| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Geometric loci and constraints |
| Difficulty | Standard +0.8 This question requires understanding that if AB is a diameter, then angle APB = 90°, leading to the dot product condition AP·BP = 0. While the vector arithmetic is straightforward, recognizing this geometric property and setting up the constraint equation requires solid conceptual understanding beyond routine calculation, making it moderately challenging. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((4\mathbf{i}+2\mathbf{j}-5\mathbf{k})-(3\mathbf{i}+2\mathbf{j}) = \mathbf{i}-5\mathbf{k}\) or \((3\mathbf{i}+2\mathbf{j})-(4\mathbf{i}+2\mathbf{j}-5\mathbf{k}) = 5\mathbf{k}-\mathbf{i}\) | M1 | \(\mathbf{b}-\mathbf{a}\) or \(\mathbf{a}-\mathbf{b}\) attempted using \(\mathbf{i},\mathbf{j},\mathbf{k}\) or column vectors. May be implied by calculation seen |
| \(AB=\sqrt{26}\) or \(5.10\) (3 sf) or \(5.1\) | A1 | Correct answer, no working: M1A1. Mark(s) cannot be gained retrospectively in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(26=(p-3)^2+4+9+(p-4)^2+4+4\) | M1 | Attempt \(AB^2=BP^2+PA^2\) (involving \(p\)) ft their \(AB\) |
| Attempt \(\ | PC\ | ^2=(\text{their radius})^2\) |
| Attempt \(\overrightarrow{PA}\cdot\overrightarrow{PB}=0\) | M1 | or \(((3-p)\mathbf{i}+2\mathbf{j}+3\mathbf{k}){\cdot}((4-p)\mathbf{i}+2\mathbf{j}-2\mathbf{k})=0\) |
| \(p^2-7p+10=0\) oe or \(\left(p-\frac{7}{2}\right)^2=\frac{9}{4}\) | A1f | Correct simplified equation, ft their (a) |
| \(p=2\) or \(5\) | A1f | ft only their (a) |
# Question 2:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4\mathbf{i}+2\mathbf{j}-5\mathbf{k})-(3\mathbf{i}+2\mathbf{j}) = \mathbf{i}-5\mathbf{k}$ or $(3\mathbf{i}+2\mathbf{j})-(4\mathbf{i}+2\mathbf{j}-5\mathbf{k}) = 5\mathbf{k}-\mathbf{i}$ | M1 | $\mathbf{b}-\mathbf{a}$ or $\mathbf{a}-\mathbf{b}$ attempted using $\mathbf{i},\mathbf{j},\mathbf{k}$ or column vectors. May be implied by calculation seen |
| $AB=\sqrt{26}$ or $5.10$ (3 sf) or $5.1$ | A1 | Correct answer, no working: M1A1. Mark(s) cannot be gained retrospectively in (b) |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $26=(p-3)^2+4+9+(p-4)^2+4+4$ | M1 | Attempt $AB^2=BP^2+PA^2$ (involving $p$) ft their $AB$ |
| Attempt $\|PC\|^2=(\text{their radius})^2$ | M1 | or $\left(\frac{7}{2}-p\right)^2+4+\frac{1}{4}=\frac{13}{2}$ |
| Attempt $\overrightarrow{PA}\cdot\overrightarrow{PB}=0$ | M1 | or $((3-p)\mathbf{i}+2\mathbf{j}+3\mathbf{k}){\cdot}((4-p)\mathbf{i}+2\mathbf{j}-2\mathbf{k})=0$ |
| $p^2-7p+10=0$ oe or $\left(p-\frac{7}{2}\right)^2=\frac{9}{4}$ | A1f | Correct simplified equation, ft their (a) |
| $p=2$ or $5$ | A1f | ft only their (a) |
---2 The points $A$ and $B$ have position vectors $3 \mathbf { i } + 2 \mathbf { j }$ and $4 \mathbf { i } + 2 \mathbf { j } - 5 \mathbf { k }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the length of $A B$.
Point $P$ has position vector $p \mathbf { i } - 3 \mathbf { k }$, where $p$ is a constant. $P$ lies on the circumference of a circle of which $A B$ is a diameter.
\item Find the two possible values of $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2022 Q2 [5]}}