| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard three-part vector line question requiring equating components for intersection (yielding three equations in two parameters), using the dot product for perpendicularity, then solving simultaneous equations. While it involves multiple steps, each technique is routine for A-level Further Maths students and follows a predictable structure with no novel insight required. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Express general point of \(l\) or \(m\) in component form, i.e. \((3-\lambda, -2+2\lambda, 1+\lambda)\) or \((4+a\mu, 4+b\mu, 2-\mu)\) | B1 | |
| Equate components and eliminate either \(\lambda\) or \(\mu\) from a pair of equations | M1 | |
| Eliminate the other parameter and obtain an equation in \(a\) and \(b\) | M1 | |
| Obtain the given answer | A1 | [4] |
| (ii) Using the correct process equate the scalar product of the direction vectors to zero | M1* | |
| Obtain \(-a + 2b - 1 = 0\), or equivalent | A1 | |
| Solve simultaneous equations for \(a\) or for \(b\) | M1(dep*) | |
| Obtain \(a = 3, b = 2\) | A1 | [4] |
| (iii) Substitute found values in component equations and solve for \(\lambda\) or for \(\mu\) | M1 | |
| Obtain answer \(i + 2j + 3k\) from either \(\lambda = 2\) or from \(\mu = -1\) | A1 | [2] |
**(i)** Express general point of $l$ or $m$ in component form, i.e. $(3-\lambda, -2+2\lambda, 1+\lambda)$ or $(4+a\mu, 4+b\mu, 2-\mu)$ | B1 |
Equate components and eliminate either $\lambda$ or $\mu$ from a pair of equations | M1 |
Eliminate the other parameter and obtain an equation in $a$ and $b$ | M1 |
Obtain the given answer | A1 | [4]
**(ii)** Using the correct process equate the scalar product of the direction vectors to zero | M1* |
Obtain $-a + 2b - 1 = 0$, or equivalent | A1 |
Solve simultaneous equations for $a$ or for $b$ | M1(dep*) |
Obtain $a = 3, b = 2$ | A1 | [4]
**(iii)** Substitute found values in component equations and solve for $\lambda$ or for $\mu$ | M1 |
Obtain answer $i + 2j + 3k$ from either $\lambda = 2$ or from $\mu = -1$ | A1 | [2]
---9 The lines $l$ and $m$ have equations $\mathbf { r } = 3 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } + \lambda ( - \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$ and $\mathbf { r } = 4 \mathbf { i } + 4 \mathbf { j } + 2 \mathbf { k } + \mu ( a \mathbf { i } + b \mathbf { j } - \mathbf { k } )$ respectively, where $a$ and $b$ are constants.\\
(i) Given that $l$ and $m$ intersect, show that
$$2 a - b = 4 .$$
(ii) Given also that $l$ and $m$ are perpendicular, find the values of $a$ and $b$.\\
(iii) When $a$ and $b$ have these values, find the position vector of the point of intersection of $l$ and $m$.
\hfill \mbox{\textit{CAIE P3 2012 Q9 [10]}}