| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.3 This is a standard 3D vectors question requiring midpoint vectors, angle between lines using dot product, and perpendicular distance from point to line. While it involves multiple parts and 3D visualization, each step uses routine A-level techniques without requiring novel insight. The perpendicular distance formula is a standard application, making this slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(\overrightarrow{ON} = 2\mathbf{j} + \mathbf{k}\) | B1 | |
| Use \(\overrightarrow{CM} = \overrightarrow{OM} - \overrightarrow{OC}\) | M1 | |
| Obtain \(\overrightarrow{CM} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Carry out correct process for evaluating scalar product of \(\overrightarrow{ON}\) and \(\overrightarrow{CM}\) | M1 | |
| Using correct process for moduli, divide scalar product by product of moduli and evaluate inverse cosine | M1 | |
| Obtain answer 72.7° or 1.27 radians | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| EITHER Taking general point \(P\) of \(ON\) to have position vector \(\lambda(2\mathbf{j}+\mathbf{k})\), form equation in \(\lambda\) by either equating scalar product of \(\overrightarrow{MP}\) and \(\overrightarrow{ON}\) to zero, or applying Pythagoras in triangle \(OPM\), or setting derivative of \( | \overrightarrow{MP} | ^2\) or \( |
| Solve and obtain \(\lambda = \frac{4}{5}\) | A1 | |
| Substitute for \(\lambda\) and calculate \(MP\) | M1 | |
| Obtain the given answer | A1) | AG |
| OR Use \(\frac{\overrightarrow{OM}\cdot\overrightarrow{ON}}{ | \overrightarrow{ON} | }\) to find projection \(OQ\) of \(OM\) on \(ON\) |
| Obtain \(OQ = \frac{4}{\sqrt{5}}\) | A1 | |
| Use Pythagoras in triangle \(OMQ\) to find \(MQ\) | M1 | |
| Obtain the given answer | A1) | AG |
| OR Using relevant scalar product, find cosine of angle \(MON\) or angle \(ONM\) | (M1 | |
| Obtain \(\cos MON = \frac{4}{5}\) or \(\cos ONM = \frac{3}{\sqrt{5}}\) | A1 | |
| Use trig to find length of perpendicular | M1 | |
| Obtain the given answer | A1) | AG |
| Total | 4 |
# Question 8:
## Part 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $\overrightarrow{ON} = 2\mathbf{j} + \mathbf{k}$ | B1 | |
| Use $\overrightarrow{CM} = \overrightarrow{OM} - \overrightarrow{OC}$ | M1 | |
| Obtain $\overrightarrow{CM} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$ | A1 | |
| **Total** | **3** | |
## Part 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Carry out correct process for evaluating scalar product of $\overrightarrow{ON}$ and $\overrightarrow{CM}$ | M1 | |
| Using correct process for moduli, divide scalar product by product of moduli and evaluate inverse cosine | M1 | |
| Obtain answer 72.7° or 1.27 radians | A1 | |
| **Total** | **3** | |
## Part 8(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| **EITHER** Taking general point $P$ of $ON$ to have position vector $\lambda(2\mathbf{j}+\mathbf{k})$, form equation in $\lambda$ by either equating scalar product of $\overrightarrow{MP}$ and $\overrightarrow{ON}$ to zero, or applying Pythagoras in triangle $OPM$, or setting derivative of $|\overrightarrow{MP}|^2$ or $|\overrightarrow{MP}|$ to zero | (M1 | |
| Solve and obtain $\lambda = \frac{4}{5}$ | A1 | |
| Substitute for $\lambda$ and calculate $MP$ | M1 | |
| Obtain the given answer | A1) | AG |
| **OR** Use $\frac{\overrightarrow{OM}\cdot\overrightarrow{ON}}{|\overrightarrow{ON}|}$ to find projection $OQ$ of $OM$ on $ON$ | (M1 | |
| Obtain $OQ = \frac{4}{\sqrt{5}}$ | A1 | |
| Use Pythagoras in triangle $OMQ$ to find $MQ$ | M1 | |
| Obtain the given answer | A1) | AG |
| **OR** Using relevant scalar product, find cosine of angle $MON$ or angle $ONM$ | (M1 | |
| Obtain $\cos MON = \frac{4}{5}$ or $\cos ONM = \frac{3}{\sqrt{5}}$ | A1 | |
| Use trig to find length of perpendicular | M1 | |
| Obtain the given answer | A1) | AG |
| **Total** | **4** | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{c1eee696-3d7f-410a-91a8-fa902309c117-14_485_716_262_676}
In the diagram, $O A B C$ is a pyramid in which $O A = 2$ units, $O B = 4$ units and $O C = 2$ units. The edge $O C$ is vertical, the base $O A B$ is horizontal and angle $A O B = 90 ^ { \circ }$. Unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are parallel to $O A$, $O B$ and $O C$ respectively. The midpoints of $A B$ and $B C$ are $M$ and $N$ respectively.\\
(a) Express the vectors $\overrightarrow { O N }$ and $\overrightarrow { C M }$ in terms of $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(b) Calculate the angle between the directions of $\overrightarrow { O N }$ and $\overrightarrow { C M }$.\\
(c) Show that the length of the perpendicular from $M$ to $O N$ is $\frac { 3 } { 5 } \sqrt { 5 }$.\\
\hfill \mbox{\textit{CAIE P3 2020 Q8 [10]}}