Higher June 2024 Paper 2R Q25

EdexcelHigherCurrent spec6 marksVectors

25 \(OPQR\) is a parallelogram.

Parallelogram OPQR with M on PQ and N on RQ; lines MR and ON cross at Y

Diagram NOT accurately drawn

\(\overrightarrow{OP} = 2\mathbf{a} \quad\) and \(\quad \overrightarrow{OR} = 3\mathbf{b}\)

The point \(M\) lies on \(PQ\) such that \(PM = \dfrac{1}{4}PQ\)

The point \(N\) lies on \(RQ\) such that \(RN = \dfrac{4}{5}RQ\)

(a) Find, in terms of \(\mathbf{a}\) and \(\mathbf{b}\), giving your answers in simplest form

(i) \(\overrightarrow{ON}\) (1)

(ii) \(\overrightarrow{MR}\) (1)

\(MR\) and \(ON\) intersect at the point \(Y\)

Given that

\(OY = k \times ON\)

(b) use a vector method to find the value of \(k\) (4)

Mark scheme (a)

Scheme	Marks
(i) Answer: \(\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}\)	B1oe
(ii) Answer: \(\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}\)	B1oe
	(2)

Notes

B1oe: must be in simplest form

eg \(1.6\mathbf{a} + 3\mathbf{b}\) or \(\dfrac{8\mathbf{a} + 15\mathbf{b}}{5}\)

B1oe: must be in simplest form

eg \(2.25\mathbf{b} - 2\mathbf{a}\) or \(\dfrac{9\mathbf{b} - 8\mathbf{a}}{4}\)

Mark scheme (b)

Scheme	Marks
eg \(\left(\overrightarrow{OM} =\right) 2\mathbf{a} + \dfrac{3}{4}\mathbf{b}\) oe or \(\left(\overrightarrow{OY} =\right) k\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right)\) oe or \(\left(\overrightarrow{YN} =\right)(1 - k)\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right)\) oe or \(\left(\overrightarrow{OY} =\right) 2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \lambda\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\) oe or \(\left(\overrightarrow{OY} =\right) 3\mathbf{b} - \mu\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\) oe	M1ft
eg \(\left(\overrightarrow{OY} =\right) k\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right)\) oe and \(2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \lambda\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\) oe or \(\left(\overrightarrow{OY} =\right) k\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right)\) oe and \(3\mathbf{b} - \mu\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\) oe or \(\left(\overrightarrow{OM} =\right) 2\mathbf{a} + \dfrac{3}{4}\mathbf{b}\) oe and \(k\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right) - \lambda\left(\text{``}{-2\mathbf{a} + \dfrac{9}{4}\mathbf{b}}\text{''}\right)\) oe or \(\left(\overrightarrow{YN} =\right)(1 - k)\left(\text{``}{\dfrac{8}{5}\mathbf{a} + 3\mathbf{b}}\text{''}\right)\) oe and \(-\lambda\left(\text{``}{-2\mathbf{a} + \dfrac{9}{4}\mathbf{b}}\text{''}\right) + \dfrac{3}{4}(3\mathbf{b}) - \dfrac{1}{5}(2\mathbf{a})\) oe or \(\left(\overrightarrow{OY} =\right) 2\mathbf{a} + \dfrac{3}{4}\mathbf{b} + \lambda\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\) and \(3\mathbf{b} - \mu\left(\text{``}{\dfrac{9}{4}\mathbf{b} - 2\mathbf{a}}\text{''}\right)\)	M1ft
eg \(3k = \dfrac{3}{4} + \dfrac{9}{4}\left(1 - \dfrac{4}{5}k\right)\) oe or \(3k = 3 - \dfrac{9}{4}\left(\dfrac{4}{5}k\right)\) oe or \(4k = 3\left(\dfrac{5 - 4k}{5}\right) + 1\) oe or \(\lambda = 0.5\) oe or \(\mu = 0.5\) oe	M1
Question requires a complete vector method* to be awarded marks* Answer: \(\dfrac{5}{8}\)	A1oe
	(4)
	(6 marks)

Notes

M1ft: ft their answers in (a)
for a correct expression for a vector eg \(\overrightarrow{OM}\) or \(\overrightarrow{OY}\) or \(\overrightarrow{YN}\)
Students may use other variations eg \(\overrightarrow{MO}\) or \(\overrightarrow{YO}\) or \(\overrightarrow{NY}\)
For all M marks
Allow any letter for \(k\) eg \(n\), \(\lambda\)
Allow any letter for \(\lambda\) eg \(\mu\)

M1ft: for 2 independent expressions for the same vector (may be embedded in a correct equation)

M1: a correct equation for \(k\) or the correct value of \(\lambda\) or \(\mu\) (cannot assume that \(Y\) is the midpoint of \(MR\))

A1oe: dep on M2