdot\overrightarrow{BC}}{|\overrightarrow{AC}||\overrightarrow{BC}|} = \dfrac{0+40+16}{\sqrt{80}\cdot\sqrt{50}}$ | M1 | Applies dot product formula between their $\overrightarrow{AC}$ and $\overrightarrow{BC}$ |
| $\cos ACB = \dfrac{56}{\sqrt{4000}} \Rightarrow ACB = 27.7°\ (3\text{ sf})$ | A1 | Anything that rounds to $27.7$ |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $ACB = \dfrac{1}{2}(\sqrt{80})(\sqrt{50})\sin 27.69...° = 14.696...$ | M1 | |
| $\approx 14.7$ | A1 | Anything that rounds to $14.7$ |
# Question 6:
## Part (c) — Alternative Method 1 (Direction Vectors)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d}_1 = \begin{pmatrix}0\\2\\-1\end{pmatrix}$, $\mathbf{d}_2 = \begin{pmatrix}3\\-5\\4\end{pmatrix}$ | | |
| $\cos\theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{\begin{pmatrix}0\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-5\\4\end{pmatrix}}{\sqrt{(0)^2+(2)^2+(-1)^2}\cdot\sqrt{(3)^2+(-5)^2+(4)^2}}$ | M2 | Applies dot product formula between $\mathbf{d}_1$ and $\mathbf{d}_2$ |
| $\cos\theta = \frac{0-10-4}{\sqrt{5}\cdot\sqrt{50}} = \frac{-7\sqrt{10}}{25} \Rightarrow \theta = 152.3054385...$ | | |
| Angle $ACB = 180 - 152.3054385... = 27.69446145... = 27.7°$ (3 sf) | A1 | Anything that rounds to 27.7 |
## Part (c) — Alternative Method 2 (Cosine Rule)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}1\\10\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}0\\8\\-4\end{pmatrix}$ and $\overrightarrow{BC}=\begin{pmatrix}1\\10\\-1\end{pmatrix}-\begin{pmatrix}4\\5\\3\end{pmatrix}=\begin{pmatrix}-3\\5\\-4\end{pmatrix}$ | M1 | An attempt to find both vectors $\overrightarrow{AC}$ or $\overrightarrow{CA}$ and $\overrightarrow{BC}$ or $\overrightarrow{CB}$ |
| Also $\overrightarrow{AB}=\begin{pmatrix}4\\5\\3\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}3\\3\\0\end{pmatrix}$ | | |
| $|\overrightarrow{AC}|=\sqrt{80}$, $|\overrightarrow{BC}|=\sqrt{50}$, $|\overrightarrow{AB}|=\sqrt{18}$ | | |
| $(\sqrt{18})^2 = (\sqrt{80})^2 + (\sqrt{50})^2 - 2(\sqrt{80})(\sqrt{50})\cos\theta$ | M1 oe | Applies cosine rule the correct way round |
| $\cos\theta = \frac{7\sqrt{10}}{25} \Rightarrow \theta = 27.69446145... = 27.7°$ (3 sf) | A1 | Anything that rounds to 27.7 |
## Part (c) — Alternative Method 3 (Vector Cross Product)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC}=\begin{pmatrix}0\\8\\-4\end{pmatrix}$ and $\overrightarrow{BC}=\begin{pmatrix}-3\\5\\-4\end{pmatrix}$ | M1 | An attempt to find both vectors $\overrightarrow{AC}$ or $\overrightarrow{CA}$ and $\overrightarrow{BC}$ or $\overrightarrow{CB}$ |
| $\overrightarrow{AC}\times\overrightarrow{BC}=\begin{pmatrix}0\\8\\-4\end{pmatrix}\times\begin{pmatrix}-3\\5\\-4\end{pmatrix}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&8&-4\\-3&5&-4\end{vmatrix}=24\mathbf{i}+12\mathbf{j}+24\mathbf{k}$ | M1 | Full method for applying vector cross product formula between $\overrightarrow{AC}$ or $\overrightarrow{CA}$ and their $\overrightarrow{BC}$ or $\overrightarrow{CB}$ |
| $\sin ACB = \frac{\sqrt{(24)^2+(12)^2+(12)^2}}{\sqrt{(0)^2+(8)^2+(-4)^2}\cdot\sqrt{(-3)^2+(5)^2+(-4)^2}}$ | | |
| $\sin ACB = \frac{\sqrt{864}}{\sqrt{80}\cdot\sqrt{50}}=\frac{3\sqrt{15}}{25} \Rightarrow \theta = 27.69446145... = 27.7°$ (3 sf) | A1 | Anything that rounds to 27.7 |
---6. With respect to a fixed origin, the point $A$ with position vector $\mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }$ lies on the line $l _ { 1 }$ with equation
$$\mathbf { r } = \left( \begin{array} { l }
1 \\
2 \\
3
\end{array} \right) + \lambda \left( \begin{array} { r }
0 \\
2 \\
- 1
\end{array} \right) , \quad \text { where } \lambda \text { is a scalar parameter, }$$
and the point $B$ with position vector $4 \mathbf { i } + p \mathbf { j } + 3 \mathbf { k }$, where $p$ is a constant, lies on the line $l _ { 2 }$ with equation
$$\mathbf { r } = \left( \begin{array} { l }
7 \\
0 \\
7
\end{array} \right) + \mu \left( \begin{array} { r }
3 \\
- 5 \\
4
\end{array} \right) , \quad \text { where } \mu \text { is a scalar parameter. }$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $p$.
\item Show that $l _ { 1 }$ and $l _ { 2 }$ intersect and find the position vector of their point of intersection, $C$.
\item Find the size of the angle $A C B$, giving your answer in degrees to 3 significant figures.
\item Find the area of the triangle $A B C$, giving your answer to 3 significant figures.\\
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\section*{Question 6 continued}
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\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2014 Q6 [10]}}