1MA0 Higher March 2013 Paper 1 Q26
26
Diagram NOT accurately drawn
\(OAYB\) is a quadrilateral.
\(\overrightarrow{OA} = 3\mathbf{a}\)
\(\overrightarrow{OB} = 6\mathbf{b}\)
(a) Express \(\overrightarrow{AB}\) in terms of a and b. (1)
\(X\) is the point on \(AB\) such that \(AX : XB = 1 : 2\)
and \(\overrightarrow{BY} = 5\mathbf{a} - \mathbf{b}\)
*(b) Prove that \(\overrightarrow{OX} = \frac{2}{5}\overrightarrow{OY}\) (4)
| Answer | Mark | Notes |
|---|---|---|
| \(6\mathbf{b} - 3\mathbf{a}\) | 1 | B1 for \(6\mathbf{b} - 3\mathbf{a}\) oe |
| Answer | Mark | Notes |
|---|---|---|
| 4 | M1 for \(\overrightarrow{AX} = \frac{1}{3}\overrightarrow{AB}\) or \(\frac{1}{3}\)‘\((6\mathbf{b} - 3\mathbf{a})\)’ or ft to \(2\mathbf{b} - \mathbf{a}\) M1 for \(\overrightarrow{OY} = \overrightarrow{OB} + \overrightarrow{BY} = 6\mathbf{b} + 5\mathbf{a} - \mathbf{b}\) (\(= 5\mathbf{b} + 5\mathbf{a}\)) oe M1 for \(\overrightarrow{OX} = 3\mathbf{a} + \)‘\(2\mathbf{b} - \mathbf{a}\)’\(\; = 2\mathbf{a} + 2\mathbf{b}\) oe Or \(\overrightarrow{OX} = 6\mathbf{b} - \frac{2}{3}\)‘\((6\mathbf{b} - 3\mathbf{a})\)’ (\(= 2\mathbf{a} + 2\mathbf{b}\)) oe C1 for \(\frac{2}{5}\overrightarrow{OY} = \frac{2}{5} \times 5(\mathbf{a} + \mathbf{b}) = 2(\mathbf{a} + \mathbf{b}) = \overrightarrow{OX}\) |