| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Equal length conditions |
| Difficulty | Moderate -0.8 This is a straightforward vectors question requiring basic operations: (i) finding magnitudes of two vectors, setting them equal, and solving a simple equation; (ii) using the standard scalar product formula to find an angle. Both parts are routine applications of standard techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{AB}=\mathbf{OB}-\mathbf{OA}=\begin{pmatrix}-1\\2\\p+4\end{pmatrix}\) | B1 | Ignore labels. Allow BA or BC |
| \(\mathbf{CB}=\mathbf{OB}-\mathbf{OC}=\begin{pmatrix}-4\\5\\p-2\end{pmatrix}\) | B1 | |
| \(1+4+(p+4)^2=16+25+(p-2)^2\) | M1 | |
| \(p=2\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{AB}\cdot\mathbf{CB}=4+10-5=9\) | M1 | Use of \(x_1x_2+y_1y_2+z_1z_2\) |
| \( | \mathbf{AB} | =\sqrt{1+4+25}=\sqrt{30},\ |
| \(\cos ABC=\frac{9}{\sqrt{30}\sqrt{42}}\) or \(\frac{9}{6\sqrt{35}}\) | M1 | Allow one of AB, CB reversed — but award A0 |
| \(ABC=75.3°\) or \(1.31\) rads (ignore reflex angle \(285°\)) | A1 [4] |
## Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{AB}=\mathbf{OB}-\mathbf{OA}=\begin{pmatrix}-1\\2\\p+4\end{pmatrix}$ | B1 | Ignore labels. Allow **BA** or **BC** |
| $\mathbf{CB}=\mathbf{OB}-\mathbf{OC}=\begin{pmatrix}-4\\5\\p-2\end{pmatrix}$ | B1 | |
| $1+4+(p+4)^2=16+25+(p-2)^2$ | M1 | |
| $p=2$ | A1 [4] | |
## Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{AB}\cdot\mathbf{CB}=4+10-5=9$ | M1 | Use of $x_1x_2+y_1y_2+z_1z_2$ |
| $|\mathbf{AB}|=\sqrt{1+4+25}=\sqrt{30},\ |\mathbf{CB}|=\sqrt{16+25+1}=\sqrt{42}$ | M1 | Product of moduli |
| $\cos ABC=\frac{9}{\sqrt{30}\sqrt{42}}$ or $\frac{9}{6\sqrt{35}}$ | M1 | Allow one of **AB**, **CB** reversed — but award A0 |
| $ABC=75.3°$ or $1.31$ rads (ignore reflex angle $285°$) | A1 [4] | |
---9 The position vectors of $A , B$ and $C$ relative to an origin $O$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
2 \\
3 \\
- 4
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { c }
1 \\
5 \\
p
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
5 \\
0 \\
2
\end{array} \right) ,$$
where $p$ is a constant.\\
(i) Find the value of $p$ for which the lengths of $A B$ and $C B$ are equal.\\
(ii) For the case where $p = 1$, use a scalar product to find angle $A B C$.
\hfill \mbox{\textit{CAIE P1 2016 Q9 [8]}}