| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: show lines are skew |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine techniques: finding a line equation from two points, checking if lines intersect by solving simultaneous equations, and using perpendicularity conditions. Part (c) requires more steps but follows predictable methods (dot product = 0). Slightly easier than average due to straightforward computational nature with no novel insight required. |
| 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 |
|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix}-3\\6\\1\end{pmatrix} - \begin{pmatrix}-4\\1\\3\end{pmatrix} = \begin{pmatrix}1\\5\\-2\end{pmatrix} \therefore \mathbf{r} = \begin{pmatrix}-4\\1\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\5\\-2\end{pmatrix}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(3-2\lambda = 9+\mu\) (3) | B1 |
| \(2 \times (1) + (3): -5 = 15+5\mu, \quad \mu = -4, \lambda = -1\) | M1 A1 |
| sub. (2): \(1-5 = -7+12,\) not true \(\therefore\) do not intersect | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OC} = \begin{pmatrix}1\\-7-3\mu\\9+\mu\end{pmatrix}, \quad \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}6+2\mu\\-13-3\mu\\8+\mu\end{pmatrix}\) | M1 A1 | |
| \(\therefore \begin{pmatrix}1\\5\\-2\end{pmatrix} \cdot \begin{pmatrix}6+2\mu\\-13-3\mu\\8+\mu\end{pmatrix} = 0, \quad 6+2\mu-65-15\mu-16-2\mu = 0\) | M1 A1 | |
| \(\mu = -5 \therefore \overrightarrow{OC} = \begin{pmatrix}-7\\8\\4\end{pmatrix}\) | M1 A1 | (13 marks) |
## (a)
$\overrightarrow{AB} = \begin{pmatrix}-3\\6\\1\end{pmatrix} - \begin{pmatrix}-4\\1\\3\end{pmatrix} = \begin{pmatrix}1\\5\\-2\end{pmatrix} \therefore \mathbf{r} = \begin{pmatrix}-4\\1\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\5\\-2\end{pmatrix}$ | M1 A1 |
## (b)
$-4+\lambda = 3+2\mu$ (1)
$1+5\lambda = -7-3\mu$ (2)
$3-2\lambda = 9+\mu$ (3) | B1 |
$2 \times (1) + (3): -5 = 15+5\mu, \quad \mu = -4, \lambda = -1$ | M1 A1 |
sub. (2): $1-5 = -7+12,$ not true $\therefore$ do not intersect | M1 A1 |
## (c)
$\overrightarrow{OC} = \begin{pmatrix}1\\-7-3\mu\\9+\mu\end{pmatrix}, \quad \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}6+2\mu\\-13-3\mu\\8+\mu\end{pmatrix}$ | M1 A1 |
$\therefore \begin{pmatrix}1\\5\\-2\end{pmatrix} \cdot \begin{pmatrix}6+2\mu\\-13-3\mu\\8+\mu\end{pmatrix} = 0, \quad 6+2\mu-65-15\mu-16-2\mu = 0$ | M1 A1 |
$\mu = -5 \therefore \overrightarrow{OC} = \begin{pmatrix}-7\\8\\4\end{pmatrix}$ | M1 A1 | **(13 marks)**
---Relative to a fixed origin, the points $A$ and $B$ have position vectors $\begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 6 \\ 1 \end{pmatrix}$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for the line $l_1$ which passes through $A$ and $B$. [2]
\end{enumerate}
The line $l_2$ has vector equation
$$\mathbf{r} = \begin{pmatrix} 3 \\ -7 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that lines $l_1$ and $l_2$ do not intersect. [5]
\item Find the position vector of the point $C$ on $l_2$ such that $\angle ABC = 90°$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [13]}}