CAIE FP1 2009 November — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionNovember
Marks6
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TopicVectors: Cross Product & Distances
TypeShortest distance between two skew lines
DifficultyStandard +0.8 This is a Further Maths question requiring the shortest distance formula for skew lines, involving cross products and dot products. While the formula itself is standard, students must correctly set up direction vectors, find the connecting vector, compute a cross product, and solve an equation involving θ. This requires solid vector manipulation skills and is more demanding than typical A-level pure maths questions, but remains a standard application of the skew lines formula.
Spec4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line
2 Relative to an origin \(O\), the points \(A , B , C\) have position vectors $$\mathbf { i } , \quad \mathbf { j } + \mathbf { k } , \quad \mathbf { i } + \mathbf { j } + \theta \mathbf { k }$$ respectively. The shortest distance between the lines \(A B\) and \(O C\) is \(\frac { 1 } { \sqrt { 2 } }\). Find the value of \(\theta\).
AnswerMarks Guidance
Evaluates some relevant vector product, e.g., \((i - j - k) \times (i + j + 0k) = (1 - \theta)i - (1 + \theta)j + 2k\)M1A1
\(\mathbf{p} = [(1-\theta)i - (1+\theta)j + 2k] \cdot i\sqrt{2\theta^2 + 6}\) M1
\(=(1-\theta)/\sqrt{2\theta^2 + 6} \)
Puts \(p = 1/\sqrt{2}\) to obtain a horizontal equation such as \(2\theta^2 + 6 = 2(\theta - 1)^2\)M1
\(\Rightarrow ... \Rightarrow \theta = -1\)A1 
Evaluates some relevant vector product, e.g., $(i - j - k) \times (i + j + 0k) = (1 - \theta)i - (1 + \theta)j + 2k$ | M1A1 |
$\mathbf{p} = [(1-\theta)i - (1+\theta)j + 2k] \cdot i|\sqrt{2\theta^2 + 6}$ | M1 |
$= |(1-\theta)/\sqrt{2\theta^2 + 6}|$ | A1 |
Puts $p = 1/\sqrt{2}$ to obtain a horizontal equation such as $2\theta^2 + 6 = 2(\theta - 1)^2$ | M1 |
$\Rightarrow ... \Rightarrow \theta = -1$ | A1 |
2 Relative to an origin $O$, the points $A , B , C$ have position vectors

$$\mathbf { i } , \quad \mathbf { j } + \mathbf { k } , \quad \mathbf { i } + \mathbf { j } + \theta \mathbf { k }$$

respectively. The shortest distance between the lines $A B$ and $O C$ is $\frac { 1 } { \sqrt { 2 } }$. Find the value of $\theta$.

\hfill \mbox{\textit{CAIE FP1 2009 Q2 [6]}}
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