| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from origin to line |
| Difficulty | Standard +0.8 This is a substantial multi-part 3D vectors question requiring: (a) finding a line equation from two points, (b) verifying a point lies on the line, (c) finding the perpendicular from origin to line (requiring dot product = 0), and (d) calculating triangle area using cross product. Part (c) is non-routine as students must set up and solve the perpendicularity condition. The combination of techniques and the geometric insight needed for parts (c) and (d) place this above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10g Problem solving with vectors: in geometry4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| \((a)\) \(\overrightarrow{AB} = \begin{pmatrix} 10 \\ -15 \\ 5 \end{pmatrix} \therefore \vec{r} = \begin{pmatrix} 3 \\ 9 \\ -7 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\) | M1 A1 | |
| \((b)\) \(3 + 2\lambda = 9 \therefore \lambda = 3\) | M1 | |
| when \(\lambda = 3, \vec{r} = \begin{pmatrix} 3 \\ 9 \\ -7 \end{pmatrix} + 3\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 9 \\ 0 \\ -4 \end{pmatrix} \therefore (9, 0, -4)\) lies on \(l\) | A1 | |
| \((c)\) \(\overrightarrow{OD} = \begin{pmatrix} 3+2\lambda \\ 9-3\lambda \\ -7+\lambda \end{pmatrix} \therefore \begin{pmatrix} 3+2\lambda \\ 9-3\lambda \\ -7+\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 0\) | M1 | |
| \(6 + 4\lambda - 27 + 9\lambda - 7 + \lambda = 0\) | A1 | |
| \(\lambda = 2 \therefore \overrightarrow{OD} = \begin{pmatrix} 7 \\ 3 \\ -5 \end{pmatrix}, D(7, 3, -5)\) | M1 A1 | |
| \((d)\) \(AB = \sqrt{100+225+25} = \sqrt{350}, OD = \sqrt{49+9+25} = \sqrt{83}\) | M1 | |
| area \(= \frac{1}{2} \times \sqrt{350} \times \sqrt{83} = 85.2\) (3sf) | M1 A1 | (11 marks) |
$(a)$ $\overrightarrow{AB} = \begin{pmatrix} 10 \\ -15 \\ 5 \end{pmatrix} \therefore \vec{r} = \begin{pmatrix} 3 \\ 9 \\ -7 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ | M1 A1 |
$(b)$ $3 + 2\lambda = 9 \therefore \lambda = 3$ | M1 |
when $\lambda = 3, \vec{r} = \begin{pmatrix} 3 \\ 9 \\ -7 \end{pmatrix} + 3\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 9 \\ 0 \\ -4 \end{pmatrix} \therefore (9, 0, -4)$ lies on $l$ | A1 |
$(c)$ $\overrightarrow{OD} = \begin{pmatrix} 3+2\lambda \\ 9-3\lambda \\ -7+\lambda \end{pmatrix} \therefore \begin{pmatrix} 3+2\lambda \\ 9-3\lambda \\ -7+\lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 0$ | M1 |
$6 + 4\lambda - 27 + 9\lambda - 7 + \lambda = 0$ | A1 |
$\lambda = 2 \therefore \overrightarrow{OD} = \begin{pmatrix} 7 \\ 3 \\ -5 \end{pmatrix}, D(7, 3, -5)$ | M1 A1 |
$(d)$ $AB = \sqrt{100+225+25} = \sqrt{350}, OD = \sqrt{49+9+25} = \sqrt{83}$ | M1 |
area $= \frac{1}{2} \times \sqrt{350} \times \sqrt{83} = 85.2$ (3sf) | M1 A1 | (11 marks)4. The points $A$ and $B$ have coordinates $( 3,9 , - 7 )$ and $( 13 , - 6 , - 2 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find, in vector form, an equation for the line $l$ which passes through $A$ and $B$.
\item Show that the point $C$ with coordinates $( 9,0 , - 4 )$ lies on $l$.
The point $D$ is the point on $l$ closest to the origin, $O$.
\item Find the coordinates of $D$.
\item Find the area of triangle $O A B$ to 3 significant figures.\\
4. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [11]}}