| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of parallelogram or trapezium using vectors |
| Difficulty | Standard +0.3 This is a structured multi-part question on 3D vectors and lines that guides students through standard techniques: identifying intersection points, finding angles between lines using dot products, calculating position vectors, and using perpendicularity conditions. While it has many parts (6 total), each individual step is routine and clearly signposted, requiring only direct application of standard C4 formulas without novel problem-solving insight. The scaffolding makes it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(A: (-6, 4, -1)\) | Accept vector forms |
| (b) | \(\begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} = 12 + 4 + 3 = \sqrt{4^2 + (-1)^2 + 3^2}\sqrt{3^2 + (-4)^2 + 1^2}\cos\theta\) then \(\cos\theta = \frac{19}{26}\) | M1 A1; awrt 0.73 A1 |
| (c) | \(X: (10, 0, 11)\) | Accept vector forms |
| (d) | \(\overrightarrow{AX} = \begin{pmatrix} 10 \\ 0 \\ 11 \end{pmatrix} - \begin{pmatrix} -6 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 16 \\ -4 \\ 12 \end{pmatrix}\) | Either order |
| (e) | \(\left | \overrightarrow{AX}\right |
| (f) | Use of correct right angled triangle: \(\frac{\left | \overrightarrow{AX}\right |
(a) | $A: (-6, 4, -1)$ | Accept vector forms | B1 | (1 mark)
(b) | $\begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} = 12 + 4 + 3 = \sqrt{4^2 + (-1)^2 + 3^2}\sqrt{3^2 + (-4)^2 + 1^2}\cos\theta$ then $\cos\theta = \frac{19}{26}$ | M1 A1; awrt 0.73 A1 | (3 marks)
(c) | $X: (10, 0, 11)$ | Accept vector forms | B1 | (1 mark)
(d) | $\overrightarrow{AX} = \begin{pmatrix} 10 \\ 0 \\ 11 \end{pmatrix} - \begin{pmatrix} -6 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 16 \\ -4 \\ 12 \end{pmatrix}$ | Either order | M1; cao A1 | (2 marks)
(e) | $\left|\overrightarrow{AX}\right| = \sqrt{16^2 + (-4)^2 + 12^2} = \sqrt{416} = \sqrt{16 \times 26} = 4\sqrt{26}$ | M1; Do not penalise if consistent incorrect signs in (d) A1 | (2 marks)
(f) | Use of correct right angled triangle: $\frac{\left|\overrightarrow{AX}\right|}{d} = \cos\theta$ then $d = \frac{4\sqrt{26}}{\frac{19}{26}} \approx 27.9$ | M1; M1; awrt 27.9 A1 | (3 marks)
**Total: [12]**
---4. The line $l _ { 1 }$ has vector equation
$$\mathbf { r } = \left( \begin{array} { c }
- 6 \\
4 \\
- 1
\end{array} \right) + \lambda \left( \begin{array} { c }
4 \\
- 1 \\
3
\end{array} \right)$$
and the line $l _ { 2 }$ has vector equation
$$\mathbf { r } = \left( \begin{array} { c }
- 6 \\
4 \\
- 1
\end{array} \right) + \mu \left( \begin{array} { c }
3 \\
- 4 \\
1
\end{array} \right)$$
where $\lambda$ and $\mu$ are parameters.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $A$ and the acute angle between $l _ { 1 }$ and $l _ { 2 }$ is $\theta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $A$.
\item Find the value of $\cos \theta$.
The point $X$ lies on $l _ { 1 }$ where $\lambda = 4$.
\item Find the coordinates of $X$.
\item Find the vector $\overrightarrow { A X }$.
\item Hence, or otherwise, show that $| \overrightarrow { A X } | = 4 \sqrt { } 26$.
The point $Y$ lies on $l _ { 2 }$. Given that the vector $\overrightarrow { Y X }$ is perpendicular to $l _ { 1 }$,
\item find the length of $A Y$, giving your answer to 3 significant figures.
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2010 Q4 [12]}}