AQA Further Paper 2 2019 June — Question 11 8 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicular distance point to line
DifficultyChallenging +1.2 This is a standard Further Maths question on finding the shortest distance between skew lines. While it requires converting between line forms, finding direction vectors, and applying the skew lines distance formula (using scalar triple product), these are well-practiced techniques at FM level. The 8 marks reflect multiple computational steps rather than conceptual difficulty. Slightly above average due to being FM content and requiring careful vector manipulation.
Spec4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line

The line \(L_1\) has equation $$\frac{x-2}{3} = \frac{y+4}{8} = \frac{4z-5}{5}$$ The line \(L_2\) has equation $$\left(\mathbf{r} - \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \mathbf{0}$$ Find the shortest distance between the two lines, giving your answer to three significant figures. [8 marks]

Question 11:
AnswerMarks
11Obtains a position vector
of a point on L .
AnswerMarks Guidance
1AO2.5 B1
       
r = −4 +µ 8 = −4 +µ 32
1        
       
5 4 5 4 5 4  5 
−2 2
   
r = 0 +λ 1
2    
   
 3  3
 2  −2  4 
     
−4 − 0 = −4
     
     
5 4  3   −7 4
12 2  91 
     
32 × 1 = −26
     
     
 5  3  −52
4
91
−4
� �.�−26�
7
− −52 43
4
=
1 3 √ 6 9 =√56.197659...=5.18 3sf
Obtains a direction vector
for L
1
AnswerMarks Guidance
ISWAO1.1b B1
Obtains a position vector
of a point on L and a
2
direction vector for L
AnswerMarks Guidance
2AO2.5 B1
Obtains a correct vector
between the two lines.
Follow through from their
AnswerMarks Guidance
position vectorsAO1.1b B1F
Calculates the vector
product of their two
direction vectors
or
calculates the scalar
product of the general
vector between the lines
with both direction
vectors to obtain a pair of
AnswerMarks Guidance
simultaneous equationsAO3.1a M1
Obtains the correct vector
product
or
obtains a correct pair of
AnswerMarks Guidance
simultaneous equationsAO1.1b A1
Uses their vector product
or the solutions to their
simultaneous equations
to calculate the shortest
distance between the two
AnswerMarks Guidance
linesAO3.1a M1
Obtains the correct
shortest distance.
Allow awrt 5.18
AnswerMarks Guidance
Accept exact answerAO1.1b A1
Total8
QMarking Instructions AO
Question 11:
11 | Obtains a position vector
of a point on L .
1 | AO2.5 | B1 |  2   3   2  12
       
r = −4 +µ 8 = −4 +µ 32
1        
       
5 4 5 4 5 4  5 
−2 2
   
r = 0 +λ 1
2    
   
 3  3
 2  −2  4 
     
−4 − 0 = −4
     
     
5 4  3   −7 4
12 2  91 
     
32 × 1 = −26
     
     
 5  3  −52
4
91
−4
� �.�−26�
7
− −52 43
4
=
1 3 √ 6 9 =√56.197659...=5.18 3sf
Obtains a direction vector
for L
1
ISW | AO1.1b | B1
Obtains a position vector
of a point on L and a
2
direction vector for L
2 | AO2.5 | B1
Obtains a correct vector
between the two lines.
Follow through from their
position vectors | AO1.1b | B1F
Calculates the vector
product of their two
direction vectors
or
calculates the scalar
product of the general
vector between the lines
with both direction
vectors to obtain a pair of
simultaneous equations | AO3.1a | M1
Obtains the correct vector
product
or
obtains a correct pair of
simultaneous equations | AO1.1b | A1
Uses their vector product
or the solutions to their
simultaneous equations
to calculate the shortest
distance between the two
lines | AO3.1a | M1
Obtains the correct
shortest distance.
Allow awrt 5.18
Accept exact answer | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
The line $L_1$ has equation
$$\frac{x-2}{3} = \frac{y+4}{8} = \frac{4z-5}{5}$$

The line $L_2$ has equation
$$\left(\mathbf{r} - \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}\right) \times \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} = \mathbf{0}$$

Find the shortest distance between the two lines, giving your answer to three significant figures.
[8 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q11 [8]}}