Higher November 2017 Paper 6 Q16
16 ABCD is a parallelogram.
Not to scale
\(\overrightarrow{\mathrm{BD}}\) = \(\mathbf{a}\) and \(\overrightarrow{\mathrm{AD}}\) = \(\mathbf{b}\).
F is the midpoint of BC.
G is the midpoint of DC.
AE = 3EB.
(a) Write down simplified expressions in terms of \(\mathbf{a}\) and \(\mathbf{b}\) for
(i) \(\overrightarrow{\mathrm{AB}}\), [1]
(ii) \(\overrightarrow{\mathrm{EB}}\). [1]
(b) Show that \(\overrightarrow{\mathrm{EF}}\) \(= \dfrac{1}{4}(3\mathbf{b} - \mathbf{a})\). [2]
(c) Prove that \(\overrightarrow{\mathrm{EF}}\) and \(\overrightarrow{\mathrm{AG}}\) are parallel. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(\mathbf{b} - \mathbf{a}\) | 1 | ||
| (ii) \(\dfrac{1}{4}(\mathbf{b} - \mathbf{a})\) or \(\dfrac{1}{4}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\) | 1 | FT from (a)(i) | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\overrightarrow{\mathrm{EF}}\) = \(\overrightarrow{\mathrm{EB}}\) + \(\overrightarrow{\mathrm{BF}}\) = \(\dfrac{1}{4}(\mathbf{b} - \mathbf{a}) + \dfrac{1}{2}\mathbf{b}\) leading to \(\dfrac{1}{4}(3\mathbf{b} - \mathbf{a})\) as given. | 2 | M1 for their part (a)(ii) + \(\dfrac{1}{2}\mathbf{b}\) oe | (a)(ii) must be in terms of \(\mathbf{a}\) and \(\mathbf{b}\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\overrightarrow{\mathrm{AG}}\) \(= \dfrac{3}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\) \(\overrightarrow{\mathrm{AG}}\) = 2\(\overrightarrow{\mathrm{EF}}\) oe so are parallel. | 3 | B2 for \(\overrightarrow{\mathrm{AG}}\) \(= \dfrac{3}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\) or M1 for \(\mathbf{b} + \dfrac{1}{2}\)(their part (a)(i)) oe | Allow vectors found in reverse throughout eg. \(\overrightarrow{\mathrm{GA}}\) instead of \(\overrightarrow{\mathrm{AG}}\) Condone “AG and EF are multiples of each other” Full marks dependent on both AG and EF in correct simplified forms |