| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Triangle and parallelogram problems |
| Difficulty | Moderate -0.8 This is a straightforward application of vector addition for parallelograms (part a) and scalar multiplication with magnitude (part b). Both parts require only routine procedures: using the parallelogram property that opposite sides are equal vectors, then finding a unit vector and scaling it. No problem-solving insight or novel approach needed—purely mechanical application of standard A-level vector techniques. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{BA}=(2\mathbf{i}+10\mathbf{j}+9\mathbf{k})+(-3\mathbf{i}+4\mathbf{j}-5\mathbf{k})\) or \(\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{BC}=(\mathbf{i}+7\mathbf{j}-2\mathbf{k})+(-2\mathbf{i}+7\mathbf{j}+6\mathbf{k})\) | M1 | Complete method for finding position vector of \(D\) |
| \(\overrightarrow{OD}=-\mathbf{i}+14\mathbf{j}+4\mathbf{k}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB}=3\mathbf{i}-4\mathbf{j}+5\mathbf{k} \Rightarrow | \overrightarrow{AB} | =\sqrt{(3)^2+(-4)^2+(5)^2}=\sqrt{50}=5\sqrt{2}\) |
| As \( | \overrightarrow{AX} | =10\sqrt{2}\) then \( |
| \(\overrightarrow{OX}=7\mathbf{i}-\mathbf{j}+8\mathbf{k}\) only | A1 | Correct answer only |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{BA}=(2\mathbf{i}+10\mathbf{j}+9\mathbf{k})+(-3\mathbf{i}+4\mathbf{j}-5\mathbf{k})$ or $\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{BC}=(\mathbf{i}+7\mathbf{j}-2\mathbf{k})+(-2\mathbf{i}+7\mathbf{j}+6\mathbf{k})$ | M1 | Complete method for finding position vector of $D$ |
| $\overrightarrow{OD}=-\mathbf{i}+14\mathbf{j}+4\mathbf{k}$ | A1 | Correct answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB}=3\mathbf{i}-4\mathbf{j}+5\mathbf{k} \Rightarrow |\overrightarrow{AB}|=\sqrt{(3)^2+(-4)^2+(5)^2}=\sqrt{50}=5\sqrt{2}$ | M1 | Complete attempt to find $|\overrightarrow{AB}|$ or $|\overrightarrow{BA}|$ |
| As $|\overrightarrow{AX}|=10\sqrt{2}$ then $|\overrightarrow{AX}|=2|\overrightarrow{AB}| \Rightarrow \overrightarrow{AX}=2\overrightarrow{AB}$ $\overrightarrow{OX}=\overrightarrow{OA}+2\overrightarrow{AB}=(\mathbf{i}+7\mathbf{j}-2\mathbf{k})+2(3\mathbf{i}-4\mathbf{j}+5\mathbf{k})$ or $\overrightarrow{OX}=\overrightarrow{OB}+\overrightarrow{AB}=(4+3\mathbf{j}+3\mathbf{k})+(3\mathbf{i}-4\mathbf{j}+5\mathbf{k})$ | M1 | Complete process for finding position vector of $X$ |
| $\overrightarrow{OX}=7\mathbf{i}-\mathbf{j}+8\mathbf{k}$ only | A1 | Correct answer only |
---\begin{enumerate}
\item Relative to a fixed origin $O$,\\
the point $A$ has position vector $\mathbf { i } + 7 \mathbf { j } - 2 \mathbf { k }$,\\
the point $B$ has position vector $4 \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k }$,\\
and the point $C$ has position vector $2 \mathbf { i } + 10 \mathbf { j } + 9 \mathbf { k }$.\\
Given that $A B C D$ is a parallelogram,\\
(a) find the position vector of point $D$.
\end{enumerate}
The vector $\overrightarrow { A X }$ has the same direction as $\overrightarrow { A B }$.\\
Given that $| \overrightarrow { A X } | = 10 \sqrt { 2 }$,\\
(b) find the position vector of $X$.
\hfill \mbox{\textit{Edexcel Paper 1 Q4 [5]}}