| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Challenging +1.2 This is a Further Maths question on vector products and scalar triple product for volume, which elevates it above standard A-level. Part (a) requires computing a cross product with unknowns (straightforward algebra), part (b)(i) applies the scalar triple product formula for tetrahedron volume (standard technique once learned), and part (b)(ii) asks for geometric interpretation (recognizing parallel planes). While it requires multiple FM techniques, each step follows standard procedures without requiring novel insight or complex problem-solving. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04f Line-plane intersection: find point4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 1 2 2p−3q |
| Answer | Marks |
|---|---|
| p = 7, q = 4 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | At least 2 entries correct |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | i | Use of 16 (a b) • c = ()7 oe |
| Answer | Marks |
|---|---|
| | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.1 | Tetrahedron formula used; allow = +7 or −7 only |
| Answer | Marks |
|---|---|
| ii | (One/Two) plane(s) … |
| Answer | Marks |
|---|---|
| (at the correct suitable distance) | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.5 | Must be clearly the plane(s) C belongs to (not OAB) |
Question 3:
3 | (a) | 1 2 2p−3q
a b = p3= 2q−2
q 2 3−2p
2 p − 3 q 2
Solving for p, q in 2 q − 2 = 6
3 − 2 p − 1 1
p = 7, q = 4 | B1
M1
A1
[3] | 3.1a
1.1
1.1 | At least 2 entries correct
Equating their a b to the RH vector and solving
attempt
No need to check for consistency of the unused
component
(b)
(b) | i | Use of 16 (a b) • c = ()7 oe
2 d
6 • e = 42 2d + 6e – 11f = 42
−11 f
| M1
A1
[2] | 1.2
1.1 | Tetrahedron formula used; allow = +7 or −7 only
CAO (up to non-zero multiples)
Allow 2 d 6 e 1 1 f 4 2 .
ii | (One/Two) plane(s) …
… parallel to the plane of OAB
(at the correct suitable distance) | M1
A1
[2] | 2.2a
2.5 | Must be clearly the plane(s) C belongs to (not OAB)
Distance not required
3 The points $A$ and $B$ have position vectors $\mathbf { a } = \mathbf { i } + \mathrm { pj } + \mathrm { q } \mathbf { k }$ and $\mathbf { b } = 2 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k }$ respectively, relative to the origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $p$ and the value of $q$ for which $\mathbf { a } \times \mathbf { b } = 2 \mathbf { i } + 6 \mathbf { j } - 1 \mathbf { 1 } \mathbf { k }$.
\item The point $C$ has coordinates ( $d , e , f$ ) and the tetrahedron $O A B C$ has volume 7.
\begin{enumerate}[label=(\roman*)]
\item Using the values of $p$ and $q$ found in part (a), find the possible relationships between $d , e$ and $f$.
\item Explain the geometrical significance of these relationships.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2023 Q3 [7]}}