| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2017 |
| Session | December |
| Marks | 7 |
| Topic | Vector Product and Surfaces |
| Type | Area of triangle using vector product |
| Difficulty | Standard +0.8 This is a Further Maths question requiring vector product calculation with large coordinates (computational care needed), simplifying to surd form, then geometric reasoning about spherical vs planar triangles. The multi-step nature, coordinate complexity, and conceptual understanding of spherical geometry place it moderately above average difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\Delta = \frac{1}{2} | \vec{AB} \times \vec{AC} | \); \(\vec{AB} \times \vec{AC} = 14 \times 96 \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}\) |
| ⟹ Area \(\Delta = 672\sqrt{21}\) | M1 A1 M1 B1 A1 [5] | 1.1a 1.1 1.1a 1.1 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\Delta = \frac{1}{2}(AB)(AC)\sin(\angle BAC) = \frac{1}{2}(14\sqrt{5})(96\sqrt{5})\frac{\sqrt{21}}{5} = 672\sqrt{21}\) | M1A1 M1 M1 A1 | |
| (ii) (a) The sphere is sufficiently large for T to be approximately flat | B1 [1] | 3.5b |
| (b) Under-estimate, since the surface of the sphere curves outwards from the plane triangle | B1 [1] | 3.5a |
**(i)** $\vec{AB} = \begin{pmatrix} 28 \\ -14 \\ 0 \end{pmatrix}$ or $14 \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$ and $\vec{AC} = \begin{pmatrix} 96 \\ 0 \\ -192 \end{pmatrix}$ or $96 \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$
Area $\Delta = \frac{1}{2} |\vec{AB} \times \vec{AC}|$; $\vec{AB} \times \vec{AC} = 14 \times 96 \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix}$
⟹ Area $\Delta = 672\sqrt{21}$ | M1 A1 M1 B1 A1 [5] | 1.1a 1.1 1.1a 1.1 1.1 | Finding any two suitable vectors; Correct; Attempted; For a correct vector product; cao from fully supported working
**Or** $\vec{AB} = \begin{pmatrix} 28 \\ -14 \\ 0 \end{pmatrix}$ or $14 \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$ and $\vec{AC} = \begin{pmatrix} 96 \\ 0 \\ -192 \end{pmatrix}$ or $96 \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$
$\cos(\angle BAC) = \frac{(b - a) \cdot (c - a)}{(AB)(AC)} = \frac{14 \times 96 \times (2 + 0 + 0)}{14\sqrt{5} \times 96\sqrt{5}} = \frac{2}{5}$
Area $\Delta = \frac{1}{2}(AB)(AC)\sin(\angle BAC) = \frac{1}{2}(14\sqrt{5})(96\sqrt{5})\frac{\sqrt{21}}{5} = 672\sqrt{21}$ | M1A1 M1 M1 A1
**(ii)** **(a)** The sphere is sufficiently large for T to be approximately flat | B1 [1] | 3.5b | Accept any reasonable suggestion
**(b)** Under-estimate, since the surface of the sphere curves outwards from the plane triangle | B1 [1] | 3.5a | Accept any valid reason7 The points $A ( 0,35,120 ) , B ( 28,21,120 )$ and $C ( 96,35 , - 72 )$ lie on the sphere $S$, with centre $O$ and radius 125 . Triangle $A B C$ is denoted by $\triangle$.
\begin{enumerate}[label=(\roman*)]
\item Find, in simplest surd form, the area of $\Delta$.
The points $A , B$ and $C$ also form a spherical triangle, $T$, on the surface of $S$. Each 'side' of $T$ is the shorter arc of the circle, centre $O$ and radius 125, which passes through two of the given vertices of $T$. In order to find an approximation to the area of the spherical triangle, $T$ is being modelled by $\Delta$.
\item (a) State the assumption being made in using this model.\\
(b) Say, giving a reason, whether the model gives an under-estimate or an over-estimate of the area of $T$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q7 [7]}}