| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a standard three-part vectors question testing collinearity (using scalar multiples), perpendicularity (dot product = 0), and equal lengths (magnitude formula). All parts use routine techniques with straightforward algebraic manipulation, making it slightly easier than average for A-level. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}2\\3\\1\end{pmatrix}\), \(\overrightarrow{AC}\begin{pmatrix}3\\p-2\\q+3\end{pmatrix}\), \(\overrightarrow{BC}\begin{pmatrix}1\\p-5\\q+2\end{pmatrix}\) | B1B1 | Any 2 of 3 relevant vectors |
| \(\rightarrow p = 6\frac{1}{2}\) and \(q = -1\frac{1}{2}\) | B1 B1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6 + 3p - 6 + q + 3 = 0\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2 = 0\) |
| \(\rightarrow q = -3p - 3\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AB^2 = 4+9+1\), \(AC^2 = 9+1+(q+3)^2\) | M1 | For attempt at either |
| \(\rightarrow (q+3)^2 = 4\) | A1 A1 | |
| \(\rightarrow q = -1\) or \(-5\) | [3] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}2\\3\\1\end{pmatrix}$, $\overrightarrow{AC}\begin{pmatrix}3\\p-2\\q+3\end{pmatrix}$, $\overrightarrow{BC}\begin{pmatrix}1\\p-5\\q+2\end{pmatrix}$ | **B1B1** | Any 2 of 3 relevant vectors |
| $\rightarrow p = 6\frac{1}{2}$ and $q = -1\frac{1}{2}$ | **B1 B1** [4] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6 + 3p - 6 + q + 3 = 0$ | **M1** | Use of $x_1x_2 + y_1y_2 + z_1z_2 = 0$ |
| $\rightarrow q = -3p - 3$ | **A1** [2] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB^2 = 4+9+1$, $AC^2 = 9+1+(q+3)^2$ | **M1** | For attempt at either |
| $\rightarrow (q+3)^2 = 4$ | **A1 A1** | |
| $\rightarrow q = -1$ or $-5$ | [3] | |
---7 Relative to an origin $O$, the position vectors of points $A , B$ and $C$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
0 \\
2 \\
- 3
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
2 \\
5 \\
- 2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
3 \\
p \\
q
\end{array} \right)$$
(i) In the case where $A B C$ is a straight line, find the values of $p$ and $q$.\\
(ii) In the case where angle $B A C$ is $90 ^ { \circ }$, express $q$ in terms of $p$.\\
(iii) In the case where $p = 3$ and the lengths of $A B$ and $A C$ are equal, find the possible values of $q$.
\hfill \mbox{\textit{CAIE P1 2015 Q7 [9]}}