| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring routine techniques: finding a line equation from two points, using a direction vector as a normal to find a plane equation, then finding an intersection point and distance. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks |
|---|---|
| (i) Use correct method to form a vector equation for \(AB\) | M1 |
| Obtain a correct equation, e.g. \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + \lambda(2\mathbf{i} - 2\mathbf{j} + \mathbf{k})\) or \(\mathbf{r} = 3\mathbf{i} + \mathbf{k} + \mu(2\mathbf{i} - 2\mathbf{j} + \mathbf{k})\) | A1 |
| [2] | |
| (ii) Using a direction vector for \(AB\) and a relevant point, obtain an equation for \(m\) in any form | M1 |
| Obtain answer \(2x - 2y + z = 4\), or equivalent | A1 |
| [2] | |
| (iii) Express general point of \(AB\) in component form, e.g. \((1 + 2\lambda, 2 - 2\lambda, \lambda)\) or \((3 + 2\mu, -2\mu, 1 + \mu)\) | B1♦ |
| Substitute in equation of \(m\) and solve for \(\lambda\) or for \(\mu\) | M1 |
| Obtain final answer \(\frac{2}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}\) for the position vector of \(N\), from \(\lambda = \frac{2}{3}\) or \(\mu = -\frac{1}{3}\) | A1 |
| Carry out a correct method for finding \(CN\) | M1 |
| Obtain the given answer \(\sqrt{13}\) | A1 |
| [5] |
**(i)** Use correct method to form a vector equation for $AB$ | M1 |
Obtain a correct equation, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + \lambda(2\mathbf{i} - 2\mathbf{j} + \mathbf{k})$ or $\mathbf{r} = 3\mathbf{i} + \mathbf{k} + \mu(2\mathbf{i} - 2\mathbf{j} + \mathbf{k})$ | A1 |
| [2] |
**(ii)** Using a direction vector for $AB$ and a relevant point, obtain an equation for $m$ in any form | M1 |
Obtain answer $2x - 2y + z = 4$, or equivalent | A1 |
| [2] |
**(iii)** Express general point of $AB$ in component form, e.g. $(1 + 2\lambda, 2 - 2\lambda, \lambda)$ or $(3 + 2\mu, -2\mu, 1 + \mu)$ | B1♦ |
Substitute in equation of $m$ and solve for $\lambda$ or for $\mu$ | M1 |
Obtain final answer $\frac{2}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}$ for the position vector of $N$, from $\lambda = \frac{2}{3}$ or $\mu = -\frac{1}{3}$ | A1 |
Carry out a correct method for finding $CN$ | M1 |
Obtain the given answer $\sqrt{13}$ | A1 |
| [5] |
[The f.t. is on the direction vector for $AB$.]7 The points $A , B$ and $C$ have position vectors, relative to the origin $O$, given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
1 \\
2 \\
0
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { l }
3 \\
0 \\
1
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
1 \\
1 \\
4
\end{array} \right)$$
The plane $m$ is perpendicular to $A B$ and contains the point $C$.\\
(i) Find a vector equation for the line passing through $A$ and $B$.\\
(ii) Obtain the equation of the plane $m$, giving your answer in the form $a x + b y + c z = d$.\\
(iii) The line through $A$ and $B$ intersects the plane $m$ at the point $N$. Find the position vector of $N$ and show that $C N = \sqrt { } ( 13 )$.
\hfill \mbox{\textit{CAIE P3 2015 Q7 [9]}}