OCR Further Pure Core AS 2020 November — Question 7 6 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyStandard +0.3 This is a straightforward Further Maths vector question requiring cross product to find perpendicular vector, then using the geometric property that the normal vector dotted with any point on the plane of intersection gives the same value. All steps are routine applications of standard techniques with no novel insight required, making it slightly easier than average even for Further Maths.
Spec4.04g Vector product: a x b perpendicular vector
The equations of two intersecting lines are $$\mathbf{r} = \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \quad \mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}$$ where \(a\) is a constant.
  1. Find a vector, \(\mathbf{b}\), which is perpendicular to both lines. [2]
  2. Show that \(\mathbf{b} \cdot \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} = \mathbf{b} \cdot \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}\). [2]
  3. Hence, or otherwise, find the value of \(a\). [2]
Question 7:
AnswerMarks Guidance
7(a) 2 −3
   
b= 2 × 1
   
   
1 −1
−3
 
b= −1
 
 
AnswerMarks
 8 M1
A11.1a
1.1Cross product either way round.
Allow inclusion of λ and/or µ for M1
only.
BC
or any non-zero numerical multiple.
AnswerMarks Guidance
Alternative methodM1 p, q or r could be any non-zero
number
p 2 −3
     
b= q and b. 2 =0andb. 1 =0
     
     
r 1 −1
AnswerMarks
eg p = 1 and 2q + r = –2 and q – r = 3 leading toA1
 3 
1 
b= 1
 
3 
−8
[2]
M1
p, q or r could be any non-zero
number
A1
Alternative method
Find where the two lines meet and obtain a = -6 (as
below)
−3 −12 −3 2
       
−1 . a and −1 . 0
       
       
 8   −1   8  5
⇒36−a−8 and −6+40
36 - -6 -8
= 34
= -6 + 40
AnswerMarks
So dot products are equalThis can be awarded 2 marks in part c
as long as some comment is made in
part c (such as “a=-6” or “see above”)
Correct formulation of both their dot
products using their b
AnswerMarks
Conclusion neededIf using this method in part b then the
intersection of the lines AND
formulation of the dot products needs
to happen for M1
M1
A1
[2]
AnswerMarks
(c)−3 −12 −3 2
       
−1 . a = −1 . 0 ⇒36−a−8=−6+40
       
       
 8   −1   8  5
AnswerMarks
a = –6M1ft
A1 ft1.1a
1.1Correct formation of both dot
products using their bStating a=-6 fine for 2 marks
Also allow “see above”
No marks if blank, even if a = -6 seen
in (b)
AnswerMarks Guidance
Alternative methodM1ft
–12 + 2λ = 2 – 3µ & –1 + λ = 5 – µForming x and z equations using their
=>λ = 4, µ = 2b and solving for λ and µ
a = µ – 2λ = –6A1 ft From y equation
[2]
Alternative method
Find where the two lines meet and obtain a = -6 (as
below)
−3 −12 −3 2
       
−1 . a and −1 . 0
       
       
 8   −1   8  5
⇒36−a−8 and −6+40
36 - -6 -8
= 34
= -6 + 40
So dot products are equal
This can be awarded 2 marks in part c
as long as some comment is made in
part c (such as “a=-6” or “see above”)
Correct formulation of both their dot
products using their b
Conclusion needed
If using this method in part b then the
intersection of the lines AND
formulation of the dot products needs
to happen for M1
M1ft
Question 7:
7 | (a) | 2 −3
   
b= 2 × 1
   
   
1 −1
−3
 
b= −1
 
 
 8  | M1
A1 | 1.1a
1.1 | Cross product either way round.
Allow inclusion of λ and/or µ for M1
only.
BC
or any non-zero numerical multiple.
Alternative method | M1 | p, q or r could be any non-zero
number
p 2 −3
     
b= q and b. 2 =0andb. 1 =0
     
     
r 1 −1
eg p = 1 and 2q + r = –2 and q – r = 3 leading to | A1
 3 
1 
b= 1
 
3 
−8
[2]
M1
p, q or r could be any non-zero
number
A1
Alternative method
Find where the two lines meet and obtain a = -6 (as
below)
−3 −12 −3 2
       
−1 . a and −1 . 0
       
       
 8   −1   8  5
⇒36−a−8 and −6+40
36 - -6 -8
= 34
= -6 + 40
So dot products are equal | This can be awarded 2 marks in part c
as long as some comment is made in
part c (such as “a=-6” or “see above”)
Correct formulation of both their dot
products using their b
Conclusion needed | If using this method in part b then the
intersection of the lines AND
formulation of the dot products needs
to happen for M1
M1
A1
[2]
(c) | −3 −12 −3 2
       
−1 . a = −1 . 0 ⇒36−a−8=−6+40
       
       
 8   −1   8  5
a = –6 | M1ft
A1 ft | 1.1a
1.1 | Correct formation of both dot
products using their b | Stating a=-6 fine for 2 marks
Also allow “see above”
No marks if blank, even if a = -6 seen
in (b)
Alternative method | M1ft
–12 + 2λ = 2 – 3µ & –1 + λ = 5 – µ | Forming x and z equations using their
=>λ = 4, µ = 2 | b and solving for λ and µ
a = µ – 2λ = –6 | A1 ft | From y equation
[2]
Alternative method
Find where the two lines meet and obtain a = -6 (as
below)
−3 −12 −3 2
       
−1 . a and −1 . 0
       
       
 8   −1   8  5
⇒36−a−8 and −6+40
36 - -6 -8
= 34
= -6 + 40
So dot products are equal
This can be awarded 2 marks in part c
as long as some comment is made in
part c (such as “a=-6” or “see above”)
Correct formulation of both their dot
products using their b
Conclusion needed
If using this method in part b then the
intersection of the lines AND
formulation of the dot products needs
to happen for M1
M1ft
The equations of two intersecting lines are
$$\mathbf{r} = \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \quad \mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}$$
where $a$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item Find a vector, $\mathbf{b}$, which is perpendicular to both lines. [2]

\item Show that $\mathbf{b} \cdot \begin{pmatrix} -12 \\ a \\ -1 \end{pmatrix} = \mathbf{b} \cdot \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}$. [2]

\item Hence, or otherwise, find the value of $a$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core AS 2020 Q7 [6]}}
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