| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line-plane intersection and related angle/perpendicularity |
| Difficulty | Moderate -0.3 Part (i) involves direct substitution to verify a point lies on the plane and checking orthogonality with the normal vector—both routine procedures. Part (ii) requires finding a plane perpendicular to the given plane through a point, which is a standard application of the dot product condition for perpendicular normals. While this is Further Maths content, the question requires only straightforward application of learned techniques with no problem-solving insight, making it slightly easier than an average A-level question. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | Substituting x(cid:32)4, y(cid:32)1,z(cid:32)2 into LHS of equation of |
| Answer | Marks |
|---|---|
| (cid:169)2(cid:185) | E1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | (cid:167)1(cid:183) |
| Answer | Marks |
|---|---|
| = 9 | E1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| M | N |
| E | Many other solutions are |
Question 7:
7 | (i) | Substituting x(cid:32)4, y(cid:32)1,z(cid:32)2 into LHS of equation of
plane,
3(cid:117)4(cid:16)5(cid:117)1(cid:14)2
(cid:32)12(cid:16)5(cid:14)2
(cid:32)9
(cid:32)RHS
(cid:167) 3 (cid:183) (cid:167)1(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:16)5 . 1
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) 1 (cid:185) (cid:169)2(cid:185)
(cid:32)3(cid:117)1(cid:16)5(cid:117)1(cid:14)2
(cid:32)0
(cid:167)1(cid:183)
(cid:168) (cid:184)
Normal vector for plane is perpendicular to 1 , so
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)2(cid:185)
E
(cid:167)1(cid:183)
(cid:168) (cid:184)
1 is in the plane.
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)2(cid:185) | E1
M1
A1
I
E1
C
[4] | 1.1
3.1a
M
1.1
2.4 | N
E
7 | (ii) | (cid:167)1(cid:183)
(cid:168) (cid:184)
1 is a suitable normal vector for second plane,
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)2(cid:185)
[because it is perpendicular to normal vector of (cid:51)].
x(cid:14)y(cid:14)2z(cid:32)...
= 9 | E1
M1
A1
[3] | 3.1a
1.1
1.1
M | N
E | Many other solutions are
possible e.g.
(cid:167)5(cid:183)
(cid:168) (cid:184)
Show 3 is perpendicular to
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169)0(cid:185)
(cid:167) 3 (cid:183)
(cid:168) (cid:184)
(cid:16)5 using scalar product E1
(cid:168) (cid:184)
(cid:168) (cid:184)
(cid:169) 1 (cid:185)
5x(cid:14)3y(cid:32)... M1
= 23 A1The plane $\Pi$ has equation $3x - 5y + z = 9$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\Pi$ contains
\begin{itemize}
\item the point $(4,1,2)$
\end{itemize}
and
\begin{itemize}
\item the vector $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.
\end{itemize}
[4]
\item Determine the equation of a plane which is perpendicular to $\Pi$ and which passes through $(4,1,2)$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS Q7 [7]}}