CAIE P1 2019 June — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeVector geometry in 3D shapes
DifficultyModerate -0.8 This is a straightforward 3D vectors question requiring basic position vector manipulation and a standard scalar product calculation for an angle. All steps are routine: finding vectors by addition/subtraction, then applying the cosine formula. The geometric setup is clearly described, and each part builds mechanically on the previous one with no problem-solving insight required.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication

\includegraphics{figure_6} The diagram shows a solid figure \(ABCDEF\) in which the horizontal base \(ABC\) is a triangle right-angled at \(A\). The lengths of \(AB\) and \(AC\) are 8 units and 4 units respectively and \(M\) is the mid-point of \(AB\). The point \(D\) is 7 units vertically above \(A\). Triangle \(DEF\) lies in a horizontal plane with \(DE\), \(DF\) and \(FE\) parallel to \(AB\), \(AC\) and \(CB\) respectively and \(N\) is the mid-point of \(FE\). The lengths of \(DE\) and \(DF\) are 4 units and 2 units respectively. Unit vectors \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) are parallel to \(\overrightarrow{AB}\), \(\overrightarrow{AC}\) and \(\overrightarrow{AD}\) respectively.
  1. Find \(\overrightarrow{MF}\) in terms of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\). [1]
  2. Find \(\overrightarrow{FN}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\). [1]
  3. Find \(\overrightarrow{MN}\) in terms of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\). [1]
  4. Use a scalar product to find angle \(FMN\). [4]

Question 6:

AnswerMarks Guidance
6(i)MF = ‒4i + 2j + 7k B1
1

AnswerMarks Guidance
6(ii)FN = 2i ‒ j B1
1

AnswerMarks Guidance
6(iii)MN = ‒2i + j + 7k B1
1
AnswerMarks Guidance
QuestionAnswer Marks

AnswerMarks Guidance
6(iv)MF.MN = 8 + 2 + 49 = 59 *M1
MF× MN
+/−59
cosFMN =
AnswerMarks Guidance
69× 54DM1 All linked correctly. Note 69× 54= 9 46
FMN = 14.9° or 0.259A1 Do not allow if exactly 1 vector is reversed – even if adjusted finally
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
--- 6(i) ---
6(i) | MF = ‒4i + 2j + 7k | B1
1
--- 6(ii) ---
6(ii) | FN = 2i ‒ j | B1
1
--- 6(iii) ---
6(iii) | MN = ‒2i + j + 7k | B1 | FT on their (MF + FN)
1
Question | Answer | Marks | Guidance
--- 6(iv) ---
6(iv) | MF.MN = 8 + 2 + 49 = 59 | *M1 | MF.MN or FM.NM but allow if one is reversed (implied by ‒59)
|MF| × |MN| = 42 +22 +72 × 22 +12 +72 | *DM1 | Product of modulus. At least one methodically correct
+/−59
cosFMN =
69× 54 | DM1 | All linked correctly. Note 69× 54= 9 46
FMN = 14.9° or 0.259 | A1 | Do not allow if exactly 1 vector is reversed – even if adjusted finally
4
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows a solid figure $ABCDEF$ in which the horizontal base $ABC$ is a triangle right-angled at $A$. The lengths of $AB$ and $AC$ are 8 units and 4 units respectively and $M$ is the mid-point of $AB$. The point $D$ is 7 units vertically above $A$. Triangle $DEF$ lies in a horizontal plane with $DE$, $DF$ and $FE$ parallel to $AB$, $AC$ and $CB$ respectively and $N$ is the mid-point of $FE$. The lengths of $DE$ and $DF$ are 4 units and 2 units respectively. Unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ respectively.

\begin{enumerate}[label=(\roman*)]
\item Find $\overrightarrow{MF}$ in terms of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$. [1]

\item Find $\overrightarrow{FN}$ in terms of $\mathbf{i}$ and $\mathbf{j}$. [1]

\item Find $\overrightarrow{MN}$ in terms of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$. [1]

\item Use a scalar product to find angle $FMN$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2019 Q6 [7]}}