| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Geometric interpretation of vector product |
| Difficulty | Standard +0.3 This question tests basic understanding of the geometric meaning of vector product (parallel vectors) and applies it to line equations. While it's a Further Maths topic, it requires only conceptual recall and straightforward explanation rather than calculation or problem-solving, making it easier than average. |
| Spec | 8.04a Vector product: definition, magnitude/direction, component form8.04d Significance of a x b = 0: and line equations using vector product |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | x and y are parallel |
| Answer | Marks | Guidance |
|---|---|---|
| = 0 (since x, y 0) sin = 0 = 0 (or ) and x | y |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | r = a + t d r – a = t d (r – a) | |
| Then, by (a), (r – a) d = 0 | M1 | |
| A1 | 2.1 | |
| 2.2a | (Since one vector a multiple of the other) |
| Answer | Marks |
|---|---|
| r = a + t d r – a = t d and d both sides | M1 |
| Conclusion follows from d d = 0 | A1 |
Question 3:
3 | (a) | x and y are parallel
x y = xy sin u (where u is a unit vector)
= 0 (since x, y 0) sin = 0 = 0 (or ) and x || y | B1
E1
[2] | 1.2
2.4
(b) | r = a + t d r – a = t d (r – a) || d
Then, by (a), (r – a) d = 0 | M1
A1 | 2.1
2.2a | (Since one vector a multiple of the other)
No statement required that neither vector is zero
Condone lack of -ness to explanation
Alternative method
r = a + t d r – a = t d and d both sides | M1
Conclusion follows from d d = 0 | A1
[2]3 The non-zero vectors $\mathbf { x }$ and $\mathbf { y }$ are such that $\mathbf { x } \times \mathbf { y } = \mathbf { 0 }$.
\begin{enumerate}[label=(\alph*)]
\item Explain the geometrical significance of this statement.
\item Use your answer to part (a) to explain how the line equation $\mathbf { r } = \mathbf { a } + t \mathbf { d }$ can be written in the form $( \mathbf { r } - \mathbf { a } ) \times \mathbf { d } = \mathbf { 0 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q3 [4]}}