| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine techniques: finding a direction vector by subtraction, writing a line equation, then using the perpendicularity condition (dot product = 0) to find a parameter value. While it involves multiple steps (5 marks typical), each step follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}8\\3\\4\end{pmatrix} - \begin{pmatrix}10\\2\\3\end{pmatrix} = \begin{pmatrix}-2\\1\\1\end{pmatrix}\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}10\\2\\3\end{pmatrix} + t\begin{pmatrix}-2\\1\\1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}8\\3\\4\end{pmatrix} + t\begin{pmatrix}-2\\1\\1\end{pmatrix}\) | M1 A1ft | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{CP} = \begin{pmatrix}10-2t\\2+t\\3+t\end{pmatrix} - \begin{pmatrix}3\\12\\3\end{pmatrix} = \begin{pmatrix}7-2t\\t-10\\t\end{pmatrix}\) | M1 A1 | |
| \(\begin{pmatrix}7-2t\\t-10\\t\end{pmatrix} \cdot \begin{pmatrix}-2\\1\\1\end{pmatrix} = -14+4t+t-10+t = 0\) | M1 | Setting \(\overrightarrow{CP} \cdot\) direction \(= 0\) |
| \(t = 4\) | A1 | |
| Position vector of \(P\) is \(\begin{pmatrix}10-8\\2+4\\3+4\end{pmatrix} = \begin{pmatrix}2\\6\\7\end{pmatrix}\) | M1 A1 | (6) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}8\\3\\4\end{pmatrix} - \begin{pmatrix}10\\2\\3\end{pmatrix} = \begin{pmatrix}-2\\1\\1\end{pmatrix}$ | M1 A1 | **(2)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \begin{pmatrix}10\\2\\3\end{pmatrix} + t\begin{pmatrix}-2\\1\\1\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}8\\3\\4\end{pmatrix} + t\begin{pmatrix}-2\\1\\1\end{pmatrix}$ | M1 A1ft | **(2)** |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{CP} = \begin{pmatrix}10-2t\\2+t\\3+t\end{pmatrix} - \begin{pmatrix}3\\12\\3\end{pmatrix} = \begin{pmatrix}7-2t\\t-10\\t\end{pmatrix}$ | M1 A1 | |
| $\begin{pmatrix}7-2t\\t-10\\t\end{pmatrix} \cdot \begin{pmatrix}-2\\1\\1\end{pmatrix} = -14+4t+t-10+t = 0$ | M1 | Setting $\overrightarrow{CP} \cdot$ direction $= 0$ |
| $t = 4$ | A1 | |
| Position vector of $P$ is $\begin{pmatrix}10-8\\2+4\\3+4\end{pmatrix} = \begin{pmatrix}2\\6\\7\end{pmatrix}$ | M1 A1 | **(6)** |\begin{enumerate}
\item Relative to a fixed origin $O$, the point $A$ has position vector $( 10 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } )$, and the point $B$ has position vector $( 8 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } )$.
\end{enumerate}
The line $l$ passes through the points $A$ and $B$.\\
(a) Find the vector $\overrightarrow { A B }$.\\
(b) Find a vector equation for the line $l$.
The point $C$ has position vector $( 3 \mathbf { i } + 12 \mathbf { j } + 3 \mathbf { k } )$.\\
The point $P$ lies on $l$. Given that the vector $\overrightarrow { C P }$ is perpendicular to $l$,\\
(c) find the position vector of the point $P$.\\
\hfill \mbox{\textit{Edexcel C4 2012 Q8 [10]}}