CAIE P3 2011 June — Question 9 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2011
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.3 This is a standard two-part vectors question requiring routine techniques: (i) finding the angle between planes using the dot product of normal vectors, and (ii) finding the line of intersection by solving simultaneous equations and expressing in vector form. Both are textbook procedures 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

9 Two planes have equations \(x + 2 y - 2 z = 7\) and \(2 x + y + 3 z = 5\).
  1. Calculate the acute angle between the planes.
  2. Find a vector equation for the line of intersection of the planes.

AnswerMarks Guidance
(i) State or imply a correct normal vector to either plane, e.g. \(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) or \(2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\)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 the final answer \(79.75°\) (or \(1.39\) radians)A1 [4]
(ii) EITHER:
AnswerMarks
Carry out a method for finding a point on the lineM1
Obtain such a point, e.g. \((1, 3, 0)\)A1
EITHER: State two correct equations for the direction vector \((a, b, c)\) of the line, e.g. \(a + 2b - 2c = 0\) and \(2a + b + 3c = 0\)B1
Solve for one ratio, e.g. \(a : b\)M1
Obtain \(a : b : c = 8 : -7 : -3\), or equivalentA1
State a correct final answer, e.g. \(\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})\)A1√
OR1:
AnswerMarks
Obtain a second point on the line, e.g. \(\left(0, \frac{31}{8}, \frac{3}{8}\right)\)A1
Subtract position vectors to find a direction vectorM1
Obtain \(\mathbf{i} - \frac{7}{8}\mathbf{j} - \frac{3}{8}\mathbf{k}\), or equivalentA1
State a correct final answer, e.g. \(\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda\left(-\frac{7}{8}\mathbf{j} - \frac{3}{8}\mathbf{k}\right)\)A1√
OR2:
AnswerMarks
Attempt to calculate the vector product of two normalsM1
Obtain two correct componentsA1
Obtain \(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}\), or equivalentA1
State a correct final answer, e.g. \(\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})\)A1√
OR3:
AnswerMarks
Express one variable in terms of a secondM1
Obtain a correct simplified expression, e.g. \(x = (31 - 8y) / 7\)A1
Express the first variable in terms of a thirdM1
Obtain a correct simplified expression, e.g. \(x = (3 - 3z) / 3\)A1
Form a vector equation of the lineM1
State a correct final answer, e.g. \(\mathbf{r} = \frac{31}{8}\mathbf{i} + \frac{3}{8}\mathbf{k} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})\)A1√
OR4:
AnswerMarks Guidance
Express one variable in terms of a secondM1
Obtain a correct simplified expression, e.g. \(y = (31 - 7x) / 7\)A1
Express the third variable in terms of the secondM1
Obtain a correct simplified expression, e.g. \(z = (3 - 3x) / 8\)A1
Form a vector equation of the lineM1
State a correct final answer, e.g. \(\mathbf{r} = \frac{31}{8}\mathbf{i} + \frac{3}{8}\mathbf{k} + \lambda(-8\mathbf{i} + 7\mathbf{j} + 3\mathbf{k})\)A1√ [6]
[The f.t. is dependent on all M marks having been earned.]
**(i)** State or imply a correct normal vector to either plane, e.g. $\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$ or $2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ | 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 the final answer $79.75°$ (or $1.39$ radians) | A1 | [4]

**(ii) EITHER:**

Carry out a method for finding a point on the line | M1 |

Obtain such a point, e.g. $(1, 3, 0)$ | A1 |

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

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

Obtain $a : b : c = 8 : -7 : -3$, or equivalent | A1 |

State a correct final answer, e.g. $\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})$ | A1√ |

**OR1:**

Obtain a second point on the line, e.g. $\left(0, \frac{31}{8}, \frac{3}{8}\right)$ | A1 |

Subtract position vectors to find a direction vector | M1 |

Obtain $\mathbf{i} - \frac{7}{8}\mathbf{j} - \frac{3}{8}\mathbf{k}$, or equivalent | A1 |

State a correct final answer, e.g. $\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda\left(-\frac{7}{8}\mathbf{j} - \frac{3}{8}\mathbf{k}\right)$ | A1√ |

**OR2:**

Attempt to calculate the vector product of two normals | M1 |

Obtain two correct components | A1 |

Obtain $8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k}$, or equivalent | A1 |

State a correct final answer, e.g. $\mathbf{r} = \mathbf{i} + 3\mathbf{j} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})$ | A1√ |

**OR3:**

Express one variable in terms of a second | M1 |

Obtain a correct simplified expression, e.g. $x = (31 - 8y) / 7$ | A1 |

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

Obtain a correct simplified expression, e.g. $x = (3 - 3z) / 3$ | A1 |

Form a vector equation of the line | M1 |

State a correct final answer, e.g. $\mathbf{r} = \frac{31}{8}\mathbf{i} + \frac{3}{8}\mathbf{k} + \lambda(8\mathbf{i} - 7\mathbf{j} - 3\mathbf{k})$ | A1√ |

**OR4:**

Express one variable in terms of a second | M1 |

Obtain a correct simplified expression, e.g. $y = (31 - 7x) / 7$ | A1 |

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

Obtain a correct simplified expression, e.g. $z = (3 - 3x) / 8$ | A1 |

Form a vector equation of the line | M1 |

State a correct final answer, e.g. $\mathbf{r} = \frac{31}{8}\mathbf{i} + \frac{3}{8}\mathbf{k} + \lambda(-8\mathbf{i} + 7\mathbf{j} + 3\mathbf{k})$ | A1√ | [6]

[The f.t. is dependent on all M marks having been earned.]
9 Two planes have equations $x + 2 y - 2 z = 7$ and $2 x + y + 3 z = 5$.\\
(i) Calculate the acute angle between the planes.\\
(ii) Find a vector equation for the line of intersection of the planes.

\hfill \mbox{\textit{CAIE P3 2011 Q9 [10]}}