Higher June 2018 Paper 2R Q23

EdexcelHigherCurrent spec5 marksVectors

23 \(OAB\) is a triangle.

\(\overrightarrow{OA} = \mathbf{a}\) \(\overrightarrow{OB} = \mathbf{b}\)

\(C\) is the midpoint of \(OA\).
\(D\) is the point on \(AB\) such that \(AD : DB = 3 : 1\)
\(E\) is the point such that \(\overrightarrow{OB} = 2\overrightarrow{BE}\)

Using a vector method, prove that the points \(C\), \(D\) and \(E\) lie on the same straight line.

(5)

Sketch of triangle OAB with C on OA, D on AB, and OB extended to E — Mark scheme

Scheme	Marks
\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\) or \(\overrightarrow{BA} = \mathbf{a} - \mathbf{b}\)	M1
\(\overrightarrow{AD} = \dfrac{3}{4}(\mathbf{b} - \mathbf{a})\) or \(\overrightarrow{DA} = \dfrac{3}{4}(\mathbf{a} - \mathbf{b})\) or \(\overrightarrow{DB} = \dfrac{1}{4}(\mathbf{b} - \mathbf{a})\) or \(\overrightarrow{BD} = \dfrac{1}{4}(\mathbf{a} - \mathbf{b})\)	M1
\(\overrightarrow{CD} = \dfrac{1}{2}\mathbf{a} + \dfrac{3}{4}(\mathbf{b} - \mathbf{a}) \left(= \dfrac{3}{4}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\right)\) or \(\overrightarrow{DE} = \dfrac{1}{4}(\mathbf{b} - \mathbf{a}) + \dfrac{1}{2}\mathbf{b} \left(= \dfrac{3}{4}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\right)\) or \(\overrightarrow{CE} = -\dfrac{1}{2}\mathbf{a} + \mathbf{b} + \dfrac{1}{2}\mathbf{b} \left(= \dfrac{3}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\right)\)	M1
	M1
Correct simplified vectors for two of \(\overrightarrow{CD}\), \(\overrightarrow{DE}\), \(\overrightarrow{CE}\) with a correct explanation	A1
	(5)
	(5 marks)

Notes