| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a straightforward multi-part vector question testing standard techniques: showing collinearity via parallel vectors (part a), finding ratios (part b), verifying perpendicularity using dot product (part c), and calculating triangle area using ½|AB×AD| (part d). All parts follow routine procedures with no novel insight required, making it slightly easier than average for C4. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{AB} = (5\mathbf{i} - 4\mathbf{j}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (3\mathbf{i} - 3\mathbf{j} - 6\mathbf{k})\) | B1 | |
| \(\overrightarrow{AC} = (7\mathbf{i} - 6\mathbf{j} - 4\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (5\mathbf{i} - 5\mathbf{j} - 10\mathbf{k}) = \frac{5}{3}\overrightarrow{AB}\) | M1 | |
| \(\therefore \overrightarrow{AC}\) is parallel to \(\overrightarrow{AB}\), also common point \(\therefore\) single straight line | A1 | |
| (b) \(3:2\) | B1 | |
| (c) \(\overrightarrow{AD} = (3\mathbf{i} + \mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\) | B1 | |
| \(\overrightarrow{BD} = (3\mathbf{i} + \mathbf{j} + 4\mathbf{k}) - (5\mathbf{i} - 4\mathbf{j}) = (-2\mathbf{i} + 5\mathbf{j} + 4\mathbf{k})\) | B1 | |
| \(\overrightarrow{AD} \cdot \overrightarrow{BD} = -2 + 10 - 8 = 0 \therefore\) perpendicular | M1 A1 | |
| (d) \(= \frac{1}{2} \times \sqrt{1 + 4 + 4} \times \sqrt{4 + 25 + 16} = \frac{1}{2} \times 3 \times 3\sqrt{5} = \frac{9}{2}\sqrt{5}\) | M2 A1 | (11) |
**(a)** $\overrightarrow{AB} = (5\mathbf{i} - 4\mathbf{j}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (3\mathbf{i} - 3\mathbf{j} - 6\mathbf{k})$ | B1 |
$\overrightarrow{AC} = (7\mathbf{i} - 6\mathbf{j} - 4\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (5\mathbf{i} - 5\mathbf{j} - 10\mathbf{k}) = \frac{5}{3}\overrightarrow{AB}$ | M1 |
$\therefore \overrightarrow{AC}$ is parallel to $\overrightarrow{AB}$, also common point $\therefore$ single straight line | A1 |
**(b)** $3:2$ | B1 |
**(c)** $\overrightarrow{AD} = (3\mathbf{i} + \mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + 6\mathbf{k}) = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})$ | B1 |
$\overrightarrow{BD} = (3\mathbf{i} + \mathbf{j} + 4\mathbf{k}) - (5\mathbf{i} - 4\mathbf{j}) = (-2\mathbf{i} + 5\mathbf{j} + 4\mathbf{k})$ | B1 |
$\overrightarrow{AD} \cdot \overrightarrow{BD} = -2 + 10 - 8 = 0 \therefore$ perpendicular | M1 A1 |
**(d)** $= \frac{1}{2} \times \sqrt{1 + 4 + 4} \times \sqrt{4 + 25 + 16} = \frac{1}{2} \times 3 \times 3\sqrt{5} = \frac{9}{2}\sqrt{5}$ | M2 A1 | (11)
---5. Relative to a fixed origin, the points $A , B$ and $C$ have position vectors ( $2 \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ ), $( 5 \mathbf { i } - 4 \mathbf { j } )$ and $( 7 \mathbf { i } - 6 \mathbf { j } - 4 \mathbf { k } )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $A , B$ and $C$ all lie on a single straight line.
\item Write down the ratio $A B : B C$
The point $D$ has position vector $( 3 \mathbf { i } + \mathbf { j } + 4 \mathbf { k } )$.
\item Show that $A D$ is perpendicular to $B D$.
\item Find the exact area of triangle $A B D$.\\
5. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [11]}}