| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Moderate -0.3 This is a standard C4 vectors question testing routine techniques: finding a direction vector, writing a line equation, using perpendicularity (dot product = 0), and calculating distance. All steps are straightforward applications of learned methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{AB} = -2\mathbf{i}+2\mathbf{j}-\mathbf{k} - (\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = -3\mathbf{i}+5\mathbf{j}-3\mathbf{k}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\mathbf{r} = \mathbf{i}-3\mathbf{j}+2\mathbf{k} + \lambda(-3\mathbf{i}+5\mathbf{j}-3\mathbf{k})\) | M1 A1ft | (2) |
| or \(\mathbf{r} = -2\mathbf{i}+2\mathbf{j}-\mathbf{k} + \lambda(-3\mathbf{i}+5\mathbf{j}-3\mathbf{k})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overrightarrow{AC} = 2\mathbf{i}+p\mathbf{j}-4\mathbf{k} - (\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = \mathbf{i}+(p+3)\mathbf{j}-6\mathbf{k}\) | B1 | or \(\overrightarrow{CA}\) |
| \(\overrightarrow{AC}\cdot\overrightarrow{AB} = \begin{pmatrix}1\\p+3\\-6\end{pmatrix}\cdot\begin{pmatrix}-3\\5\\-3\end{pmatrix} = 0\) | M1 | |
| \(-3+5p+15+18=0\) | ||
| \(p = -6\) | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(AC^2 = (2-1)^2 + (-6+3)^2 + (-4-2)^2 \;(=46)\) | M1 | |
| \(AC = \sqrt{46}\) | A1 | accept awrt 6.8 (2) |
| Total | [10] |
# Question 4:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = -2\mathbf{i}+2\mathbf{j}-\mathbf{k} - (\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = -3\mathbf{i}+5\mathbf{j}-3\mathbf{k}$ | M1 A1 | **(2)** |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \mathbf{i}-3\mathbf{j}+2\mathbf{k} + \lambda(-3\mathbf{i}+5\mathbf{j}-3\mathbf{k})$ | M1 A1ft | **(2)** |
| or $\mathbf{r} = -2\mathbf{i}+2\mathbf{j}-\mathbf{k} + \lambda(-3\mathbf{i}+5\mathbf{j}-3\mathbf{k})$ | | |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AC} = 2\mathbf{i}+p\mathbf{j}-4\mathbf{k} - (\mathbf{i}-3\mathbf{j}+2\mathbf{k}) = \mathbf{i}+(p+3)\mathbf{j}-6\mathbf{k}$ | B1 | or $\overrightarrow{CA}$ |
| $\overrightarrow{AC}\cdot\overrightarrow{AB} = \begin{pmatrix}1\\p+3\\-6\end{pmatrix}\cdot\begin{pmatrix}-3\\5\\-3\end{pmatrix} = 0$ | M1 | |
| $-3+5p+15+18=0$ | | |
| $p = -6$ | M1 A1 | **(4)** |
## Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $AC^2 = (2-1)^2 + (-6+3)^2 + (-4-2)^2 \;(=46)$ | M1 | |
| $AC = \sqrt{46}$ | A1 | accept awrt 6.8 **(2)** |
| **Total** | **[10]** | |\begin{enumerate}
\item Relative to a fixed origin $O$, the point $A$ has position vector $\mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k }$ and the point $B$ has position vector $- 2 \mathbf { i } + 2 \mathbf { j } - \mathbf { k }$. The points $A$ and $B$ lie on a straight line $l$.\\
(a) Find $\overrightarrow { A B }$.\\
(b) Find a vector equation of $l$.
\end{enumerate}
The point $C$ has position vector $2 \mathbf { i } + p \mathbf { j } - 4 \mathbf { k }$ with respect to $O$, where $p$ is a constant. Given that $A C$ is perpendicular to $l$, find\\
(c) the value of $p$,\\
(d) the distance $A C$.\\
\hfill \mbox{\textit{Edexcel C4 2011 Q4 [10]}}