| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing routine matrix inversion (using cofactors/adjugate method), solving a system via matrix multiplication, and finding angle between planes using dot product of normals. All techniques are algorithmic with no novel insight required, though the 3×3 inverse calculation is somewhat lengthy. Slightly above average difficulty due to being Further Maths content and multi-part nature. |
| Spec | 4.03o Inverse 3x3 matrix4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | 0 1 −1 |
| Answer | Marks | Guidance |
|---|---|---|
| −1 5 −3 | B1 | |
| [1] | 1.1 | BC |
| 4 | (b) | x 0 1 −119 |
| Answer | Marks |
|---|---|
| x=4−k, y =6+5k, z =1−3k | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1a |
| 1.1,1.1 | Allow SCB2 for correct |
| Answer | Marks |
|---|---|
| elimination | 4−k |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (c) | DR |
| Answer | Marks |
|---|---|
| ⇒θ=40.2° or 0.702 rads | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1 | soi |
| Answer | Marks |
|---|---|
| 40° or 0.70 or better | 21 |
Question 4:
4 | (a) | 0 1 −1
M−1 = 2 −8 5
−1 5 −3 | B1
[1] | 1.1 | BC
4 | (b) | x 0 1 −119
y = 2 −8 5 4
z −1 5 −3 k
x=4−k, y =6+5k, z =1−3k | M1
A2,1,0
[3] | 1.1a
1.1,1.1 | Allow SCB2 for correct
answer found using
elimination | 4−k
condone 6+5k
1−3k
4 | (c) | DR
normals are i+2j+3k and −i+j+2k
1×(−1)+2×1+3×2
cosθ=
12 +22 +32 (−1)2 +12 +22
7
=
14 6
⇒θ=40.2° or 0.702 rads | B1
M1
A1
A1
[4] | 1.2
3.1a
1.1
1.1 | soi
correct expression
40° or 0.70 or better | 21
oe eg
6
mark final answer4
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { M } ^ { - 1 }$, where $\mathbf { M } = \left( \begin{array} { r r r } 1 & 2 & 3 \\ - 1 & 1 & 2 \\ - 2 & 1 & 2 \end{array} \right)$.
\item Hence find, in terms of the constant $k$, the point of intersection of the planes
$$\begin{aligned}
x + 2 y + 3 z & = 19 \\
- x + y + 2 z & = 4 \\
- 2 x + y + 2 z & = k
\end{aligned}$$
\item In this question you must show detailed reasoning.
Find the acute angle between the planes $x + 2 y + 3 z = 19$ and $- x + y + 2 z = 4$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2019 Q4 [8]}}