| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 7 |
| Topic | Vector Product and Surfaces |
| Type | Perpendicular distance from point to line |
| Difficulty | Challenging +1.2 This is a Further Maths vector product question requiring proof of algebraic identities and geometric application. Part (i)(a) uses distributivity of cross product (routine), part (i)(b) requires recognizing both vectors are perpendicular to AB (standard), and part (ii) applies the area formula for triangles using cross products—a well-established technique. While it requires multiple steps and understanding of vector geometry, these are standard Further Maths methods without requiring novel insight. |
| Spec | 1.10g Problem solving with vectors: in geometry8.04a Vector product: definition, magnitude/direction, component form8.04b Vector product properties: anti-commutative and distributive8.04c Areas using vector product: triangles and parallelograms |
| Answer | Marks | Guidance |
|---|---|---|
| M1, A1 | 1.1a, 1.1 | Use of the distributive property; AG 0 must be seen |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.2 | |
| A1 | 1.2 | Must note equality |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 3.1a | Noting the vector product form for triangle area |
| M1 | 1.1 | Noting the '\(\frac{1}{2} \times\) base \(\times\) height' form for triangle area |
| A1 | 2.2a | AG Legitimately established |
| [3] |
## Part (i)(a)
**Answer:** $a \times (b - a) = a \times b - a \times a = a \times b - 0 = a \times b$
| **M1, A1** | **1.1a, 1.1** | Use of the distributive property; AG 0 must be seen |
| **[2]** | | |
## Part (i)(b)
**Answer:** $b \times (b - a) = (0) - b \times a$ by a similar use of the distributive property
$= a \times b$ by the anti-commutative property
$= a \times (b - a)$
| **M1** | **1.2** | |
| **A1** | **1.2** | Must note equality |
| **[2]** | | |
## Part (ii)
**Answer:** Area $\triangle ABC = \frac{1}{2} \mid a \times b \mid$
Area $\triangle ABC = \frac{1}{2} (AB) \cdot \mid OD \mid$
Equating the two $\Rightarrow \mid OD \mid = \frac{\mid a \times b \mid}{\mid b - a \mid}$
| **M1** | **3.1a** | Noting the vector product form for triangle area |
| **M1** | **1.1** | Noting the '$\frac{1}{2} \times$ base $\times$ height' form for triangle area |
| **A1** | **2.2a** | AG Legitimately established |
| **[3]** | | |
---5 The points $A$ and $B$ have position vectors $\mathbf { a }$ and $\mathbf { b }$ respectively, relative to a fixed origin $O$.
\begin{enumerate}[label=(\roman*)]
\item (a) Prove that $\mathbf { a } \times ( \mathbf { b } - \mathbf { a } ) = \mathbf { a } \times \mathbf { b }$.\\
(b) Determine the relationship between $\mathbf { a } \times ( \mathbf { b } - \mathbf { a } )$ and $\mathbf { b } \times ( \mathbf { b } - \mathbf { a } )$.
\item The point $D$ is on the line $A B$. $O D$ is perpendicular to $A B$. By considering the area of triangle $O A B$, show\\
that $| O D | = \frac { | \mathbf { a } \times \mathbf { b } | } { | \mathbf { b } - \mathbf { a } | }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q5 [7]}}