4MA0 Higher January 2016 Paper 3 Q23
23
Diagram NOT accurately drawn
\(OAB\) is a triangle.
\(P\) is the point on \(OA\) such that \(OP : PA = 2 : 1\)
\(C\) is the point such that \(B\) is the midpoint of \(OC\).
\(M\) is the midpoint of \(AB\).
\(\overrightarrow{OA} = 6\mathbf{a}\) \(\overrightarrow{OB} = 4\mathbf{b}\)
Show that \(PMC\) is a straight line.
(5)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OP}\) = 4a or \(\overrightarrow{PO}\) = 4a or \(\overrightarrow{PA}\) = 2a or \(\overrightarrow{AP}\) = −2a | M1 |
| \(\overrightarrow{AB}\) = 4b − 6a oe or \(\overrightarrow{AM}\) = 2b − 3a oe or \(\overrightarrow{MB}\) = 2b − 3a oe | M1 |
| \(\overrightarrow{PM}\) = 2b − a oe or \(\overrightarrow{MC}\) = 6b − 3a oe or \(\overrightarrow{PC}\) = 8b − 4a oe | M1 |
| \(\overrightarrow{PM}\) = 2b − a oe AND \(\overrightarrow{MC}\) = 6b − 3a oe or \(\overrightarrow{PM}\) = 2b − a oe AND \(\overrightarrow{PC}\) = 8b − 4a oe or \(\overrightarrow{PC}\) = 8b − 4a oe AND \(\overrightarrow{MC}\) = 6b − 3a oe | A1 |
| Show | A1 |
| (5) |
Notes
M1: for correct use of ratio to find a relevant vector
NB: may be seen on diagram
M1: indep
correct vector equation for \(\overrightarrow{PM}\) or \(\overrightarrow{MC}\) or \(\overrightarrow{PC}\) in terms of a and/or b
A1: Completion with final reasons given using simplified vectors