| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Shortest distance from point to line |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question testing routine techniques: finding line equations, magnitudes, angles using dot product, perpendicular distance formula, and triangle area. All parts follow textbook methods with straightforward arithmetic. Part (e) requires combining earlier results but involves no novel insight. Slightly above average due to length and the final synthesis step, but well within typical C4 scope. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}10\\14\\-4\end{pmatrix} - \begin{pmatrix}8\\13\\-2\end{pmatrix} = \begin{pmatrix}2\\1\\-2\end{pmatrix}\) or \(\overrightarrow{BA} = \begin{pmatrix}-2\\-1\\2\end{pmatrix}\) | M1 | |
| \(\mathbf{r} = \begin{pmatrix}8\\13\\-2\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}10\\14\\-4\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}\) | M1 A1ft | accept equivalents (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = \begin{pmatrix}10\\14\\-4\end{pmatrix} - \begin{pmatrix}9\\9\\6\end{pmatrix} = \begin{pmatrix}1\\5\\-10\end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix}-1\\-5\\10\end{pmatrix}\) | ||
| \(CB = \sqrt{1^2 + 5^2 + (-10)^2} = \sqrt{126}\ (= 3\sqrt{14} \approx 11.2)\) | M1 A1 | awrt 11.2 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\overrightarrow{CB}\cdot\overrightarrow{AB} = \ | \overrightarrow{CB}\ | \ |
| \((\pm)(2 + 5 + 20) = \sqrt{126}\sqrt{9}\cos\theta\) | M1 A1 | |
| \(\cos\theta = \frac{3}{\sqrt{14}} \Rightarrow \theta \approx 36.7°\) | A1 | awrt 36.7° (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{d}{\sqrt{126}} = \sin\theta\) | M1 A1ft | |
| \(d = 3\sqrt{5}\ (\approx 6.7)\) | A1 | awrt 6.7 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(BX^2 = BC^2 - d^2 = 126 - 45 = 81\) | M1 | |
| Area \(CBX = \frac{1}{2} \times BX \times d = \frac{1}{2}\times 9 \times 3\sqrt{5} = \frac{27\sqrt{5}}{2}\ (\approx 30.2)\) | M1 A1 | awrt 30.1 or 30.2 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Area \(CBX = \frac{1}{2}\times d \times BC \sin\angle XCB\) | M1 | |
| \(= \frac{1}{2}\times 3\sqrt{5}\times\sqrt{126}\sin(90-36.7)°\) | M1 | sine of correct angle |
| \(\approx 30.2\) | A1 | \(\frac{27\sqrt{5}}{2}\), awrt 30.1 or 30.2 (3) |
## Question 7:
**Part (a):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}10\\14\\-4\end{pmatrix} - \begin{pmatrix}8\\13\\-2\end{pmatrix} = \begin{pmatrix}2\\1\\-2\end{pmatrix}$ or $\overrightarrow{BA} = \begin{pmatrix}-2\\-1\\2\end{pmatrix}$ | M1 | |
| $\mathbf{r} = \begin{pmatrix}8\\13\\-2\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}10\\14\\-4\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-2\end{pmatrix}$ | M1 A1ft | accept equivalents (3) |
**Part (b):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = \begin{pmatrix}10\\14\\-4\end{pmatrix} - \begin{pmatrix}9\\9\\6\end{pmatrix} = \begin{pmatrix}1\\5\\-10\end{pmatrix}$ or $\overrightarrow{BC} = \begin{pmatrix}-1\\-5\\10\end{pmatrix}$ | | |
| $CB = \sqrt{1^2 + 5^2 + (-10)^2} = \sqrt{126}\ (= 3\sqrt{14} \approx 11.2)$ | M1 A1 | awrt 11.2 (2) |
**Part (c):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\overrightarrow{CB}\cdot\overrightarrow{AB} = \|\overrightarrow{CB}\|\|\overrightarrow{AB}\|\cos\theta$ | | |
| $(\pm)(2 + 5 + 20) = \sqrt{126}\sqrt{9}\cos\theta$ | M1 A1 | |
| $\cos\theta = \frac{3}{\sqrt{14}} \Rightarrow \theta \approx 36.7°$ | A1 | awrt 36.7° (3) |
**Part (d):**
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{d}{\sqrt{126}} = \sin\theta$ | M1 A1ft | |
| $d = 3\sqrt{5}\ (\approx 6.7)$ | A1 | awrt 6.7 (3) |
**Part (e):**
| Working | Marks | Notes |
|---------|-------|-------|
| $BX^2 = BC^2 - d^2 = 126 - 45 = 81$ | M1 | |
| Area $CBX = \frac{1}{2} \times BX \times d = \frac{1}{2}\times 9 \times 3\sqrt{5} = \frac{27\sqrt{5}}{2}\ (\approx 30.2)$ | M1 A1 | awrt 30.1 or 30.2 (3) |
**Alternative for (e):**
| Working | Marks | Notes |
|---------|-------|-------|
| Area $CBX = \frac{1}{2}\times d \times BC \sin\angle XCB$ | M1 | |
| $= \frac{1}{2}\times 3\sqrt{5}\times\sqrt{126}\sin(90-36.7)°$ | M1 | sine of correct angle |
| $\approx 30.2$ | A1 | $\frac{27\sqrt{5}}{2}$, awrt 30.1 or 30.2 (3) |
**Total: [14]**
---7. Relative to a fixed origin $O$, the point $A$ has position vector $( 8 \mathbf { i } + 13 \mathbf { j } - 2 \mathbf { k } )$, the point $B$ has position vector ( $10 \mathbf { i } + 14 \mathbf { j } - 4 \mathbf { k }$ ), and the point $C$ has position vector $( 9 \mathbf { i } + 9 \mathbf { j } + 6 \mathbf { k } )$.
The line $l$ passes through the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find a vector equation for the line $l$.
\item Find $| \overrightarrow { C B } |$.
\item Find the size of the acute angle between the line segment $C B$ and the line $l$, giving your answer in degrees to 1 decimal place.
\item Find the shortest distance from the point $C$ to the line $l$.
The point $X$ lies on $l$. Given that the vector $\overrightarrow { C X }$ is perpendicular to $l$,
\item find the area of the triangle $C X B$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q7 [14]}}