4MA0 Higher June 2015 Paper 3R Q18
18
Diagram NOT accurately drawn
\(PQR\) is a triangle.
The midpoint of \(PQ\) is \(W\).
\(X\) is the point on \(QR\) such that \(QX : XR = 2 : 1\)
\(PRY\) is a straight line.
\(\overrightarrow{PW} = \mathbf{a}\) \(\overrightarrow{PR} = \mathbf{b}\)
(a) Find, in terms of a and b,
(i) \(\overrightarrow{QR}\)
(ii) \(\overrightarrow{QX}\)
(iii) \(\overrightarrow{WX}\)
(3)\(R\) is the midpoint of the straight line \(PRY\).
(b) Use a vector method to show that \(WXY\) is a straight line. (2)
| Scheme | Marks |
|---|---|
| (i) \(\mathbf{b} - 2\mathbf{a}\) | B1 |
| (ii) \(\dfrac{2}{3}\mathbf{b} - \dfrac{4}{3}\mathbf{a}\) | B1 |
| (iii) \(\dfrac{2}{3}\mathbf{b} - \dfrac{1}{3}\mathbf{a}\) | B1 |
| (3) |
Notes
B1: oe eg. \(\dfrac{2}{3}(-2\mathbf{a} + \mathbf{b})\)
Allow ft from (i)
B1: oe. eg. \(\mathbf{a} + \dfrac{2}{3}(-2\mathbf{a} + \mathbf{b})\)
Allow ft from (ii)
| Scheme | Marks |
|---|---|
| M1 | |
| shown | A1 |
| (2) |
Notes
M1: for \(\overrightarrow{WY} = -\mathbf{a} + 2\mathbf{b}\) oe or \(\overrightarrow{XY} = \dfrac{2}{3}(-\mathbf{a} + 2\mathbf{b})\) oe
Allow ft from (a)
A1: for conclusion using correct vectors
eg. \(\overrightarrow{WY} = 2\mathbf{b} - \mathbf{a}\) \(\overrightarrow{XY} = \dfrac{2}{3}(-\mathbf{a} + 2\mathbf{b})\)
\(\overrightarrow{XY} = \dfrac{2}{3}\overrightarrow{WY}\)