| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question testing routine techniques: substituting a line into a plane equation, using the dot product formula for angles, and finding a direction vector satisfying two perpendicularity conditions. Part (c) requires finding a vector perpendicular to both the line direction and plane normal (a cross product), but this is a textbook application. All parts follow standard algorithms with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 1 3 2 |
| Answer | Marks |
|---|---|
| 3 −2 7 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Substituting the expression for a |
| Answer | Marks |
|---|---|
| equation in λ | Condone coordinates |
| (b) | 2 3 |
| Answer | Marks |
|---|---|
| (φ = 90° – 85.48…° = ) awrt 4.51° | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | a.b |
| Answer | Marks |
|---|---|
| answer | a.b |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | 4 |
| Answer | Marks |
|---|---|
| 1 −19 | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Method shown or at least two |
Question 3:
3 | (a) | 1 3 2
−3 +λ 2 . −5 =4
3 −2 −3
2 + 15 – 9 + λ(6 – 10 + 6) = 4
8 + 2λ = 4 => 2λ = –4 =>λ = –2 so
1 3 −5
r = −3 +−2 2 = −7
3 −2 7 | M1
M1
A1
[3] | 1.1
1.1
1.1 | Substituting the expression for a
point on the line into the equation
of the plane
Dotting out to form and solve
equation in λ | Condone coordinates
(b) | 2 3
−5 . 2
−3 −2
soi
4+25+9 9+4+4
6−10+6 2
= = =0.07868...
38 17 646
θ = awrt 85.5…° soi
(φ = 90° – 85.48…° = ) awrt 4.51° | M1
A1
A1
[3] | 1.1
1.1
1.1 | a.b
BC. Using cosθ=
a b
Can be implied by correct final
answer | a.b
May see sinφ=
a b
Or use of cross product
or 1.49… rads
or 0.0788 rads
3 | (c) | 4
λ=1⇒r= −1
1
3 2 −16
b= 2 × −5 = 5
−2 −3 −19
4 −16
So equation of l 2 is r= −1 +µ 5 oe
1 −19 | B1
M1
A1
[3] | 3.1a
2.2a
1.1 | Method shown or at least two
terms correctly evaluated
Must be r =. Allow parameter λ.3 The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 1 \\ - 3 \\ 3 \end{array} \right) + \lambda \left( \begin{array} { r } 3 \\ 2 \\ - 2 \end{array} \right)$.\\
The plane $\Pi$ has equation $\mathbf { r } \cdot \left( \begin{array} { r } 2 \\ - 5 \\ - 3 \end{array} \right) = 4$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of the point of intersection of $l _ { 1 }$ and $\Pi$.
\item Find the acute angle between $l _ { 1 }$ and $\Pi$.\\
$A$ is the point on $l _ { 1 }$ where $\lambda = 1$.\\
$l _ { 2 }$ is the line with the following properties.
\begin{itemize}
\item $l _ { 2 }$ passes through $A$
\item $l _ { 2 }$ is perpendicular to $l _ { 1 }$
\item $l _ { 2 }$ is parallel to $\Pi$
\item Find, in vector form, the equation of $l _ { 2 }$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q3 [9]}}