CAIE P1 2012 June — Question 2 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeUnit vector in given direction
DifficultyModerate -0.8 This is a straightforward two-part question testing basic vector operations: finding a unit vector (subtract vectors, find magnitude, divide) and using the dot product for perpendicularity (set dot product to zero, solve for p). Both are routine procedures with no problem-solving insight required, making it easier than average but not trivial since it requires correct execution of standard techniques.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry

Relative to an origin \(O\), the position vectors of the points \(A\), \(B\) and \(C\) are given by $$\overrightarrow{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix}.$$ Find
  1. the unit vector in the direction of \(\overrightarrow{AB}\), [3]
  2. the value of the constant \(p\) for which angle \(BOC = 90°\). [2]

\(\vec{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\), \(\vec{OB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}\), \(\vec{OC} = \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix}\)
AnswerMarks Guidance
(i) \(\vec{AB} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\); Modulus \(= \sqrt{(4 + 9 + 36)}\)B1, M1 co; Correct method for modulus
Unit Vector \(= \frac{1}{7}\begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\)A1 co for his vector \(\vec{AB}\)
(ii) \(\vec{OB} \cdot \vec{OC} = 4 + 6 - 2p = 0 \to p = 5\)M1, A1 Dot product \(= 0\); co
$\vec{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$, $\vec{OB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}$, $\vec{OC} = \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix}$

**(i)** $\vec{AB} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}$; Modulus $= \sqrt{(4 + 9 + 36)}$ | B1, M1 | co; Correct method for modulus

Unit Vector $= \frac{1}{7}\begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}$ | A1 | co for his vector $\vec{AB}$

**(ii)** $\vec{OB} \cdot \vec{OC} = 4 + 6 - 2p = 0 \to p = 5$ | M1, A1 | Dot product $= 0$; co
Relative to an origin $O$, the position vectors of the points $A$, $B$ and $C$ are given by
$$\overrightarrow{OA} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 1 \\ 3 \\ p \end{pmatrix}.$$

Find
\begin{enumerate}[label=(\roman*)]
\item the unit vector in the direction of $\overrightarrow{AB}$, [3]

\item the value of the constant $p$ for which angle $BOC = 90°$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2012 Q2 [5]}}