CAIE P1 2018 November — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeAngle between two vectors
DifficultyStandard +0.3 This is a straightforward 3D vectors question requiring coordinate setup, vector arithmetic, and dot product application. While it involves multiple steps (finding positions, computing vectors, dot product, then angle), each step uses standard techniques with no conceptual challenges. The geometric setup is clearly described, making this slightly easier than average.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors

\includegraphics{figure_7} The diagram shows a solid cylinder standing on a horizontal circular base with centre \(O\) and radius \(4\) units. Points \(A\), \(B\) and \(C\) lie on the circumference of the base such that \(AB\) is a diameter and angle \(BOC = 90°\). Points \(P\), \(Q\) and \(R\) lie on the upper surface of the cylinder vertically above \(A\), \(B\) and \(C\) respectively. The height of the cylinder is \(12\) units. The mid-point of \(CR\) is \(M\) and \(N\) lies on \(BQ\) with \(BN = 4\) units. Unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are parallel to \(OB\) and \(OC\) respectively and the unit vector \(\mathbf{k}\) is vertically upwards. Evaluate \(\overrightarrow{PN} \cdot \overrightarrow{PM}\) and hence find angle \(MPN\). [7]

Question 7:
AnswerMarks
7(cid:74)(cid:74)(cid:74)(cid:71)
PN = 8i – 8kB1
(cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
AnswerMarks Guidance
PM = 4i + 4j – 6kB2,1,0 Loses 1 mark for each component incorrect
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
SC: PN = – 8i + 8k and PM = – 4i – 4j +6k scores 2/3.
(cid:74)(cid:74)(cid:74)(cid:71)(cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
AnswerMarks Guidance
PN.PM = 32 + 0 + 48 = 80M1 Evaluates x x +y y +z z for correct vectors or one or both
1 2 1 2 1 2
reversed.
AnswerMarks Guidance
PN × PM = √128 × √68 (= 16 34 )M1 Product of their moduli – may be seen in cosine rule
√128 × √68 cos MPˆN = 80M1 All linked correctly.
Angle MPˆ N = 31.0◦ awrtA1 Answer must come directly from +ve cosine ratio.
Cosine rule not accepted as a complete method. Allow 0.540c
awrt.
Note: Correct answer from incorrect vectors scores A0 (XP)
7
AnswerMarks Guidance
QuestionAnswer Marks
Question 7:
7 | (cid:74)(cid:74)(cid:74)(cid:71)
PN = 8i – 8k | B1
(cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
PM = 4i + 4j – 6k | B2,1,0 | Loses 1 mark for each component incorrect
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
SC: PN = – 8i + 8k and PM = – 4i – 4j +6k scores 2/3.
(cid:74)(cid:74)(cid:74)(cid:71)(cid:74)(cid:74)(cid:74)(cid:74)(cid:71)
PN.PM = 32 + 0 + 48 = 80 | M1 | Evaluates x x +y y +z z for correct vectors or one or both
1 2 1 2 1 2
reversed.
PN × PM = √128 × √68 (= 16 34 ) | M1 | Product of their moduli – may be seen in cosine rule
√128 × √68 cos MPˆN = 80 | M1 | All linked correctly.
Angle MPˆ N = 31.0◦ awrt | A1 | Answer must come directly from +ve cosine ratio.
Cosine rule not accepted as a complete method. Allow 0.540c
awrt.
Note: Correct answer from incorrect vectors scores A0 (XP)
7
Question | Answer | Marks | Guidance
\includegraphics{figure_7}

The diagram shows a solid cylinder standing on a horizontal circular base with centre $O$ and radius $4$ units. Points $A$, $B$ and $C$ lie on the circumference of the base such that $AB$ is a diameter and angle $BOC = 90°$. Points $P$, $Q$ and $R$ lie on the upper surface of the cylinder vertically above $A$, $B$ and $C$ respectively. The height of the cylinder is $12$ units. The mid-point of $CR$ is $M$ and $N$ lies on $BQ$ with $BN = 4$ units.

Unit vectors $\mathbf{i}$ and $\mathbf{j}$ are parallel to $OB$ and $OC$ respectively and the unit vector $\mathbf{k}$ is vertically upwards.

Evaluate $\overrightarrow{PN} \cdot \overrightarrow{PM}$ and hence find angle $MPN$. [7]

\hfill \mbox{\textit{CAIE P1 2018 Q7 [7]}}