Higher January 2020 Paper 2 Q23
23
Diagram NOT accurately drawn
\(ABC\) is a triangle.
The midpoint of \(BC\) is \(M\).
\(P\) is a point on \(AM\).
\(\overrightarrow{AB} = 4\mathbf{a}\)
\(\overrightarrow{AC} = 2\mathbf{b}\)
\(\overrightarrow{AP} = \dfrac{3}{2}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\)
Find the ratio \(AP : PM\)
(3)
| Scheme | Marks |
|---|---|
\(\overrightarrow{PM} = -\dfrac{3}{2}\mathbf{a} - \dfrac{3}{4}\mathbf{b} + 4\mathbf{a} + \dfrac{1}{2}(2\mathbf{b} - 4\mathbf{a}) \left(= \dfrac{1}{2}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\right)\) \(\overrightarrow{AM} = 4\mathbf{a} + \dfrac{1}{2}(2\mathbf{b} - 4\mathbf{a})\;(= 2\mathbf{a} + \mathbf{b})\) \(\overrightarrow{AM} = 2\mathbf{b} + \dfrac{1}{2}(4\mathbf{a} - 2\mathbf{b})\;(= 2\mathbf{a} + \mathbf{b})\) \(\overrightarrow{MA} = \dfrac{1}{2}(2\mathbf{b} - 4\mathbf{a}) - 2\mathbf{b}\;(= -2\mathbf{a} - \mathbf{b})\) \(\overrightarrow{MA} = \dfrac{1}{2}(4\mathbf{a} - 2\mathbf{b}) - 4\mathbf{a}\;(= -2\mathbf{a} - \mathbf{b})\) | M1 |
\((AP : PM =)\; \left|\dfrac{3}{2}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right| : \left|\dfrac{1}{2}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\right|\) oe \((AP : AM =)\; \left|\dfrac{3}{2}\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right| : \left|2\mathbf{a} + \mathbf{b}\right|\;(= 3 : 4)\) oe \((AM : PM =)\; \left|2\mathbf{a} + \mathbf{b}\right| : \left|\dfrac{1}{2}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\right|\;(= 4 : 1)\) oe \(AP = 3PM\) oe eg \(\dfrac{3}{2}\mathbf{a} + \dfrac{3}{4}\mathbf{b} = 3\left(\dfrac{1}{2}\mathbf{a} + \dfrac{1}{4}\mathbf{b}\right)\) oe \(AM = \dfrac{4}{3}AP\) oe \(AM = 4PM\) oe | M1 |
| 3 : 1 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for finding \(\overrightarrow{PM}\) or \(\overrightarrow{AM}\) or \(\overrightarrow{MA}\)
M1: For use of a correct ratio or fraction linking
\(AP\) and \(PM\) or
\(AP\) and \(AM\) or
\(AM\) and \(PM\)
(in either order)
vectors must be in form \(p\mathbf{a} + q\mathbf{b}\)