Question 6 - A-Level Maths

OCR FP3 2015 June — Question 6 7 marks

Exam Board	OCR
Module	FP3 (Further Pure Mathematics 3)
Year	2015
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Shortest distance between two skew lines
Difficulty	Standard +0.8 This is a standard Further Maths question on skew lines requiring the formula d = \|(a₂-a₁)·(b₁×b₂)\|/\|b₁×b₂\|. While it involves multiple steps (extracting direction vectors, computing cross product, dot product, and magnitudes), it's a direct application of a known technique without requiring geometric insight or proof. Slightly above average difficulty due to being Further Maths content with computational complexity, but routine for FP3 students.
Spec	4.04h Shortest distances: between parallel lines and between skew lines 4.04i Shortest distance: between a point and a line

6 Find the shortest distance between the lines with equations $$\frac { x - 1 } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 5 } { - 1 } \quad \text { and } \quad \frac { x - 3 } { 4 } = \frac { y - 1 } { - 2 } = \frac { z + 1 } { 3 } .$$

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
$\begin{pmatrix}2\\3\\-1\end{pmatrix} \times \begin{pmatrix}4\\-2\\3\end{pmatrix} = \begin{pmatrix}7\\-10\\-16\end{pmatrix}$	M1*	Direction vectors of lines
M1dep*	Vector product
A1	condone 1 error
$\begin{pmatrix}3\\1\\-1\end{pmatrix} - \begin{pmatrix}1\\-2\\5\end{pmatrix} = \begin{pmatrix}2\\3\\-6\end{pmatrix}$	M1	Vector between lines
A1
\(\left	\frac{\begin{pmatrix}2\\3\\-6\end{pmatrix} \cdot \begin{pmatrix}7\\-10\\-16\end{pmatrix}}{\sqrt{7^2+10^2+16^2}}\right	= \frac{80}{\sqrt{405}} = \frac{16\sqrt{5}}{9}\)
A1	or 3.98
[7]

$\begin{pmatrix}2\\3\\-1\end{pmatrix} \times \begin{pmatrix}4\\-2\\3\end{pmatrix} = \begin{pmatrix}7\\-10\\-16\end{pmatrix}$ | M1* | Direction vectors of lines
| M1dep* | Vector product
| A1 | condone 1 error
$\begin{pmatrix}3\\1\\-1\end{pmatrix} - \begin{pmatrix}1\\-2\\5\end{pmatrix} = \begin{pmatrix}2\\3\\-6\end{pmatrix}$ | M1 | Vector between lines
| A1 |
$\left|\frac{\begin{pmatrix}2\\3\\-6\end{pmatrix} \cdot \begin{pmatrix}7\\-10\\-16\end{pmatrix}}{\sqrt{7^2+10^2+16^2}}\right| = \frac{80}{\sqrt{405}} = \frac{16\sqrt{5}}{9}$ | M1 | Component of their vector in their direction
| A1 | or 3.98
| [7] |

Show LaTeX source

6 Find the shortest distance between the lines with equations

$$\frac { x - 1 } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 5 } { - 1 } \quad \text { and } \quad \frac { x - 3 } { 4 } = \frac { y - 1 } { - 2 } = \frac { z + 1 } { 3 } .$$

\hfill \mbox{\textit{OCR FP3 2015 Q6 [7]}}

This paper (8 questions)

View full paper

Q1 8 Q2 8 Q3 11 Q4 9 Q5 8 Q6 7 Q7 9 Q8 12

Answer	Marks	Guidance
\(\begin{pmatrix}2\\3\\-1\end{pmatrix} \times \begin{pmatrix}4\\-2\\3\end{pmatrix} = \begin{pmatrix}7\\-10\\-16\end{pmatrix}\)	M1*	Direction vectors of lines
M1dep*	Vector product
A1	condone 1 error
\(\begin{pmatrix}3\\1\\-1\end{pmatrix} - \begin{pmatrix}1\\-2\\5\end{pmatrix} = \begin{pmatrix}2\\3\\-6\end{pmatrix}\)	M1	Vector between lines
A1
\(\left	\frac{\begin{pmatrix}2\\3\\-6\end{pmatrix} \cdot \begin{pmatrix}7\\-10\\-16\end{pmatrix}}{\sqrt{7^2+10^2+16^2}}\right	= \frac{80}{\sqrt{405}} = \frac{16\sqrt{5}}{9}\)
A1	or 3.98
[7]