CAIE P1 2014 November — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2014
SessionNovember
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. Students need to find vectors by adding/subtracting position vectors (routine), then apply the cosine formula with dot product (standard technique). The pyramid setup is clearly described with no geometric insight required, making this easier than average for A-level.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication

\includegraphics{figure_7} The diagram shows a pyramid \(OABCX\). The horizontal square base \(OABC\) has side 8 units and the centre of the base is \(D\). The top of the pyramid, \(X\), is vertically above \(D\) and \(XD = 10\) units. The mid-point of \(OX\) is \(M\). The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are parallel to \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) respectively and the unit vector \(\mathbf{k}\) is vertically upwards.
  1. Express the vectors \(\overrightarrow{AM}\) and \(\overrightarrow{AC}\) in terms of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\). [3]
  2. Use a scalar product to find angle \(MAC\). [4]

(i) \(AM = -6i + 2j + 5k\)
AnswerMarks Guidance
\(AC = -8i + 8j\)B2,1, B1 [3] co –1 each error; co
(ii) \(AM.AC = 48 + 16 = 64\)
\(64 = \sqrt{128}\sqrt{65}\cos\theta\)
AnswerMarks Guidance
\(\rightarrow \theta = 45.4°\)M1, M1, M1, A1 [4] Use of \(x_1y_1 + \) etc. with suitable vectors; Product of moduli. Correct link.; co
**(i)** $AM = -6i + 2j + 5k$

$AC = -8i + 8j$ | B2,1, B1 [3] | co –1 each error; co

**(ii)** $AM.AC = 48 + 16 = 64$

$64 = \sqrt{128}\sqrt{65}\cos\theta$

$\rightarrow \theta = 45.4°$ | M1, M1, M1, A1 [4] | Use of $x_1y_1 + $ etc. with suitable vectors; Product of moduli. Correct link.; co
\includegraphics{figure_7}

The diagram shows a pyramid $OABCX$. The horizontal square base $OABC$ has side 8 units and the centre of the base is $D$. The top of the pyramid, $X$, is vertically above $D$ and $XD = 10$ units. The mid-point of $OX$ is $M$. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are parallel to $\overrightarrow{OA}$ and $\overrightarrow{OC}$ respectively and the unit vector $\mathbf{k}$ is vertically upwards.

\begin{enumerate}[label=(\roman*)]
\item Express the vectors $\overrightarrow{AM}$ and $\overrightarrow{AC}$ in terms of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$. [3]
\item Use a scalar product to find angle $MAC$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2014 Q7 [7]}}