| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Equal length conditions |
| Difficulty | Moderate -0.3 This is a straightforward two-part vectors question requiring basic operations: (a) uses the relation AB=BD to find D by vector addition (routine), and (b) solves |AC|=4 using magnitude formula leading to a simple quadratic. Both parts are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b 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 | Marks | Guidance |
| \(\overrightarrow{OD} = \overrightarrow{OB} + \overrightarrow{BD} = \overrightarrow{OB} + \overrightarrow{AB}\) or \(\overrightarrow{OD} = 2\overrightarrow{OB} - \overrightarrow{OA}\) or \(\overrightarrow{OD} = \overrightarrow{OA} + 2\overrightarrow{AB}\) leading to \(= \begin{pmatrix}4\\-2\\3\end{pmatrix} + \begin{pmatrix}2\\-5\\7\end{pmatrix}\) | M1 | 3.1a - Complete applied strategy to find a vector expression for \(\overrightarrow{OD}\). M1 implied by at least two correct components. Give M0 for subtracting wrong way. |
| \(= \begin{pmatrix}6\\-7\\10\end{pmatrix}\) or \(6\mathbf{i} - 7\mathbf{j} + 10\mathbf{k}\) | A1 | 1.1b - Finding coordinates \((6,-7,10)\) without reference to correct position vectors is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((a-2)^2 + (5-3)^2 + (-2--4)^2\) | M1 | 1.1b - Finds difference between \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\), then squares and adds each of the 3 components. Ignore labelling. |
| \(\left\{ | \overrightarrow{AC} | = 4 \Rightarrow\right\}\ (a-2)^2 + (5-3)^2 + (-2--4)^2 = (4)^2 \Rightarrow (a-2)^2 = 8 \Rightarrow a = \ldots\) or \(a^2 - 4a - 4 = 0 \Rightarrow a = \ldots\) |
| (as \(a < 0 \Rightarrow\)) \(a = 2 - 2\sqrt{2}\) (or \(a = 2 - \sqrt{8}\)) | A1 | 1.1b - Obtains only one exact value. Writing \(a = 2 \pm 2\sqrt{2}\) without evidence of rejecting \(a = 2+2\sqrt{2}\) is A0. Allow exact alternatives such as \(2-\sqrt{8}\) or \(\frac{4-\sqrt{32}}{2}\). Writing \(a = -0.828...\) without reference to correct exact value is A0. |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OD} = \overrightarrow{OB} + \overrightarrow{BD} = \overrightarrow{OB} + \overrightarrow{AB}$ or $\overrightarrow{OD} = 2\overrightarrow{OB} - \overrightarrow{OA}$ or $\overrightarrow{OD} = \overrightarrow{OA} + 2\overrightarrow{AB}$ leading to $= \begin{pmatrix}4\\-2\\3\end{pmatrix} + \begin{pmatrix}2\\-5\\7\end{pmatrix}$ | M1 | 3.1a - Complete applied strategy to find a vector expression for $\overrightarrow{OD}$. M1 implied by at least two correct components. Give M0 for subtracting wrong way. |
| $= \begin{pmatrix}6\\-7\\10\end{pmatrix}$ or $6\mathbf{i} - 7\mathbf{j} + 10\mathbf{k}$ | A1 | 1.1b - Finding coordinates $(6,-7,10)$ without reference to correct position vectors is A0 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(a-2)^2 + (5-3)^2 + (-2--4)^2$ | M1 | 1.1b - Finds difference between $\overrightarrow{OA}$ and $\overrightarrow{OC}$, then squares and adds each of the 3 components. Ignore labelling. |
| $\left\{|\overrightarrow{AC}| = 4 \Rightarrow\right\}\ (a-2)^2 + (5-3)^2 + (-2--4)^2 = (4)^2 \Rightarrow (a-2)^2 = 8 \Rightarrow a = \ldots$ or $a^2 - 4a - 4 = 0 \Rightarrow a = \ldots$ | dM1 | 2.1 - Complete method of correctly applying Pythagoras on $|\overrightarrow{AC}|=4$ and using correct method to solve quadratic. Condone at least one of awrt 4.8 or awrt $-0.83$ for dM1. |
| (as $a < 0 \Rightarrow$) $a = 2 - 2\sqrt{2}$ (or $a = 2 - \sqrt{8}$) | A1 | 1.1b - Obtains **only one** exact value. Writing $a = 2 \pm 2\sqrt{2}$ without evidence of rejecting $a = 2+2\sqrt{2}$ is A0. Allow exact alternatives such as $2-\sqrt{8}$ or $\frac{4-\sqrt{32}}{2}$. Writing $a = -0.828...$ without reference to correct exact value is A0. |\begin{enumerate}
\item Relative to a fixed origin $O$,\\
the point $A$ has position vector $( 2 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k } )$,\\
the point $B$ has position vector ( $4 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k }$ ),\\
and the point $C$ has position vector ( $a \mathbf { i } + 5 \mathbf { j } - 2 \mathbf { k }$ ), where $a$ is a constant and $a < 0 D$ is the point such that $\overrightarrow { A B } = \overrightarrow { B D }$.\\
(a) Find the position vector of $D$.
\end{enumerate}
Given $| \overrightarrow { A C } | = 4$\\
(b) find the value of $a$.
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q2 [5]}}