CAIE P1 2017 June — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2017
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePosition vectors and magnitudes
DifficultyModerate -0.8 This is a straightforward multi-part vectors question requiring only standard techniques: finding a position vector using ratio division (routine calculation), computing magnitude (direct formula application), and checking perpendicularity via dot product (standard test). All parts are textbook exercises with no problem-solving insight needed, making it easier than average.
Spec1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors
Relative to an origin \(O\), the position vectors of points \(A\) and \(B\) are given by $$\overrightarrow{OA} = \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix} \text{ and } \overrightarrow{OB} = \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix}$$ The point \(P\) lies on \(AB\) and is such that \(\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB}\).
  1. Find the position vector of \(P\). [3]
  2. Find the distance \(OP\). [1]
  3. Determine whether \(OP\) is perpendicular to \(AB\). Justify your answer. [2]
Question 4:
AnswerMarks
4(i) 5  5  0 
uuur uuur ( uuur)      
OB−OA = AB = 4 − 1 = 3
     
     
AnswerMarks
−3 3 −6B1
5  0  5
uuur   1   
OP= 1 + 3 = 2
     
3
     
AnswerMarks Guidance
3 −6 1M1 A1 uuur
If OP not scored in (i) can score SR B1 if seen correct in (ii). Other
equivalent methods possible
AnswerMarks
Total:3
AnswerMarks Guidance
4(ii)Distance OP = 52 +22 +12 = 30 or 5.48 B1 FT
FT on theirOP from (i)
AnswerMarks
Total:1
AnswerMarks
4(iii)u u ur uu ur u u ur uu ur =( )( )
Attempt A B .O P . Can score as part of A B .O P AB OP cosθ
uuur2 uuur2 uuur2
AnswerMarks Guidance
Rare ALT: Pythagoras OP + AP =5+30= OAM1 uuur uuur uuur uuur
Allow any combination of AB. PO etc. and also if AP or PB used instead of
uuur
AB giving 2‒2 = 0 & 4‒4 = 0 respectively. Allow notation × instead of . .
AnswerMarks Guidance
(0 + 6 ‒ 6) = 0 hence perpendicular. (Accept 90º)A1 FT If result not zero then 'Not perpendicular' can score A1FT if value is 'correct'
uuur uuur
for their values of AB,OP etc. from (i).
AnswerMarks Guidance
Total:2
QuestionAnswer Marks 
Question 4:
--- 4(i) ---
4(i) |  5  5  0 
uuur uuur ( uuur)      
OB−OA = AB = 4 − 1 = 3
     
     
−3 3 −6 | B1
5  0  5
uuur   1   
OP= 1 + 3 = 2
     
3
     
3 −6 1 | M1 A1 | uuur
If OP not scored in (i) can score SR B1 if seen correct in (ii). Other
equivalent methods possible
Total: | 3
--- 4(ii) ---
4(ii) | Distance OP = 52 +22 +12 = 30 or 5.48 | B1 FT | uuur
FT on theirOP from (i)
Total: | 1
--- 4(iii) ---
4(iii) | u u ur uu ur u u ur uu ur =( )( )
Attempt A B .O P . Can score as part of A B .O P AB OP cosθ
uuur2 uuur2 uuur2
Rare ALT: Pythagoras OP + AP =5+30= OA | M1 | uuur uuur uuur uuur
Allow any combination of AB. PO etc. and also if AP or PB used instead of
uuur
AB giving 2‒2 = 0 & 4‒4 = 0 respectively. Allow notation × instead of . .
(0 + 6 ‒ 6) = 0 hence perpendicular. (Accept 90º) | A1 FT | If result not zero then 'Not perpendicular' can score A1FT if value is 'correct'
uuur uuur
for their values of AB,OP etc. from (i).
Total: | 2
Question | Answer | Marks | Guidance
Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow{OA} = \begin{pmatrix} 5 \\ 1 \\ 3 \end{pmatrix} \text{ and } \overrightarrow{OB} = \begin{pmatrix} 5 \\ 4 \\ -3 \end{pmatrix}$$

The point $P$ lies on $AB$ and is such that $\overrightarrow{AP} = \frac{3}{4}\overrightarrow{AB}$.

\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $P$. [3]
\item Find the distance $OP$. [1]
\item Determine whether $OP$ is perpendicular to $AB$. Justify your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2017 Q4 [6]}}
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