| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to line |
| Difficulty | Standard +0.3 This is a straightforward 3D vectors question requiring standard techniques: finding a position vector using midpoint formula, calculating an angle using dot product, and finding perpendicular distance using the cross product formula. All methods are routine for Further Maths students, though the 3D visualization and multi-step nature elevate it slightly above average A-level difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(3\mathbf{i}+2\mathbf{j}+\mathbf{k}\) | B1 | Accept coordinates in place of position vector |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AM}\) or \(\overrightarrow{AP}\) correct soi | B1 | \(\overrightarrow{AM}=2\mathbf{j}+\mathbf{k}\), or \(\overrightarrow{AP}=-2\mathbf{i}+\mathbf{j}+2\mathbf{k}\) |
| Carry out correct process for evaluating scalar product of \(\overrightarrow{AM}\) and \(\overrightarrow{AP}\) | M1 | or \(\overrightarrow{MA}\) and \(\overrightarrow{PA}\): \(0+2+2\) |
| Using correct process for moduli, divide scalar product by product of moduli and obtain inverse cosine | M1 | For their vectors: \(\theta=\cos^{-1}\left(\frac{4}{3\sqrt{5}}\right)\) |
| Obtain answer \(53.4°\) or \(0.932^c\) | A1 | |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find \(\overrightarrow{PQ}\) (or \(\overrightarrow{QP}\)) for general point \(Q\) on line through \(O\) and \(M\) | B1 FT | e.g. \(\mathbf{PQ}=-(\mathbf{i}+\mathbf{j}+2\mathbf{k})+\mu(3\mathbf{i}+2\mathbf{j}+\mathbf{k})\). Follow their \(M\) |
| Calculate scalar product of \(\overrightarrow{PQ}\) and direction vector for line through \(O\) and \(M\) and equate to zero | *M1 | |
| Solve and obtain correct solution e.g. \(\mu=-\frac{1}{2}\) | A1 | |
| Carry out method to calculate \(PQ\) | DM1 | \(\sqrt{.5^2+0+1.5^2}\) |
| Obtain answer \(\frac{\sqrt{10}}{2}\) | A1 | Or exact equivalent |
| Alternative Method 1: Find \(\overrightarrow{PQ}\) for general \(Q\) on line through \(O\) and \(M\) | B1 FT | |
| Use correct method to express \(PQ^2\) (or \(PQ\)) in terms of \(\mu\) | *M1 | |
| Obtain correct equation e.g. \(PQ^2=(1+3\mu)^2+(1+2\mu)^2+(2+\mu)^2\) | A1 | |
| Carry out complete method for finding minimum | DM1 | e.g. \(6(1+3\mu)+4(1+2\mu)+2(2+\mu)=0\), \(\mu=-\frac{1}{2}\) |
| Obtain answer \(\frac{\sqrt{10}}{2}\) | A1 | Or exact equivalent |
| Alternative Method 2: State \(\overrightarrow{PA}\) in component form e.g. \(\mathbf{i}+\mathbf{j}+2\mathbf{k}\) | B1 | |
| Use scalar product to find projection of \(\overrightarrow{PA}\) on line through \(O\) and \(M\) | M1 | |
| Obtain correct answer \(\frac{7}{\sqrt{14}}\) | A1 | OE |
| Use Pythagoras to find perpendicular | M1 | \(d=\sqrt{AP^2-AQ^2}=\sqrt{1+1+2^2-\left(\frac{7}{\sqrt{14}}\right)^2}\) |
| Obtain answer \(\frac{\sqrt{10}}{2}\) | A1 | Or exact equivalent |
| Alternative Method 3: State \(\overrightarrow{PA}\) in component form e.g. \(\mathbf{i}+\mathbf{j}+2\mathbf{k}\) | B1 | |
| Calculate vector product of \(\overrightarrow{PA}\) and direction vector for line through \(O\) and \(M\) | M1 | |
| Obtain correct answer e.g. \(3\mathbf{i}-5\mathbf{j}+\mathbf{k}\) | A1 | |
| Divide modulus of product by that of direction vector | M1 | e.g. \(\frac{\sqrt{3^2+5^2+1^2}}{\sqrt{3^2+2^2+1^2}}\) |
| Obtain answer \(\frac{\sqrt{10}}{2}\) | A1 | Or exact equivalent |
| Total: 5 marks |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $3\mathbf{i}+2\mathbf{j}+\mathbf{k}$ | B1 | Accept coordinates in place of position vector |
| **Total: 1 mark** | | |
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AM}$ or $\overrightarrow{AP}$ correct soi | B1 | $\overrightarrow{AM}=2\mathbf{j}+\mathbf{k}$, or $\overrightarrow{AP}=-2\mathbf{i}+\mathbf{j}+2\mathbf{k}$ |
| Carry out correct process for evaluating scalar product of $\overrightarrow{AM}$ and $\overrightarrow{AP}$ | M1 | or $\overrightarrow{MA}$ and $\overrightarrow{PA}$: $0+2+2$ |
| Using correct process for moduli, divide scalar product by product of moduli and obtain inverse cosine | M1 | For their vectors: $\theta=\cos^{-1}\left(\frac{4}{3\sqrt{5}}\right)$ |
| Obtain answer $53.4°$ or $0.932^c$ | A1 | |
| **Total: 4 marks** | | |
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find $\overrightarrow{PQ}$ (or $\overrightarrow{QP}$) for general point $Q$ on line through $O$ and $M$ | B1 FT | e.g. $\mathbf{PQ}=-(\mathbf{i}+\mathbf{j}+2\mathbf{k})+\mu(3\mathbf{i}+2\mathbf{j}+\mathbf{k})$. Follow their $M$ |
| Calculate scalar product of $\overrightarrow{PQ}$ and direction vector for line through $O$ and $M$ and equate to zero | *M1 | |
| Solve and obtain correct solution e.g. $\mu=-\frac{1}{2}$ | A1 | |
| Carry out method to calculate $PQ$ | DM1 | $\sqrt{.5^2+0+1.5^2}$ |
| Obtain answer $\frac{\sqrt{10}}{2}$ | A1 | Or exact equivalent |
| **Alternative Method 1:** Find $\overrightarrow{PQ}$ for general $Q$ on line through $O$ and $M$ | B1 FT | |
| Use correct method to express $PQ^2$ (or $PQ$) in terms of $\mu$ | *M1 | |
| Obtain correct equation e.g. $PQ^2=(1+3\mu)^2+(1+2\mu)^2+(2+\mu)^2$ | A1 | |
| Carry out complete method for finding minimum | DM1 | e.g. $6(1+3\mu)+4(1+2\mu)+2(2+\mu)=0$, $\mu=-\frac{1}{2}$ |
| Obtain answer $\frac{\sqrt{10}}{2}$ | A1 | Or exact equivalent |
| **Alternative Method 2:** State $\overrightarrow{PA}$ in component form e.g. $\mathbf{i}+\mathbf{j}+2\mathbf{k}$ | B1 | |
| Use scalar product to find projection of $\overrightarrow{PA}$ on line through $O$ and $M$ | M1 | |
| Obtain correct answer $\frac{7}{\sqrt{14}}$ | A1 | OE |
| Use Pythagoras to find perpendicular | M1 | $d=\sqrt{AP^2-AQ^2}=\sqrt{1+1+2^2-\left(\frac{7}{\sqrt{14}}\right)^2}$ |
| Obtain answer $\frac{\sqrt{10}}{2}$ | A1 | Or exact equivalent |
| **Alternative Method 3:** State $\overrightarrow{PA}$ in component form e.g. $\mathbf{i}+\mathbf{j}+2\mathbf{k}$ | B1 | |
| Calculate vector product of $\overrightarrow{PA}$ and direction vector for line through $O$ and $M$ | M1 | |
| Obtain correct answer e.g. $3\mathbf{i}-5\mathbf{j}+\mathbf{k}$ | A1 | |
| Divide modulus of product by that of direction vector | M1 | e.g. $\frac{\sqrt{3^2+5^2+1^2}}{\sqrt{3^2+2^2+1^2}}$ |
| Obtain answer $\frac{\sqrt{10}}{2}$ | A1 | Or exact equivalent |
| **Total: 5 marks** | | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{ce3c4a9c-bf83-4d28-96e2-ef31c3673dea-16_593_780_264_685}
In the diagram, $O A B C D E F G$ is a cuboid in which $O A = 3$ units, $O C = 2$ units and $O D = 2$ units. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A , O D$ and $O C$ respectively. $M$ is the midpoint of $E F$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of $M$.\\
The position vector of $P$ is $\mathbf { i } + \mathbf { j } + 2 \mathbf { k }$.
\item Calculate angle PAM.
\item Find the exact length of the perpendicular from $P$ to the line passing through $O$ and $M$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q11 [10]}}