AQA Further Paper 1 2019 June — Question 10 8 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeFind foot of perpendicular from point to line
DifficultyStandard +0.8 This is a standard Further Maths 3D vectors question requiring finding the foot of perpendicular from a point to a line. Part (a) involves finding the line equation, setting up a perpendicular condition (dot product = 0), and solving - methodical but multi-step. Part (b) is routine distance calculation once D is found. The 6+2 mark allocation and systematic approach place it moderately above average difficulty, but it's a well-practiced technique in Further Maths.
Spec4.04h Shortest distances: between parallel lines and between skew lines
The points \(A(5, -4, 6)\) and \(B(6, -6, 8)\) lie on the line \(L\). The point \(C\) is \((15, -5, 9)\).
  1. \(D\) is the point on \(L\) that is closest to \(C\). Find the coordinates of \(D\). [6 marks]
  2. Hence find, in exact form, the shortest distance from \(C\) to \(L\). [2 marks]
Question 10:
AnswerMarks
10(a)Obtains an equation of .
Condone one error in their
direction vector. Condon𝐿𝐿e
lack of "r =" , PI by correct
AnswerMarks Guidance
vAO1.1a M1
   
r = −4 +µ −2
   
  6     2  
−10+µ
 
v = 1−2µ
 
 −3+2µ
 
 1  −10+µ
   
0= −2 � 1−2µ
   
  2     −3+2µ 
−10+µ−2+4µ−6+4µ=0
µ=2
D=( 7, −8, 10 )
Obtains a correct equation
of . Condone lack of "r ="
AnswerMarks Guidance
, PI by correct vAO1.1b A1
Ob𝐿𝐿tains their correct general
AnswerMarks Guidance
vector from line toCAO3.1a B1F
Finds scalar product of their

AnswerMarks Guidance
v and their ABAO3.1a M1
Solves to find the correctµ
AnswerMarks Guidance
for their equation.AO1.1b A1F
Finds correct DAO3.2a A1
AnswerMarks
10(b)Obtains their components of

CD, must have their correct
magnitude, but ignore sign.
AnswerMarks Guidance
Allow one error.AO1.1a M1

 
CD= −3
 
 
 1 
CD= 74
Obtains their correct CD, in
AnswerMarks Guidance
exact form.AO1.1b A1F
Total8
QMarking Instructions AO 
Question 10:
--- 10(a) ---
10(a) | Obtains an equation of .
Condone one error in their
direction vector. Condon𝐿𝐿e
lack of "r =" , PI by correct
v | AO1.1a | M1 |  5   1 
   
r = −4 +µ −2
   
  6     2  
−10+µ
 
v = 1−2µ
 
 −3+2µ
 
 1  −10+µ
   
0= −2 � 1−2µ
   
  2     −3+2µ 
−10+µ−2+4µ−6+4µ=0
µ=2
D=( 7, −8, 10 )
Obtains a correct equation
of . Condone lack of "r ="
, PI by correct v | AO1.1b | A1
Ob𝐿𝐿tains their correct general
vector from line toC | AO3.1a | B1F
Finds scalar product of their

v and their AB | AO3.1a | M1
Solves to find the correctµ
for their equation. | AO1.1b | A1F
Finds correct D | AO3.2a | A1
--- 10(b) ---
10(b) | Obtains their components of

CD, must have their correct
magnitude, but ignore sign.
Allow one error. | AO1.1a | M1 | −8

 
CD= −3
 
 
 1 
CD= 74
Obtains their correct CD, in
exact form. | AO1.1b | A1F
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
The points $A(5, -4, 6)$ and $B(6, -6, 8)$ lie on the line $L$. The point $C$ is $(15, -5, 9)$.

\begin{enumerate}[label=(\alph*)]
\item $D$ is the point on $L$ that is closest to $C$.

Find the coordinates of $D$.
[6 marks]

\item Hence find, in exact form, the shortest distance from $C$ to $L$.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2019 Q10 [8]}}
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