| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Line equation in vector product form |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring understanding of two representations of a line and vector product properties. Part (a) requires recognizing that both forms represent the same line (so **c** must lie on the line and **d** must be parallel to **b**), which is conceptually non-trivial. Part (b) involves applying the scalar triple product property, which is standard once the setup is understood. The question tests deeper understanding rather than routine calculation, placing it above average difficulty but not requiring extended multi-step reasoning or novel geometric insight. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | b and d are the direction vectors of l … |
| Answer | Marks |
|---|---|
| a – c = µb or a – c = γd or (a – c) b = 0 or (a – c) d = 0 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 2.4 | or statement that one of b, d is a multiple of the other |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | (a – c) d = 0 since d, a – c parallel (from (a)) |
| Answer | Marks |
|---|---|
| since a d is perpendicular to a | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Correct answer, fully justified |
| Answer | Marks |
|---|---|
| (a d) + ( b d) – (c d) = 0 | M1 |
| But b d = 0 since b, d parallel a d = c d | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| since a d is perpendicular to a | A1 | Correct answer, fully justified |
Question 4:
4 | (a) | b and d are the direction vectors of l …
… so b = md for some scalar m or b d=0
a and c are (the position vectors of) points on l …
a – c = µb or a – c = γd or (a – c) b = 0 or (a – c) d = 0 | B1
B1
B1
B1
[4] | 1.2
2.2a
1.2
2.4 | or statement that one of b, d is a multiple of the other
or that they are parallel
or (a – c) is a multiple of (or parallel to) b (or d)
SC1 (a + b – c ) d = 0
4 | (b) | (a – c) d = 0 since d, a – c parallel (from (a))
a d = c d
a . (c d) = a . (a d) = 0
since a d is perpendicular to a | M1
A1
A1 | 1.1
1.1
2.4 | Correct answer, fully justified
ALT. Substitution for r and use of the Distributive property
of the VP:
(a d) + ( b d) – (c d) = 0 | M1
But b d = 0 since b, d parallel a d = c d | A1
a . (c d) = a . (a d) = 0
since a d is perpendicular to a | A1 | Correct answer, fully justified
[3]
ALT. Substitution for r and use of the Distributive property
of the VP:4 The equation of line $l$ can be written in either of the following vector forms.
\begin{itemize}
\item $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$, where $\lambda \in \mathbb { R }$
\item $( \mathbf { r } - \mathbf { c } ) \times \mathbf { d } = \mathbf { 0 }$
\begin{enumerate}[label=(\alph*)]
\item Write down two equations involving the vectors $\mathbf { a , b , c }$, and d, giving reasons for your answers.
\item Determine the value of $\mathbf { a } \cdot ( \mathbf { c } \times \mathbf { d } )$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2023 Q4 [7]}}