| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Parallel and perpendicular planes |
| Difficulty | Standard +0.3 This is a straightforward application of standard techniques: (i) uses the perpendicularity condition for planes (dot product of normals equals zero), and (ii) requires substituting the line equation into each plane equation to find intersection points, then calculating distance. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply a correct normal vector to either plane, e.g. \(3i + 2j + 4k\) or \(ai + j + k\) | B1 | |
| Equate scalar product of normals to zero and obtain an equation in \(a\), e.g. \(3a + 2 + 4 = 0\) | M1 | |
| Obtain \(a = -2\) | A1 | [3] |
| (ii) Express general point of the line in component form, e.g. \((\lambda, 1 + 2\lambda, -1 + 2\lambda)\) | B1 | |
| Either substitute components in the equation of \(p\) and solve for \(\lambda\), or substitute components and the value of \(a\) in the equation of \(q\) and solve for \(\lambda\) | M1* | |
| Obtain \(\lambda = -1\) for point \(A\) | A1 | |
| Obtain \(\lambda = 2\) for point \(B\) | A1 | |
| Carry out correct process for finding the length of \(AB\) | M1(dep*) | |
| Obtain answer \(AB = 3\) | A1 | [6] |
**(i)** State or imply a correct normal vector to either plane, e.g. $3i + 2j + 4k$ or $ai + j + k$ | B1 |
Equate scalar product of normals to zero and obtain an equation in $a$, e.g. $3a + 2 + 4 = 0$ | M1 |
Obtain $a = -2$ | A1 | [3]
**(ii)** Express general point of the line in component form, e.g. $(\lambda, 1 + 2\lambda, -1 + 2\lambda)$ | B1 |
Either substitute components in the equation of $p$ and solve for $\lambda$, or substitute components and the value of $a$ in the equation of $q$ and solve for $\lambda$ | M1* |
Obtain $\lambda = -1$ for point $A$ | A1 |
Obtain $\lambda = 2$ for point $B$ | A1 |
Carry out correct process for finding the length of $AB$ | M1(dep*) |
Obtain answer $AB = 3$ | A1 | [6]
[The second M mark is dependent on both values of $\lambda$ being found by correct methods.]9 The plane $p$ has equation $3 x + 2 y + 4 z = 13$. A second plane $q$ is perpendicular to $p$ and has equation $a x + y + z = 4$, where $a$ is a constant.\\
(i) Find the value of $a$.\\
(ii) The line with equation $\mathbf { r } = \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } )$ meets the plane $p$ at the point $A$ and the plane $q$ at the point $B$. Find the length of $A B$.
\hfill \mbox{\textit{CAIE P3 2010 Q9 [9]}}