CAIE P3 2010 November — Question 7 9 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2010
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePoint on line satisfying condition
DifficultyStandard +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.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles

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 \(A B\).
  1. Find a vector equation for the line \(A B\).
  2. Find the position vector of \(P\).
  3. 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 \(a x + b y + c z = d\).

AnswerMarks
(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]
AnswerMarks
(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 formA1
Solve and obtain \(\lambda = -\frac{1}{2}\) or equivalentA1
Obtain final answer \(\overrightarrow{OP} = \frac{2}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}\), or equivalentA1
[4 marks total]
AnswerMarks
(iii) EITHER: State or imply \(\overrightarrow{OP}\) is a normal to the required planeM1
State normal vector \(2\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}\), or equivalentA1√
Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\)M1
Obtain answer \(2x + 5y + 7z = 26\), or equivalentA1
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 equivalentA1
Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\)M1(dep*)
Obtain answer \(2x + 5y + 7z = 26\), or equivalentA1
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 equivalentA1
Substitute coordinates of a relevant point in \(2x + 5y + 7z = d\) and evaluate \(d\)M1(dep*)
Obtain answer \(2x + 5y + 7z = 26\), or equivalentA1
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 equationA1√
Reduce to linear formM1(dep*)
Obtain answer \(2x + 5y + 7z = 26\), or equivalentA1
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 equivalentA1
[4 marks total]
**(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]}}