CAIE P3 2008 November — Question 7 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2008
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.3 This is a standard two-part question on planes requiring (i) angle formula using normal vectors (routine dot product calculation) and (ii) finding intersection line by solving simultaneous equations and expressing parametrically. Both are textbook procedures for Further Maths students with no novel insight required, making it slightly easier than average.
Spec4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane

7 Two planes have equations \(2 x - y - 3 z = 7\) and \(x + 2 y + 2 z = 0\).
  1. Find the acute angle between the planes.
  2. Find a vector equation for their line of intersection.

AnswerMarks Guidance
(i) State or imply a correct normal vector to either plane, e.g. \(2\mathbf{i} - \mathbf{j} - 3\mathbf{k}\), or \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\)B1
Carry out correct process for evaluating the scalar product of the two normalsM1
Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the resultM1
Obtain answer \(57.7°\) (or 1.01 radians)A1 [4]
(ii) EITHER: Carry out a complete method for finding a point on the lineM1
Obtain such a point, e.g. \((2, 0, -1)\)A1
EITHER: State two correct equations for a direction vector of the line, e.g. \(2a - b - 3c = 0\) and \(a + 2b + 2c = 0\)B1
Solve for one ratio, e.g. \(a : b\)M1
Obtain \(a : b : c = 4 : -7 : 5\), or equivalentA1
State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})\)A1√
OR: Obtain a second point on the line, e.g. \((0, \frac{7}{2}, -\frac{7}{2})\)A1
Subtract position vectors to obtain a direction vectorM1
Obtain \(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}\), or equivalentA1
State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})\)A1√
OR: Attempt to calculate the vector product of two normalsM1
Obtain two correct componentsA1
Obtain \(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}\), or equivalentA1
State a correct answer, e.g. \(\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})\)A1√
OR1: Express one variable in terms of a secondM1
Obtain a correct simplified expression, e.g. \(x = \frac{14-4y}{7}\)A1
Express the first variable in terms of a thirdM1
Obtain a correct simplified expression, e.g. \(x = \frac{14+4z}{5}\)A1
Form a vector equation for the lineM1
State a correct answer, e.g. \(\mathbf{r} = \frac{2}{7}\mathbf{i} - \frac{2}{7}\mathbf{k} + \lambda(\frac{2}{7}\mathbf{i} + \frac{2}{7}\mathbf{j} + \mathbf{k})\), or equivalentA1√
OR2: Express one variable in terms of a secondM1
Obtain a correct simplified expression, e.g. \(y = \frac{14-7x}{4}\)A1
Express the third variable in terms of the secondM1
Obtain a correct simplified expression, e.g. \(z = \frac{5x-14}{4}\)A1
Form a vector equation for the lineM1
State a correct answer, e.g. \(\mathbf{r} = \frac{2}{7}\mathbf{i} - \frac{2}{7}\mathbf{k} + \lambda(\frac{2}{7}\mathbf{i} + \frac{2}{7}\mathbf{j} + \mathbf{k})\), or equivalentA1√ [6]
[The f.t. is dependent on all M marks having been obtained.]
**(i)** State or imply a correct normal vector to either plane, e.g. $2\mathbf{i} - \mathbf{j} - 3\mathbf{k}$, or $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ | B1 |

Carry out correct process for evaluating the scalar product of the two normals | M1 |

Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result | M1 |

Obtain answer $57.7°$ (or 1.01 radians) | A1 | [4]

**(ii)** **EITHER:** Carry out a complete method for finding a point on the line | M1 |

Obtain such a point, e.g. $(2, 0, -1)$ | A1 |

**EITHER:** State two correct equations for a direction vector of the line, e.g. $2a - b - 3c = 0$ and $a + 2b + 2c = 0$ | B1 |

Solve for one ratio, e.g. $a : b$ | M1 |

Obtain $a : b : c = 4 : -7 : 5$, or equivalent | A1 |

State a correct answer, e.g. $\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})$ | A1√ |

**OR:** Obtain a second point on the line, e.g. $(0, \frac{7}{2}, -\frac{7}{2})$ | A1 |

Subtract position vectors to obtain a direction vector | M1 |

Obtain $4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$, or equivalent | A1 |

State a correct answer, e.g. $\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})$ | A1√ |

**OR:** Attempt to calculate the vector product of two normals | M1 |

Obtain two correct components | A1 |

Obtain $4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$, or equivalent | A1 |

State a correct answer, e.g. $\mathbf{r} = 2\mathbf{i} - \mathbf{k} + \lambda(4\mathbf{i} - 7\mathbf{j} + 5\mathbf{k})$ | A1√ |

**OR1:** Express one variable in terms of a second | M1 |

Obtain a correct simplified expression, e.g. $x = \frac{14-4y}{7}$ | A1 |

Express the first variable in terms of a third | M1 |

Obtain a correct simplified expression, e.g. $x = \frac{14+4z}{5}$ | A1 |

Form a vector equation for the line | M1 |

State a correct answer, e.g. $\mathbf{r} = \frac{2}{7}\mathbf{i} - \frac{2}{7}\mathbf{k} + \lambda(\frac{2}{7}\mathbf{i} + \frac{2}{7}\mathbf{j} + \mathbf{k})$, or equivalent | A1√ |

**OR2:** Express one variable in terms of a second | M1 |

Obtain a correct simplified expression, e.g. $y = \frac{14-7x}{4}$ | A1 |

Express the third variable in terms of the second | M1 |

Obtain a correct simplified expression, e.g. $z = \frac{5x-14}{4}$ | A1 |

Form a vector equation for the line | M1 |

State a correct answer, e.g. $\mathbf{r} = \frac{2}{7}\mathbf{i} - \frac{2}{7}\mathbf{k} + \lambda(\frac{2}{7}\mathbf{i} + \frac{2}{7}\mathbf{j} + \mathbf{k})$, or equivalent | A1√ | [6] |

[The f.t. is dependent on all M marks having been obtained.] |
7 Two planes have equations $2 x - y - 3 z = 7$ and $x + 2 y + 2 z = 0$.\\
(i) Find the acute angle between the planes.\\
(ii) Find a vector equation for their line of intersection.

\hfill \mbox{\textit{CAIE P3 2008 Q7 [10]}}