| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.8 This is a multi-part 3D vectors question requiring: (i) standard line equation from two points, (ii) finding a point on a line using perpendicularity condition (requires setting up and solving a scalar product equation with parameter), and (iii) finding a plane equation using two direction vectors (one from AB, one from cross product). Part (ii) requires problem-solving beyond routine application, and part (iii) involves multiple vector operations including cross products. More demanding than typical A-level questions but within standard Further Maths scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks |
|---|---|
| (i) State correct equation in any form, e.g. \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\) | B1 |
| Answer | Marks |
|---|---|
| (ii) EITHER: Equate a relevant scalar product to zero and form an equation in \(\lambda\) | M1 |
| OR 1: Equate derivative of \(OP^2\) (or \(OP\)) to zero and form an equation in \(\lambda\) | M1 |
| OR 2: Use Pythagoras in \(OAP\) or \(OBP\) and form an equation in \(\lambda\) | M1 |
| State a correct equation in any form | A1 |
| Solve and obtain \(\lambda = -\frac{1}{2}\) or equivalent | A1 |
| Obtain final answer \(\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}\), or equivalent | A1 |
| Answer | Marks |
|---|---|
| (iii) EITHER: State or imply \(\overrightarrow{OP}\) is a normal to the required plane | M1 |
| State normal vector \(2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}\), or equivalent | A1√ |
| Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\) | M1 |
| Obtain answer \(2x + 5y + 7z = 26\), or equivalent | A1 |
| OR 1: Find a vector normal to plane \(AOB\) and calculate its vector product with a direction vector for the line \(AB\) | M1* |
| Obtain answer \(2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}\), or equivalent | A1 |
| Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\) | M1(dep*) |
| Obtain answer \(2x + 5y + 7z = 26\), or equivalent | A1 |
| OR 2: Set up and solve simultaneous equations in \(a, b, c\) derived from zero scalar products of \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) with (i) a direction vector for line \(AB\), (ii) a normal to plane \(OAB\) | M1* |
| Obtain \(a : b : c = 2 : 5 : 7\), or equivalent | A1 |
| Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\) | M1(dep*) |
| Obtain answer \(2x + 5y + 7z = 26\), or equivalent | A1 |
| OR 3: With \(Q(x, y, z)\) on plane, use Pythagoras in \(OPQ\) to form an equation in \(x, y\) and \(z\) | M1* |
| Form a correct equation | A1√ |
| Reduce to linear form | M1(dep*) |
| Obtain answer \(2x + 5y + 7z = 26\), or equivalent | A1 |
| OR 4: Find a vector normal to plane \(AOB\) and form a 2-parameter equation with relevant vectors, e.g. \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) + \mu(8\mathbf{i} - 6\mathbf{j} + 2\mathbf{k})\) | M1* |
| State three correct equations in \(x, y, z, \lambda\) and \(\mu\) | A1 |
| Eliminate \(\lambda\) and \(\mu\) | M1(dep*) |
| Obtain answer \(2x + 5y + 7z = 26\), or equivalent | A1 |
**(i)** State correct equation in any form, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})$ | B1 |
[1 mark total]
**(ii)** EITHER: Equate a relevant scalar product to zero and form an equation in $\lambda$ | M1 |
OR 1: Equate derivative of $OP^2$ (or $OP$) to zero and form an equation in $\lambda$ | M1 |
OR 2: Use Pythagoras in $OAP$ or $OBP$ and form an equation in $\lambda$ | M1 |
State a correct equation in any form | A1 |
Solve and obtain $\lambda = -\frac{1}{2}$ or equivalent | A1 |
Obtain final answer $\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}$, or equivalent | A1 |
[4 marks total]
**(iii)** EITHER: State or imply $\overrightarrow{OP}$ is a normal to the required plane | M1 |
State normal vector $2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}$, or equivalent | A1√ |
Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$ | M1 |
Obtain answer $2x + 5y + 7z = 26$, or equivalent | A1 |
OR 1: Find a vector normal to plane $AOB$ and calculate its vector product with a direction vector for the line $AB$ | M1* |
Obtain answer $2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}$, or equivalent | A1 |
Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$ | M1(dep*) |
Obtain answer $2x + 5y + 7z = 26$, or equivalent | A1 |
OR 2: Set up and solve simultaneous equations in $a, b, c$ derived from zero scalar products of $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ with (i) a direction vector for line $AB$, (ii) a normal to plane $OAB$ | M1* |
Obtain $a : b : c = 2 : 5 : 7$, or equivalent | A1 |
Substitute coordinates of a relevant point in $2x + 5y + 7z = d$ and evaluate $d$ | M1(dep*) |
Obtain answer $2x + 5y + 7z = 26$, or equivalent | A1 |
OR 3: With $Q(x, y, z)$ on plane, use Pythagoras in $OPQ$ to form an equation in $x, y$ and $z$ | M1* |
Form a correct equation | A1√ |
Reduce to linear form | M1(dep*) |
Obtain answer $2x + 5y + 7z = 26$, or equivalent | A1 |
OR 4: Find a vector normal to plane $AOB$ and form a 2-parameter equation with relevant vectors, e.g. $\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} + \lambda(2\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) + \mu(8\mathbf{i} - 6\mathbf{j} + 2\mathbf{k})$ | M1* |
State three correct equations in $x, y, z, \lambda$ and $\mu$ | A1 |
Eliminate $\lambda$ and $\mu$ | M1(dep*) |
Obtain answer $2x + 5y + 7z = 26$, or equivalent | A1 |
[4 marks total]
7 With respect to the origin $O$, the points $A$ and $B$ have position vectors given by $\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k }$ and $\overrightarrow { O B } = 3 \mathbf { i } + 4 \mathbf { j }$. The point $P$ lies on the line $A B$ and $O P$ is perpendicular to $A B$.\\
(i) Find a vector equation for the line $A B$.\\
(ii) Find the position vector of $P$.\\
(iii) Find the equation of the plane which contains $A B$ and which is perpendicular to the plane $O A B$, giving your answer in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{CAIE P3 2010 Q7 [9]}}