| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring finding intersection points, calculating angles using dot product, and finding triangle area. While it involves several steps, each technique is routine for C3/C4 level with no novel insights required. The methods are direct applications of standard formulas, making it slightly easier than average. |
| 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 multiplication4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Equate j components: \(3 + 2\lambda = 9 \Rightarrow \lambda = 3\) | M1 A1 | |
| Leading to \(C = (5, 9, -1)\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Choosing correct directions or finding \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\) | M1 | |
| \(\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 5+2 = \sqrt{6} \times \sqrt{29} \times \cos\angle ABC\) | M1 A1 | Use of scalar product |
| \(\angle ACB = 57.95°\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(A=(2,3,-4)\), \(B=(-5,9,-5)\), \(\overrightarrow{AC}=\begin{pmatrix}3\\6\\3\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}10\\0\\4\end{pmatrix}\) | ||
| \(AC^2 = 3^2+6^2+3^2 = (3\sqrt{6})\), \(BC^2 = 10^2+4^2 = (2\sqrt{29})\) | M1 A1 A1 | |
| Area \(= \frac{1}{2} \cdot AC \cdot BC \cdot \sin\angle ACB = \frac{1}{2} \times 3\sqrt{6} \times 2\sqrt{29} \times \sin 57.95°\) | M1 | |
| \(= 33.5\) | A1 | (5), (12 marks) |
## Question 9:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Equate **j** components: $3 + 2\lambda = 9 \Rightarrow \lambda = 3$ | M1 A1 | |
| Leading to $C = (5, 9, -1)$ | A1 | (3) |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Choosing correct directions or finding $\overrightarrow{AC}$ and $\overrightarrow{BC}$ | M1 | |
| $\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \begin{pmatrix}5\\0\\2\end{pmatrix} = 5+2 = \sqrt{6} \times \sqrt{29} \times \cos\angle ABC$ | M1 A1 | Use of scalar product |
| $\angle ACB = 57.95°$ | A1 | (4) |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $A=(2,3,-4)$, $B=(-5,9,-5)$, $\overrightarrow{AC}=\begin{pmatrix}3\\6\\3\end{pmatrix}$, $\overrightarrow{BC}=\begin{pmatrix}10\\0\\4\end{pmatrix}$ | | |
| $AC^2 = 3^2+6^2+3^2 = (3\sqrt{6})$, $BC^2 = 10^2+4^2 = (2\sqrt{29})$ | M1 A1 A1 | |
| Area $= \frac{1}{2} \cdot AC \cdot BC \cdot \sin\angle ACB = \frac{1}{2} \times 3\sqrt{6} \times 2\sqrt{29} \times \sin 57.95°$ | M1 | |
| $= 33.5$ | A1 | (5), **(12 marks)** |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 4 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)$, where $\lambda$ is a scalar parameter.
\end{enumerate}
The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 0 \\ 9 \\ - 3 \end{array} \right) + \mu \left( \begin{array} { l } 5 \\ 0 \\ 2 \end{array} \right)$, where $\mu$ is a scalar parameter.\\
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $C$, find\\
(a) the coordinates of $C$.
The point $A$ is the point on $l _ { 1 }$ where $\lambda = 0$ and the point $B$ is the point on $l _ { 2 }$ where $\mu = - 1$\\
(b) Find the size of the angle $A C B$. Give your answer in degrees to 2 decimal places.\\
(c) Hence, or otherwise, find the area of the triangle $A B C$.\\
\hfill \mbox{\textit{Edexcel C34 Q9 [12]}}