| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Point on line satisfying distance or other condition |
| Difficulty | Standard +0.3 This is a standard C4 vectors question involving finding a point on a line satisfying a given condition (likely a distance or perpendicularity constraint). Such questions follow routine procedures: substitute the parameter, apply the condition to form an equation, solve for the parameter, then find the point. This is slightly easier than average as it's a well-practiced technique with clear steps. |
| 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 vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\overrightarrow{PA}\} = \begin{pmatrix}3\\-2\\6\end{pmatrix} - \begin{pmatrix}-p\\0\\2p\end{pmatrix} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}\) | M1 | Finds difference between \(\overrightarrow{OA}\) and \(\overrightarrow{OP}\); ignore labelling |
| Correct difference \(\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}\) or \(\begin{pmatrix}-3-p\\2\\2p-6\end{pmatrix}\) | A1 | |
| \(\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 6+2p-4-6+2p = 0\) | M1 | See notes |
| \(p = 1\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | AP | = \sqrt{4^2+(-2)^2+4^2}\) or \( |
| \(PA\) or \(AP = \sqrt{36}\) or \(6\) | A1 | cao |
| \(AB = ``6"\ \{= PA\}\) or \(PB = ``6\sqrt{2}"\ \{=\sqrt{2}\,PA\}\) | B1ft | See notes |
| \(\overrightarrow{OB} = \begin{pmatrix}3\\-2\\6\end{pmatrix} \pm 2\begin{pmatrix}2\\2\\-1\end{pmatrix}\) or \(\overrightarrow{OB} = \begin{pmatrix}13\\8\\1\end{pmatrix} - 3\begin{pmatrix}2\\2\\-1\end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix}13\\8\\1\end{pmatrix} - 7\begin{pmatrix}2\\2\\-1\end{pmatrix}\) | M1 | Uses correct method to find both possible sets of coordinates of \(B\) |
| \(\begin{pmatrix}7\\2\\4\end{pmatrix}\) and \(\begin{pmatrix}-1\\-6\\8\end{pmatrix}\) | A1 | cao; both coordinates correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((10+2\lambda)^2+(10+2\lambda)^2+(-5-\lambda)^2 = ``36"\) giving \(9\lambda^2+90\lambda+189=0\) | B1ft | Could be implied |
| \(\lambda^2+10\lambda+21=0 \Rightarrow (\lambda+3)(\lambda+7)=0 \Rightarrow \lambda=-3,-7\) | ||
| Then apply final M1 A1 | M1 A1 | As in original scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 45° = \frac{1}{\sqrt{2}} = \frac{36}{6\sqrt{9\lambda^2+90\lambda+261}}\) | M1 | For finding \( |
| \(\sqrt{36}\) or \(6\) | A1 | cao |
| \( | \overrightarrow{PB} | = \sqrt{9\lambda^2+90\lambda+261}\) |
| \(9\lambda^2+90\lambda+189=0 \Rightarrow \lambda=-3,-7\) then apply final M1 A1 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PA} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}\) and direction vector of \(l\) is \(\mathbf{d} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) | — | — |
| So, \(\ | \overrightarrow{PA}\ | = 2\ |
| Apply final M1 A1 as in the original scheme | ...M1 A1 | — |
| Note: \(\overrightarrow{PA} = 2\mathbf{d}\) with no other creditable working is M0A0B0 | — | — |
| Note: \(\overrightarrow{PA} = 2\mathbf{d}\), followed by \(\overrightarrow{OB} = \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix} \pm 2\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}\) is M1A1B1M1 and the final A1 mark is for both sets of correct coordinates | — | — |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\overrightarrow{PA}\} = \begin{pmatrix}3\\-2\\6\end{pmatrix} - \begin{pmatrix}-p\\0\\2p\end{pmatrix} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}$ | M1 | Finds difference between $\overrightarrow{OA}$ and $\overrightarrow{OP}$; ignore labelling |
| Correct difference $\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}$ or $\begin{pmatrix}-3-p\\2\\2p-6\end{pmatrix}$ | A1 | |
| $\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 6+2p-4-6+2p = 0$ | M1 | See notes |
| $p = 1$ | A1 | cso |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|AP| = \sqrt{4^2+(-2)^2+4^2}$ or $|AP| = \sqrt{(-4)^2+2^2+(-4)^2}$ | M1 | See notes |
| $PA$ or $AP = \sqrt{36}$ or $6$ | A1 | cao |
| $AB = ``6"\ \{= PA\}$ or $PB = ``6\sqrt{2}"\ \{=\sqrt{2}\,PA\}$ | B1ft | See notes |
| $\overrightarrow{OB} = \begin{pmatrix}3\\-2\\6\end{pmatrix} \pm 2\begin{pmatrix}2\\2\\-1\end{pmatrix}$ or $\overrightarrow{OB} = \begin{pmatrix}13\\8\\1\end{pmatrix} - 3\begin{pmatrix}2\\2\\-1\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}13\\8\\1\end{pmatrix} - 7\begin{pmatrix}2\\2\\-1\end{pmatrix}$ | M1 | Uses correct method to find both possible sets of coordinates of $B$ |
| $\begin{pmatrix}7\\2\\4\end{pmatrix}$ and $\begin{pmatrix}-1\\-6\\8\end{pmatrix}$ | A1 | cao; both coordinates correct |
### Way 2 (setting $AB = ``6"$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(10+2\lambda)^2+(10+2\lambda)^2+(-5-\lambda)^2 = ``36"$ giving $9\lambda^2+90\lambda+189=0$ | B1ft | Could be implied |
| $\lambda^2+10\lambda+21=0 \Rightarrow (\lambda+3)(\lambda+7)=0 \Rightarrow \lambda=-3,-7$ | | |
| Then apply final M1 A1 | M1 A1 | As in original scheme |
### Way 4 (dot product between $\overrightarrow{PA}$ and $\overrightarrow{PB}$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 45° = \frac{1}{\sqrt{2}} = \frac{36}{6\sqrt{9\lambda^2+90\lambda+261}}$ | M1 | For finding $|\overrightarrow{PA}|$ as before |
| $\sqrt{36}$ or $6$ | A1 | cao |
| $|\overrightarrow{PB}| = \sqrt{9\lambda^2+90\lambda+261}$ | B1 | oe |
| $9\lambda^2+90\lambda+189=0 \Rightarrow \lambda=-3,-7$ then apply final M1 A1 | M1 A1 | |
## Question 8(b) — Way 6 (Continued):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PA} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and direction vector of $l$ is $\mathbf{d} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ | — | — |
| So, $\|\overrightarrow{PA}\| = 2\|\mathbf{d}\|$ or $PA = 2\|\mathbf{d}\|$ | M1 A1 B1 | A correct statement relating these distances (and not vectors) |
| Apply final M1 A1 as in the original scheme | ...M1 A1 | — |
| **Note:** $\overrightarrow{PA} = 2\mathbf{d}$ with no other creditable working is M0A0B0 | — | — |
| **Note:** $\overrightarrow{PA} = 2\mathbf{d}$, followed by $\overrightarrow{OB} = \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix} \pm 2\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ is M1A1B1M1 and the final A1 mark is for both sets of correct coordinates | — | — |\begin{enumerate}
\item With respect to a fixed origin $O$, the line $l$ has equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { c }
13 \\
8 \\
1
\end{array} \right) + \lambda \left( \begin{array} { r }
2 \\
2 \\
- 1
\end{array} \right) \text {, where } \lambda \text { is a scalar parameter. }$$
The point $A$ lies on $l$ and has coordinates ( $3 , - 2,6$ ).\\
The point $P$ has position vector ( $- p \mathbf { i } + 2 p \mathbf { k }$ ) relative to $O$, where $p$ is a constant.\\
Given that vector $\overrightarrow { P A }$ is perpendicular to $l$,\\
(a) find the value of $p$.
Given also that $B$ is a point on $l$ such that $\angle B P A = 45 ^ { \circ }$,\\
(b) find the coordinates of the two possible positions of $B$.\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$
Question 8 continued
\hfill \mbox{\textit{Edexcel C4 2013 Q8 [9]}}