4MA0 Higher January 2018 Paper 4R Q17

EdexcelHigherOld spec7 marksVectors

17 \(PQRS\) is a quadrilateral.

Diagram NOT accurately drawn

\(\overrightarrow{PQ} = 6\mathbf{a}\) \(\overrightarrow{QR} = 4\mathbf{b}\) \(\overrightarrow{RS} = 2\mathbf{c}\)

\(V\) is the midpoint of \(PQ\).
\(W\) is the midpoint of \(QR\).

\(X\) is the point on \(RS\) such that \(RX = \dfrac{3}{4}RS\).

\(U\) is the point on \(PS\) such that \(PU = \dfrac{3}{4}PS\).

(a)

(i) Find \(\overrightarrow{PS}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\)

(ii) Find \(\overrightarrow{VW}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) (2)

(b) Show, by a vector method, that \(VW\) is parallel to \(UX\). (3)

Given that \(\overrightarrow{RS} = \begin{pmatrix} 6 \\ -5 \end{pmatrix}\)

Scheme	Marks
(i) Answer: \(6\mathbf{a} + 4\mathbf{b} + 2\mathbf{c}\)	B1
(ii) Answer: \(3\mathbf{a} + 2\mathbf{b}\)	B1
	(2)

Mark scheme (b)

Scheme	Marks
Eg \((\overrightarrow{UX} =)\;-\dfrac{3}{4}(6\mathbf{a} + 4\mathbf{b} + 2\mathbf{c}) + 6\mathbf{a} + 4\mathbf{b} + 1.5\mathbf{c}\) or \((\overrightarrow{UX} =)\;-4.5\mathbf{a} - 3\mathbf{b} - 1.5\mathbf{c} + 6\mathbf{a} + 4\mathbf{b} + 1.5\mathbf{c}\) or \((\overrightarrow{UX} =)\;\dfrac{1}{4}(6\mathbf{a} + 4\mathbf{b} + 2\mathbf{c} - 2\mathbf{c})\)	M1
\((\overrightarrow{UX} =)\;1.5\mathbf{a} + \mathbf{b}\)	A1
\(\overrightarrow{UX} = \dfrac{1}{2}\overrightarrow{VW}\) oe and conclusion	A1
	(3)

M1: For a correct expression for \(\overrightarrow{UX}\)

A1: For \((\overrightarrow{UX} =)\;1.5\mathbf{a} + \mathbf{b}\)

A1: NB: A correct simplified expression for \(\overrightarrow{UX}\) and \(\overrightarrow{VW}\) must be given.