| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Area of triangle using vector product |
| Difficulty | Challenging +1.2 Part (a) is a standard vector product application requiring calculation of a×b and finding its magnitude—routine for Further Maths students. Part (b) requires understanding that C lies on line AB and using the area relationship to find scalar parameters, involving more problem-solving but still following established methods. This is moderately above average difficulty due to the Further Maths context and the two-part reasoning in (b), but remains a fairly standard exercise in vector products. |
| Spec | 1.10c Magnitude and direction: of vectors4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | 12 | 10 |
| 6 | (a) | a × b = –14i + 2j + 10k |
| Use of formula Area ∆ = 1 | a × b |
| Answer | Marks |
|---|---|
| Area ∆OAB = 5 3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | A correct vector product (possibly BC) |
| Answer | Marks |
|---|---|
| (b) | (r – a) × (b – a) = 0 is the line through A and B |
| Answer | Marks | Guidance |
|---|---|---|
| Area ∆OAC = 1 | a × c | = 1 |
| Answer | Marks |
|---|---|
| = 1 | 0 + λ a × b |
| Answer | Marks |
|---|---|
| giving c = –i +3j – 2k or c = 3i + j + 4k | M1 |
| Answer | Marks |
|---|---|
| A1 | 2.2a |
| Answer | Marks |
|---|---|
| 2.1 | From this point on, work may appear |
| Answer | Marks | Guidance |
|---|---|---|
| C is on the line AB | B1 | |
| Common “base” OA means that C is either the | M1 | |
| internal or the external bisector of AB | A1 | (For half the “height”) |
| Answer | Marks |
|---|---|
| A1 | At least one must be attempted |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | A1 | |
| giving c = –i +3j – 2k or c = 3i + j + 4k | A1 | Both correct |
Question 6:
6 | 12 | 10 | 8 | 6 | 4 | 2
6 | (a) | a × b = –14i + 2j + 10k
Use of formula Area ∆ = 1 | a × b |
2
Area ∆OAB = 5 3 | B1
M1
A1
[3] | 1.1
1.1
1.1 | A correct vector product (possibly BC)
Including an attempt at a vector product
Accept alternative exact equivalents (e.g. 75)
(b) | (r – a) × (b – a) = 0 is the line through A and B
so c = a + λ(b – a) or c = (1 – λ)a + λb
Area ∆OAC = 1 | a × c | = 1 | (1 – λ)a × a + λ a × b |
2 2
= 1 | 0 + λ a × b |
2
Area ∆OAC = 1 Area ∆OAB ⇒ λ = ±1
2 2
giving c = –i +3j – 2k or c = 3i + j + 4k | M1
A1
M1
M1
A1
A1 | 2.2a
3.1a
2.1
3.1a
1.1
2.1 | From this point on, work may appear
with numerical equivalent set-out
Use of a × a = 0
Alternative method
C is on the line AB | B1
Common “base” OA means that C is either the | M1
internal or the external bisector of AB | A1 | (For half the “height”)
M1
A1 | At least one must be attempted
M1
i.e. c = 1 (a + b) or 1 (3a – b)
2 2 | A1
giving c = –i +3j – 2k or c = 3i + j + 4k | A1 | Both correct
[6]
At least one must be attempted6 The points $A$ and $B$ have position vectors $\mathbf { a } = \mathbf { i } + 2 \mathbf { j } + \mathbf { k }$ and $\mathbf { b } = - 3 \mathbf { i } + 4 \mathbf { j } - 5 \mathbf { k }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Determine the area of triangle $O A B$, giving your answer in an exact form.
The point $C$ lies on the line $( \mathbf { r } - \mathbf { a } ) \times ( \mathbf { b } - \mathbf { a } ) = \mathbf { O }$ such that the area of triangle $O A C$ is half the area of triangle $O A B$.
\item Determine the two possible position vectors of $C$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2020 Q6 [9]}}