| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Find unknown constant then intersection |
| Difficulty | Standard +0.3 This is a standard three-part vector line intersection question requiring equating components, solving simultaneous equations, and using the scalar product formula for angles. While it involves multiple steps (6+ marks typical), each technique is routine for C4 level with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(5 + \lambda = 2 - 3\mu\); \(3 - 2\lambda = -11 - 4\mu\) | B1 B1 | |
| \(\therefore \lambda + 3\mu + 3 = 0\) | ||
| \(2\lambda - 4\mu - 14 = 0\) | ||
| \(2\lambda + 6\mu + 6 = 0\) | ||
| \(10\mu + 20 = 0 \Rightarrow \mu = -2 \therefore \lambda = 3\) | M1 A1 | |
| \(\therefore\) point is \((8, -3, 4)\) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\therefore a - 10 = 4 \Rightarrow a = 14\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\cos\theta = \frac{-3+8+10}{\sqrt{9}\sqrt{25+25}}\) | M1 A1 | |
| \(= \frac{15}{3 \times 5\sqrt{2}} = \frac{1}{\sqrt{2}}\) | ||
| Angle \(= 45°\) | M1 A1 | (4 marks) |
# Question 5:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $5 + \lambda = 2 - 3\mu$; $3 - 2\lambda = -11 - 4\mu$ | B1 B1 | |
| $\therefore \lambda + 3\mu + 3 = 0$ | | |
| $2\lambda - 4\mu - 14 = 0$ | | |
| $2\lambda + 6\mu + 6 = 0$ | | |
| $10\mu + 20 = 0 \Rightarrow \mu = -2 \therefore \lambda = 3$ | M1 A1 | |
| $\therefore$ point is $(8, -3, 4)$ | A1 | (5 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\therefore a - 10 = 4 \Rightarrow a = 14$ | M1 A1 | (2 marks) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $\cos\theta = \frac{-3+8+10}{\sqrt{9}\sqrt{25+25}}$ | M1 A1 | |
| $= \frac{15}{3 \times 5\sqrt{2}} = \frac{1}{\sqrt{2}}$ | | |
| Angle $= 45°$ | M1 A1 | (4 marks) |
---5. The vector equations of two straight lines are
$$\begin{aligned}
& \mathbf { r } = 5 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } ) \quad \text { and } \\
& \mathbf { r } = 2 \mathbf { i } - 11 \mathbf { j } + a \mathbf { k } + \mu ( - 3 \mathbf { i } - 4 \mathbf { j } + 5 \mathbf { k } ) .
\end{aligned}$$
Given that the two lines intersect, find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of the point of intersection,
\item the value of the constant $a$,
\item the acute angle between the two lines.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [11]}}