CAIE P1 2015 June — Question 4 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeAngle between two vectors
DifficultyModerate -0.3 This is a straightforward vectors question requiring standard techniques: (i) uses the dot product formula for angle between vectors (routine 3-mark calculation), and (ii) involves equating magnitudes and solving a quadratic equation. Both parts are textbook exercises with no novel insight required, making it slightly easier than average for A-level.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors
Relative to the origin \(O\), the position vectors of points \(A\) and \(B\) are given by $$\overrightarrow{OA} = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} \text{ and } \overrightarrow{OB} = \begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}.$$
  1. Find the cosine of angle \(AOB\). [3]
The position vector of \(C\) is given by \(\overrightarrow{OC} = \begin{pmatrix} k \\ -2k \\ 2k - 3 \end{pmatrix}\).
  1. Given that \(AB\) and \(OC\) have the same length, find the possible values of \(k\). [4]
Question 4:
(i) ---
4
AnswerMarks
(i) 3   6   k 
uuur uuur
     
OA=  0  ,OB=  −3 ,OC = −2k 
     
 −4   2   2k −3 
OA∙ OB = 18 – 8 = 10
Modulus of OA = 5, of OB = 7
10
Angle AOB = cos −1  aef
35
10 2
→ or
AnswerMarks
35 7M1
M1
A1
AnswerMarks
[3]Use of x 1 x 2 + y 1 y 2 + z 1 z 2
All linked with modulus
cao, (if angle given, no penalty),
correct angle implies correct cosine
AnswerMarks
(ii) 3 
 
AB = b – a = −3
 
6
 
k² + 4k² + (2k – 3)² = 9 + 9 + 36
→ 9k² −12k −45( = 0)
5
→ k= 3 or k = −
AnswerMarks
3B1
M1
DM1
A1
AnswerMarks
[4]allow for a – b
Correct use of moduli using their
AB
obtains 3 term quadratic.
cao
5 (i)
(ii)
AnswerMarks
(iii)24 = r + r + rθ
24−2r
→ θ =
r
1 24r
A = r²θ = −r2 = 12r – r². aef, ag
2 2
(A=)36−(r−6)2
Greatest value of A = 36
AnswerMarks
(r = 6) → θ = 2M1
M1A1
[3]
B1 B1
[2]
B1
B1
AnswerMarks
[2](May not use (cid:1))
Attempt at s = rθ linked with 24
and r
Uses A formula with θ as f(r). cao
cao
Ft on (ii).
cao, may use calculus or the
12r−r2
discriminant on
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Cambridge International AS/A Level – May/June 20159709 11 
Question 4:
--- 4
(i) ---
4
(i) |  3   6   k 
uuur uuur
     
OA=  0  ,OB=  −3 ,OC = −2k 
     
 −4   2   2k −3 
OA∙ OB = 18 – 8 = 10
Modulus of OA = 5, of OB = 7
10
Angle AOB = cos −1  aef
35
10 2
→ or
35 7 | M1
M1
A1
[3] | Use of x 1 x 2 + y 1 y 2 + z 1 z 2
All linked with modulus
cao, (if angle given, no penalty),
correct angle implies correct cosine
(ii) |  3 
 
AB = b – a = −3
 
6
 
k² + 4k² + (2k – 3)² = 9 + 9 + 36
→ 9k² −12k −45( = 0)
5
→ k= 3 or k = −
3 | B1
M1
DM1
A1
[4] | allow for a – b
Correct use of moduli using their
AB
obtains 3 term quadratic.
cao
5 (i)
(ii)
(iii) | 24 = r + r + rθ
24−2r
→ θ =
r
1 24r
A = r²θ = −r2 = 12r – r². aef, ag
2 2
(A=)36−(r−6)2
Greatest value of A = 36
(r = 6) → θ = 2 | M1
M1A1
[3]
B1 B1
[2]
B1
B1
[2] | (May not use (cid:1))
Attempt at s = rθ linked with 24
and r
Uses A formula with θ as f(r). cao
cao
Ft on (ii).
cao, may use calculus or the
12r−r2
discriminant on
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 11
Relative to the origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow{OA} = \begin{pmatrix} 3 \\ 0 \\ -4 \end{pmatrix} \text{ and } \overrightarrow{OB} = \begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}.$$

\begin{enumerate}[label=(\roman*)]
\item Find the cosine of angle $AOB$. [3]
\end{enumerate}

The position vector of $C$ is given by $\overrightarrow{OC} = \begin{pmatrix} k \\ -2k \\ 2k - 3 \end{pmatrix}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that $AB$ and $OC$ have the same length, find the possible values of $k$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q4 [7]}}
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