| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line-plane intersection and related angle/perpendicularity |
| Difficulty | Standard +0.8 This is a substantial two-part 3D vectors question requiring multiple techniques: finding a line-plane intersection (standard), then determining a plane containing a given line with a specified angle constraint (requires understanding of normal vectors, dot product for angles, and solving a system with geometric constraints). The second part demands problem-solving beyond routine application, though the methods are within P3 scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Carry out a correct method for finding a vector equation for \(AB\) | M1 | |
| Obtain \(\mathbf{r} = 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} - \mathbf{k})\) or \(\mathbf{r} = \mu(2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + (1-\mu)(5\mathbf{i} - 2\mathbf{j} + \mathbf{k})\), or equivalent | A1 | |
| Substitute components in equation of \(p\) and solve for \(\lambda\) or for \(\mu\) | M1 | |
| Obtain \(\lambda = \frac{3}{2}\) or \(\mu = -\frac{1}{2}\) and final answer \(\frac{13}{2}\mathbf{i} - \frac{3}{2}\mathbf{j} + \frac{1}{2}\mathbf{k}\), or equivalent | A1 | [4] |
| (ii) Either equate scalar product of direction vector of \(AB\) and normal to \(q\) to zero or substitute for \(A\) and \(B\) in the equation of \(q\) and subtract expressions | M1* | |
| Obtain \(3 + b - c = 0\), or equivalent | A1 | |
| Using the correct method for the moduli, divide the scalar product of the normals to \(p\) and \(q\) by the product of their moduli and equate to \(\pm\frac{1}{2}\), or form horizontal equivalent | M1* | |
| Obtain correct equation in any form, e.g. \(\frac{1+b}{\sqrt{(1+b^2+c^2)}\sqrt{(1+1)}} = \pm\frac{1}{2}\) | A1 | |
| Solve simultaneous equations for \(b\) or for \(c\) | M1 (dep*) | |
| Obtain \(b = -4\) and \(c = -1\) | A1 | |
| Use a relevant point and obtain final answer \(x - 4y - z = 12\), or equivalent | A1 | [7] |
**(i)** Carry out a correct method for finding a vector equation for $AB$ | M1 |
Obtain $\mathbf{r} = 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} - \mathbf{k})$ or $\mathbf{r} = \mu(2\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) + (1-\mu)(5\mathbf{i} - 2\mathbf{j} + \mathbf{k})$, or equivalent | A1 |
Substitute components in equation of $p$ and solve for $\lambda$ or for $\mu$ | M1 |
Obtain $\lambda = \frac{3}{2}$ or $\mu = -\frac{1}{2}$ and final answer $\frac{13}{2}\mathbf{i} - \frac{3}{2}\mathbf{j} + \frac{1}{2}\mathbf{k}$, or equivalent | A1 | [4]
**(ii)** Either equate scalar product of direction vector of $AB$ and normal to $q$ to zero or substitute for $A$ and $B$ in the equation of $q$ and subtract expressions | M1* |
Obtain $3 + b - c = 0$, or equivalent | A1 |
Using the correct method for the moduli, divide the scalar product of the normals to $p$ and $q$ by the product of their moduli and equate to $\pm\frac{1}{2}$, or form horizontal equivalent | M1* |
Obtain correct equation in any form, e.g. $\frac{1+b}{\sqrt{(1+b^2+c^2)}\sqrt{(1+1)}} = \pm\frac{1}{2}$ | A1 |
Solve simultaneous equations for $b$ or for $c$ | M1 (dep*) |
Obtain $b = -4$ and $c = -1$ | A1 |
Use a relevant point and obtain final answer $x - 4y - z = 12$, or equivalent | A1 | [7]
(The f.t. is on $b$ and $c$.)The points $A$ and $B$ have position vectors $\mathbf{2i - 3j + 2k}$ and $\mathbf{5i - 2j + k}$ respectively. The plane $p$ has equation $x + y = 5$.
\begin{enumerate}[label=(\roman*)]
\item Find the position vector of the point of intersection of the line through $A$ and $B$ and the plane $p$. [4]
\item A second plane $q$ has an equation of the form $x + by + cz = d$, where $b$, $c$ and $d$ are constants. The plane $q$ contains the line $AB$, and the acute angle between the planes $p$ and $q$ is $60°$. Find the equation of $q$. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q10 [11]}}