CAIE FP1 2008 November — Question 11 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with plane
DifficultyChallenging +1.2 This is a multi-part Further Maths question requiring cross products for the normal vector, converting parametric to Cartesian form, finding perpendicular distance from a point to a plane (involving an inequality), and calculating the angle between planes. While it involves several techniques and is longer than typical A-level questions, each step follows standard procedures without requiring novel insight or particularly complex reasoning.
Spec4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point

11 The plane \(\Pi _ { 1 }\) has equation $$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \theta ( 2 \mathbf { j } - \mathbf { k } ) + \phi ( 3 \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } )$$ Find a vector normal to \(\Pi _ { 1 }\) and hence show that the equation of \(\Pi _ { 1 }\) can be written as \(2 x + 3 y + 6 z = 14\). The line \(l\) has equation $$\mathbf { r } = 3 \mathbf { i } + 8 \mathbf { j } + 2 \mathbf { k } + t ( 4 \mathbf { i } + 6 \mathbf { j } + 5 \mathbf { k } )$$ The point on \(l\) where \(t = \lambda\) is denoted by \(P\). Find the set of values of \(\lambda\) for which the perpendicular distance of \(P\) from \(\Pi _ { 1 }\) is not greater than 4 . The plane \(\Pi _ { 2 }\) contains \(l\) and the point with position vector \(\mathbf { i } + 2 \mathbf { j } + \mathbf { k }\). Find the acute angle between \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\).

AnswerMarks Guidance
\((2j - k) \times (3i + 2j - 2k) = -2i - 3j - 6j\) (oew)M1A1
\(\Pi_1: 2x + 3y + 6z = 14\) (AG)M1A1
Perpendicular distance, \(p\), of \(P\) from \(l\) in terms of 1 parameter, e.g.,
\(p = (1/7)2(3 + 4\lambda) + 3(8 + 6\lambda) + 6(2 + 5\lambda) - 14 \)
\(=4 + 8\lambda \)
\(p \leq 4 \Rightarrow -1 \leq \lambda \leq 0\)M1A1
\((3i + 8j + 2k) - (i + 2j + k) = 2i + 6j + k\)
\((2i + 6j + k) \times (4i + 6j + 5k) = 24i - 6j - 12k\)M1A1
\(\cos\alpha = \frac{(2i + 3j + 6k)(4i - j - 2k)}{7\sqrt{21}} = 1/\sqrt{21}\)M1
\(\alpha = 77.4°\)A1
$(2j - k) \times (3i + 2j - 2k) = -2i - 3j - 6j$ (oew) | M1A1 |

$\Pi_1: 2x + 3y + 6z = 14$ (AG) | M1A1 |

Perpendicular distance, $p$, of $P$ from $l$ in terms of 1 parameter, e.g., | |

$p = (1/7)|2(3 + 4\lambda) + 3(8 + 6\lambda) + 6(2 + 5\lambda) - 14|$ | M1 |

$= |4 + 8\lambda|$ | A1 |

$p \leq 4 \Rightarrow -1 \leq \lambda \leq 0$ | M1A1 |

$(3i + 8j + 2k) - (i + 2j + k) = 2i + 6j + k$ | |

$(2i + 6j + k) \times (4i + 6j + 5k) = 24i - 6j - 12k$ | M1A1 |

$\cos\alpha = \frac{(2i + 3j + 6k)(4i - j - 2k)}{7\sqrt{21}} = 1/\sqrt{21}$ | M1 |

$\alpha = 77.4°$ | A1 |
11 The plane $\Pi _ { 1 }$ has equation

$$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \theta ( 2 \mathbf { j } - \mathbf { k } ) + \phi ( 3 \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } )$$

Find a vector normal to $\Pi _ { 1 }$ and hence show that the equation of $\Pi _ { 1 }$ can be written as $2 x + 3 y + 6 z = 14$.

The line $l$ has equation

$$\mathbf { r } = 3 \mathbf { i } + 8 \mathbf { j } + 2 \mathbf { k } + t ( 4 \mathbf { i } + 6 \mathbf { j } + 5 \mathbf { k } )$$

The point on $l$ where $t = \lambda$ is denoted by $P$. Find the set of values of $\lambda$ for which the perpendicular distance of $P$ from $\Pi _ { 1 }$ is not greater than 4 .

The plane $\Pi _ { 2 }$ contains $l$ and the point with position vector $\mathbf { i } + 2 \mathbf { j } + \mathbf { k }$. Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.

\hfill \mbox{\textit{CAIE FP1 2008 Q11 [12]}}