CAIE FP1 2019 June — Question 3 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeCommon perpendicular to two skew lines
DifficultyChallenging +1.2 This is a standard Further Maths question on finding the common perpendicular to skew lines. It requires systematic application of dot product conditions (PQ perpendicular to both direction vectors) leading to simultaneous equations, but follows a well-established method taught in FP1. More challenging than typical A-level due to the 3D vector manipulation and algebraic complexity, but not requiring novel insight.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines
3 The lines \(l _ { 1 }\) and \(l _ { 2 }\) have equations \(\mathbf { r } = 6 \mathbf { i } + 2 \mathbf { j } + 7 \mathbf { k } + \lambda ( \mathbf { i } + \mathbf { j } )\) and \(\mathbf { r } = 4 \mathbf { i } + 4 \mathbf { j } + \mu ( - 6 \mathbf { j } + \mathbf { k } )\) respectively. The point \(P\) on \(l _ { 1 }\) and the point \(Q\) on \(l _ { 2 }\) are such that \(P Q\) is perpendicular to both \(l _ { 1 }\) and \(l _ { 2 }\). Find the position vectors of \(P\) and \(Q\).
Question 3:
AnswerMarks Guidance
\(\overrightarrow{OP} = \begin{pmatrix} 6+\lambda \\ 2+\lambda \\ 7 \end{pmatrix}, \overrightarrow{OQ} = \begin{pmatrix} 4 \\ 4-6\mu \\ \mu \end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}\)M1 A1 Finds \(\overrightarrow{PQ}\)
\(\begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 0\)M1 Uses dot product of \(\overrightarrow{PQ}\) with line directions is 0. Or, alternatively, \(\overrightarrow{PQ}\) is multiple of common perpendicular
\(-2\lambda - 6\mu = 0\)A1 Deduces one equation. CWO
\(\begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}\begin{pmatrix} 0 \\ -6 \\ 1 \end{pmatrix} = 0 \Rightarrow 6\lambda + 37\mu = 19\)A1 Deduces second equation. CWO
\(\lambda = -3,\ \mu = 1\)M1 A1 Solves simultaneous equations
\(\overrightarrow{OP} = \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}, \overrightarrow{OQ} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}\)A1 States \(\overrightarrow{OP}\) and \(\overrightarrow{OQ}\) 
## Question 3:

| $\overrightarrow{OP} = \begin{pmatrix} 6+\lambda \\ 2+\lambda \\ 7 \end{pmatrix}, \overrightarrow{OQ} = \begin{pmatrix} 4 \\ 4-6\mu \\ \mu \end{pmatrix} \Rightarrow \overrightarrow{PQ} = \begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}$ | M1 A1 | Finds $\overrightarrow{PQ}$ |
|---|---|---|
| $\begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 0$ | M1 | Uses dot product of $\overrightarrow{PQ}$ with line directions is 0. Or, alternatively, $\overrightarrow{PQ}$ is multiple of common perpendicular |
| $-2\lambda - 6\mu = 0$ | A1 | Deduces one equation. CWO |
| $\begin{pmatrix} -2-\lambda \\ 2-\lambda-6\mu \\ -7+\mu \end{pmatrix}\begin{pmatrix} 0 \\ -6 \\ 1 \end{pmatrix} = 0 \Rightarrow 6\lambda + 37\mu = 19$ | A1 | Deduces second equation. CWO |
| $\lambda = -3,\ \mu = 1$ | M1 A1 | Solves simultaneous equations |
| $\overrightarrow{OP} = \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}, \overrightarrow{OQ} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix}$ | A1 | States $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ |

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3 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = 6 \mathbf { i } + 2 \mathbf { j } + 7 \mathbf { k } + \lambda ( \mathbf { i } + \mathbf { j } )$ and $\mathbf { r } = 4 \mathbf { i } + 4 \mathbf { j } + \mu ( - 6 \mathbf { j } + \mathbf { k } )$ respectively. The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$. Find the position vectors of $P$ and $Q$.\\

\hfill \mbox{\textit{CAIE FP1 2019 Q3 [8]}}
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