OCR AS Pure 2017 Specimen — Question 4 6 marks

Exam BoardOCR
ModuleAS Pure (AS Pure Mathematics)
Year2017
SessionSpecimen
Marks6
TopicVectors Introduction & 2D
TypeVector between two points
DifficultyModerate -0.8 This is a straightforward AS-level vectors question requiring only basic operations: vector subtraction to find BC, adding vectors to find D, finding a midpoint using the standard formula, and calculating magnitude using Pythagoras. All steps are routine applications of standard formulas with no problem-solving insight required, making it easier than average.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors

The points \(A\), \(B\) and \(C\) have position vectors \(\begin{pmatrix} -2 \\ 1 \end{pmatrix}\), \(\begin{pmatrix} 2 \\ 5 \end{pmatrix}\) and \(\begin{pmatrix} 6 \\ 3 \end{pmatrix}\) respectively. \(M\) is the midpoint of \(BC\).
  1. Find the position vector of the point \(D\) such that \(\overrightarrow{BC} = \overrightarrow{AD}\). [3]
  2. Find the magnitude of \(\overrightarrow{AM}\). [3]

The points $A$, $B$ and $C$ have position vectors $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 3 \end{pmatrix}$ respectively.

$M$ is the midpoint of $BC$.

\begin{enumerate}[label=(\alph*)]
\item Find the position vector of the point $D$ such that $\overrightarrow{BC} = \overrightarrow{AD}$. [3]

\item Find the magnitude of $\overrightarrow{AM}$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR AS Pure 2017 Q4 [6]}}