| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Show lines are skew (non-intersecting) |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring routine calculations: finding vectors AB and CD, showing one is a scalar multiple of the other, calculating an angle using the dot product formula, and showing lines are skew by demonstrating no common point exists. All techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\overrightarrow{AB} = \begin{pmatrix}2\\-2\\-4\end{pmatrix}\) and \(\overrightarrow{CD} = \begin{pmatrix}2\\1\\1\end{pmatrix}\) | B1 | Or equivalent seen or implied |
| Use the correct process for calculating the modulus of both vectors to obtain \(AB\) and \(CD\) | M1 | \(AB = \sqrt{24},\; CD = \sqrt{6}\) |
| Using exact values, verify that \(AB = 2CD\) | A1 | Obtain given statement from correct work. Allow from \(BA = 2DC\), OE |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the correct process to calculate the scalar product of the relevant vectors (*their* \(\overrightarrow{AB}\) and \(\overrightarrow{CD}\)) | M1 | \(\begin{pmatrix}2\\-2\\-4\end{pmatrix}\) and \(\begin{pmatrix}2\\1\\1\end{pmatrix}\) or \(\begin{pmatrix}2\\-2\\-4\end{pmatrix}\) and \(\begin{pmatrix}4\\2\\2\end{pmatrix}\) |
| Divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result | M1 | |
| Obtain answer \(99.6°\) (or \(1.74\) radians) or better | A1 | Do not ISW if go on to subtract from \(180°\). \((99.594\ldots, 1.738\ldots)\) Accept \(260.4°\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State correct vector equations for \(AB\) and \(CD\) in any form, e.g. \(\mathbf{r} = \begin{pmatrix}2\\1\\5\end{pmatrix} + \lambda\begin{pmatrix}2\\-2\\-4\end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix}1\\1\\2\end{pmatrix} + \mu\begin{pmatrix}2\\1\\1\end{pmatrix}\) | B1ft | Follow their \(\overrightarrow{AB}\) and \(\overrightarrow{CD}\) |
| Equate at least two pairs of components of their lines and solve for \(\lambda\) or \(\mu\) | M1 | |
| Obtain correct pair of values from correct equations | A1 | |
| Verify that all three equations are not satisfied and that the lines do not intersect | A1 | CWO with conclusion, e.g. \(\frac{17}{3} \neq \frac{7}{3}\) or \(\frac{17}{3} = \frac{7}{3}\) is inconsistent or equivalent |
| Total: 4 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\overrightarrow{AB} = \begin{pmatrix}2\\-2\\-4\end{pmatrix}$ and $\overrightarrow{CD} = \begin{pmatrix}2\\1\\1\end{pmatrix}$ | B1 | Or equivalent seen or implied |
| Use the correct process for calculating the modulus of both vectors to obtain $AB$ and $CD$ | M1 | $AB = \sqrt{24},\; CD = \sqrt{6}$ |
| Using exact values, verify that $AB = 2CD$ | A1 | Obtain given statement from correct work. Allow from $BA = 2DC$, OE |
| **Total: 3** | | |
---
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the correct process to calculate the scalar product of the relevant vectors (*their* $\overrightarrow{AB}$ and $\overrightarrow{CD}$) | M1 | $\begin{pmatrix}2\\-2\\-4\end{pmatrix}$ and $\begin{pmatrix}2\\1\\1\end{pmatrix}$ or $\begin{pmatrix}2\\-2\\-4\end{pmatrix}$ and $\begin{pmatrix}4\\2\\2\end{pmatrix}$ |
| Divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result | M1 | |
| Obtain answer $99.6°$ (or $1.74$ radians) or better | A1 | Do not ISW if go on to subtract from $180°$. $(99.594\ldots, 1.738\ldots)$ Accept $260.4°$ |
| **Total: 3** | | |
---
## Question 8(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State correct vector equations for $AB$ and $CD$ in any form, e.g. $\mathbf{r} = \begin{pmatrix}2\\1\\5\end{pmatrix} + \lambda\begin{pmatrix}2\\-2\\-4\end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix}1\\1\\2\end{pmatrix} + \mu\begin{pmatrix}2\\1\\1\end{pmatrix}$ | B1ft | Follow their $\overrightarrow{AB}$ and $\overrightarrow{CD}$ |
| Equate at least two pairs of components of their lines and solve for $\lambda$ or $\mu$ | M1 | |
| Obtain correct pair of values from correct equations | A1 | |
| Verify that all three equations are not satisfied and that the lines do not intersect | A1 | CWO with conclusion, e.g. $\frac{17}{3} \neq \frac{7}{3}$ or $\frac{17}{3} = \frac{7}{3}$ is inconsistent or equivalent |
| **Total: 4** | | |
---8 With respect to the origin $O$, the position vectors of the points $A , B , C$ and $D$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
2 \\
1 \\
5
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
4 \\
- 1 \\
1
\end{array} \right) , \quad \overrightarrow { O C } = \left( \begin{array} { l }
1 \\
1 \\
2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O D } = \left( \begin{array} { l }
3 \\
2 \\
3
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that $A B = 2 C D$.
\item Find the angle between the directions of $\overrightarrow { A B }$ and $\overrightarrow { C D }$.
\item Show that the line through $A$ and $B$ does not intersect the line through $C$ and $D$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q8 [10]}}