OCR FP3 2016 June — Question 2 4 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2016
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeShortest distance between two skew lines
DifficultyStandard +0.8 This is a standard Further Maths question requiring the formula for shortest distance between skew lines using the scalar triple product: d = |(**a**₂-**a**₁)·(**b**₁×**b**₂)|/|**b**₁×**b**₂|. While it involves multiple steps (cross product calculation, dot product, magnitude), it's a direct application of a known formula with straightforward arithmetic, making it moderately above average difficulty for Further Maths students but routine within FP3.
Spec4.04i Shortest distance: between a point and a line

2 Find the shortest distance between the lines \(\mathbf { r } = \left( \begin{array} { l } 2 \\ 1 \\ 0 \end{array} \right) + \lambda \left( \begin{array} { c } 1 \\ 2 \\ - 1 \end{array} \right)\) and \(\mathbf { r } = \left( \begin{array} { c } - 1 \\ 1 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } 3 \\ 0 \\ 1 \end{array} \right)\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{pmatrix}1\\2\\-1\end{pmatrix} \times \begin{pmatrix}3\\0\\1\end{pmatrix} = \begin{pmatrix}2\\-4\\-6\end{pmatrix} = -2\begin{pmatrix}-1\\2\\3\end{pmatrix}\)M1 At least 2 correct values for cross product or method shown
A1Any multiple
\(\begin{pmatrix}2\\1\\0\end{pmatrix} - \begin{pmatrix}-1\\1\\2\end{pmatrix} = \begin{pmatrix}3\\0\\-2\end{pmatrix}\)M1
Shortest distance \(= \dfrac{\left\begin{pmatrix}3\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\3\end{pmatrix}\right }{\sqrt{1^2+2^2+3^2}}\)
\(= \dfrac{9}{\sqrt{14}}\) or \(2.41\)A1
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\2\\-1\end{pmatrix} \times \begin{pmatrix}3\\0\\1\end{pmatrix} = \begin{pmatrix}2\\-4\\-6\end{pmatrix} = -2\begin{pmatrix}-1\\2\\3\end{pmatrix}$ | M1 | At least 2 correct values for cross product or method shown |
| | A1 | Any multiple |
| $\begin{pmatrix}2\\1\\0\end{pmatrix} - \begin{pmatrix}-1\\1\\2\end{pmatrix} = \begin{pmatrix}3\\0\\-2\end{pmatrix}$ | M1 | |
| Shortest distance $= \dfrac{\left|\begin{pmatrix}3\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\3\end{pmatrix}\right|}{\sqrt{1^2+2^2+3^2}}$ | M1 | |
| $= \dfrac{9}{\sqrt{14}}$ or $2.41$ | A1 | |

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2 Find the shortest distance between the lines $\mathbf { r } = \left( \begin{array} { l } 2 \\ 1 \\ 0 \end{array} \right) + \lambda \left( \begin{array} { c } 1 \\ 2 \\ - 1 \end{array} \right)$ and $\mathbf { r } = \left( \begin{array} { c } - 1 \\ 1 \\ 2 \end{array} \right) + \mu \left( \begin{array} { l } 3 \\ 0 \\ 1 \end{array} \right)$.

\hfill \mbox{\textit{OCR FP3 2016 Q2 [4]}}