4MA0 Higher June 2017 Paper 3 Q17

EdexcelHigherOld spec6 marksVectors

Diagram NOT accurately drawn

\(PQS\) is a triangle.
\(X\) is the midpoint of \(QS\) and \(Y\) is the midpoint of \(PS\).

\(R\) is the point of intersection of \(PX\) and \(QY\).
\(V\) is a point so that \(VQXS\) is a straight line.

\(\overrightarrow{PQ} = \mathbf{a}\) \(\overrightarrow{PS} = \mathbf{b}\)

(a) Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\),

(i) \(\overrightarrow{QS}\)

(ii) \(\overrightarrow{QY}\)

(iii) \(\overrightarrow{PX}\) (3)

\(P\) has coordinates (3, 1) and \(\overrightarrow{PR} = \dfrac{2}{3}\overrightarrow{PX}\)

\(\overrightarrow{PR} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}\) and \(\overrightarrow{XV} = \begin{pmatrix} -5 \\ 4 \end{pmatrix}\)

(b) Work out the coordinates of \(V\). (3)

Mark scheme (a)

B1: for \(-\mathbf{a} + 0.5\mathbf{b}\) oe
ft from (i)

B1: for \(0.5\mathbf{a} + 0.5\mathbf{b}\) oe
(may not be simplified)
ft from (i)

Mark scheme (b)

Scheme	Marks
\(\overrightarrow{PX} = 1.5\begin{pmatrix} 4 \\ 2 \end{pmatrix}\left(= \begin{pmatrix} 6 \\ 3 \end{pmatrix}\right)\) or (7, 3) seen as coordinates for \(R\) \(\overrightarrow{PV} = 1.5\begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix}\) or \(\begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix}\) or \(\begin{pmatrix} 1 \\ 7 \end{pmatrix}\) or (\(X\)) = (3 + 1.5 × 4, 1 + 1.5 × 2) or (3 + 6, 1 + 3) or (9, 4) or \(\overrightarrow{OX} = \begin{pmatrix} 9 \\ 4 \end{pmatrix}\)	M1
\(\overrightarrow{OV} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 7 \end{pmatrix}\) or \(\begin{pmatrix} 4 \\ 8 \end{pmatrix}\) or \(V\)(“9” – 5 , “4” + 4)	M1
(4, 8)	A1
	(3)
	(6 marks)

M1: dep

A1: SC: If M0 then award
B1 for (4, \(y\)) or (\(x\), 8)