AQA Further Paper 1 2021 June — Question 11 12 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2021
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeAngle between two planes
DifficultyStandard +0.8 This is a multi-part Further Maths question on 3D vector geometry requiring several standard techniques: finding angles between lines using dot product, determining a plane equation from two lines (requiring cross product for normal vector), distance from origin to plane, and angle between planes. While it involves multiple steps and various techniques, each individual component follows standard procedures taught in Further Maths with no novel insight required. The 12-mark allocation and multi-part structure make it moderately challenging but still routine for Further Maths students.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane
The line \(L_1\) has equation \(\mathbf{r} = \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}\) The line \(L_2\) has equation \(\mathbf{r} = \begin{pmatrix} 6 \\ 4 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}\)
  1. Find the acute angle between the lines \(L_1\) and \(L_2\), giving your answer to the nearest 0.1° [3 marks]
  2. The lines \(L_1\) and \(L_2\) lie in the plane \(\Pi_1\)
    1. Find the equation of \(\Pi_1\), giving your answer in the form \(\mathbf{r} \cdot \mathbf{n} = d\) [4 marks]
    2. Hence find the shortest distance of the plane \(\Pi_1\) from the origin. [1 mark]
  3. The points \(A(4, -1, -1)\), \(B(1, 5, -7)\) and \(C(3, 4, -8)\) lie in the plane \(\Pi_2\) Find the angle between the planes \(\Pi_1\) and \(\Pi_2\), giving your answer to the nearest 0.1° [4 marks]
Question 11:
AnswerMarks
11(a)Obtains scalar (or vector)
product of the direction vectors
103o
AnswerMarks Guidance
PI by seeing AWRT1.1a M1
   
3  1 =−2
   
−1  1 
Moduli of vectors are 14 and 6
Let α be angle between lines
−2 −1
cos𝛼𝛼 = =
Angle between li√n1e4s√ 6 √21
=180−α=77.4o
Divides their scalar product (or
their magnitude of vector
product) by product of the
magnitudes
103o
AnswerMarks Guidance
PI by AWRT1.1a M1
Deduces the correct angle,
AnswerMarks Guidance
correct to at least 1dp2.2a A1
Total3
QMarking Instructions AO
AnswerMarks Guidance
11(b)(i)Finds vector product of direction
vectors3.1a M1
     
n = 3 × 1 =4 0
1      
−1  1 2
2 1
   
d = 2 • 0 =8
   
3 2
1
 
r• 0 =8
 
2
Obtains correct result (or
AnswerMarks Guidance
multiple of it)1.1b A1
Takes scalar product of their
normal vector (n ) and a point
1
AnswerMarks Guidance
in the plane2.2a M1
Obtains correct result in the
AnswerMarks Guidance
correct form (or multiple of it)1.1b A1
Total4
QMarking Instructions AO
AnswerMarks Guidance
11(b)(ii)Obtains the correct distance for
their vector equation ACF2.2a B1F
8
= √5
AnswerMarks Guidance
Total1
QMarking Instructions AO
AnswerMarks
11(c)Forms two direction vectors
from the three points and
identifies the need to take the
AnswerMarks Guidance
cross product of them3.1a M1
 
     
AB= 6 =−3 −2 , AC = 5
     
−6  2  −7
 1  −1 4
     
n = −2 × 5 = 5
2      
 2  −7 3
Let β be angle between planes
1 4
   
0 • 5
   
 2   3  2
cosβ= =
5 50 10
β=50.8o
Obtains correct result (or
AnswerMarks Guidance
multiple of it)1.1b A1
Finds scalar (or vector) product
AnswerMarks Guidance
of their normal vectors1.1a M1
Obtains correct angle
129.2o
AnswerMarks Guidance
(accept )1.1b A1
Total4
Question total12
QMarking Instructions AO 
Question 11:
--- 11(a) ---
11(a) | Obtains scalar (or vector)
product of the direction vectors
103o
PI by seeing AWRT | 1.1a | M1 | 2 −2
   
3  1 =−2
   
−1  1 
Moduli of vectors are 14 and 6
Let α be angle between lines
−2 −1
cos𝛼𝛼 = =
Angle between li√n1e4s√ 6 √21
=180−α=77.4o
Divides their scalar product (or
their magnitude of vector
product) by product of the
magnitudes
103o
PI by AWRT | 1.1a | M1
Deduces the correct angle,
correct to at least 1dp | 2.2a | A1
Total | 3
Q | Marking Instructions | AO | Marks | Typical solution
--- 11(b)(i) ---
11(b)(i) | Finds vector product of direction
vectors | 3.1a | M1 | 2 −2 1
     
n = 3 × 1 =4 0
1      
−1  1 2
2 1
   
d = 2 • 0 =8
   
3 2
1
 
r• 0 =8
 
2
Obtains correct result (or
multiple of it) | 1.1b | A1
Takes scalar product of their
normal vector (n ) and a point
1
in the plane | 2.2a | M1
Obtains correct result in the
correct form (or multiple of it) | 1.1b | A1
Total | 4
Q | Marking Instructions | AO | Marks | Typical solution
--- 11(b)(ii) ---
11(b)(ii) | Obtains the correct distance for
their vector equation ACF | 2.2a | B1F | Distance to origin
8
= √5
Total | 1
Q | Marking Instructions | AO | Marks | Typical solution
--- 11(c) ---
11(c) | Forms two direction vectors
from the three points and
identifies the need to take the
cross product of them | 3.1a | M1 | −3  1  −1
 
     
AB= 6 =−3 −2 , AC = 5
     
−6  2  −7
 1  −1 4
     
n = −2 × 5 = 5
2      
 2  −7 3
Let β be angle between planes
1 4
   
0 • 5
   
 2   3  2
cosβ= =
5 50 10
β=50.8o
Obtains correct result (or
multiple of it) | 1.1b | A1
Finds scalar (or vector) product
of their normal vectors | 1.1a | M1
Obtains correct angle
129.2o
(accept ) | 1.1b | A1
Total | 4
Question total | 12
Q | Marking Instructions | AO | Marks | Typical solution
The line $L_1$ has equation $\mathbf{r} = \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$

The line $L_2$ has equation $\mathbf{r} = \begin{pmatrix} 6 \\ 4 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$

\begin{enumerate}[label=(\alph*)]
\item Find the acute angle between the lines $L_1$ and $L_2$, giving your answer to the nearest 0.1°
[3 marks]

\item The lines $L_1$ and $L_2$ lie in the plane $\Pi_1$
\begin{enumerate}[label=(\roman*)]
\item Find the equation of $\Pi_1$, giving your answer in the form $\mathbf{r} \cdot \mathbf{n} = d$
[4 marks]

\item Hence find the shortest distance of the plane $\Pi_1$ from the origin.
[1 mark]
\end{enumerate}

\item The points $A(4, -1, -1)$, $B(1, 5, -7)$ and $C(3, 4, -8)$ lie in the plane $\Pi_2$

Find the angle between the planes $\Pi_1$ and $\Pi_2$, giving your answer to the nearest 0.1°
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2021 Q11 [12]}}
This paper (15 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 5 Q5 5 Q6 10 Q7 7 Q8 6 Q9 4 Q10 6 Q11 12 Q12 14 Q13 3 Q14 12 Q15 13