| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Minimum distance between lines or points |
| Difficulty | Standard +0.8 This is a multi-part question on minimum distance between skew lines using partial differentiation. While the setup and differentiation are standard Further Maths techniques, it requires careful algebraic manipulation, understanding of multivariable calculus, and geometric interpretation. The question is more substantial than typical A-level work but follows a guided structure, placing it moderately above average difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05d Partial differentiation: first and second order, mixed derivatives8.05f Nature of stationary points: classify using Hessian matrix |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | 8q4p3 |
| Answer | Marks |
|---|---|
| = 81q2 + 26p2 + 50 – 90pq – 58p + 90q | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Attempted soi |
| Answer | Marks |
|---|---|
| (b) | z z |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| (c) | (Diagram may consist of two skew lines; P on one, Q on the other.) |
| Answer | Marks |
|---|---|
| so not a saddle-point | E1 |
| E1 | 3.4 |
| 3.4 | Or z-p-q (-shaped) paraboloid drawn |
| Answer | Marks |
|---|---|
| z-p-q (-shaped) paraboloid OR z-p AND z-q (-shaped) parabola drawn | E1 |
| Noting surface has a minimum for each section ( 2 shown) | E1 |
| Answer | Marks |
|---|---|
| Skew lines have a minimum distance, so z must have a minimum | E1 |
| There is only one stationary point in (b), so it must be this minimum and not | E1 |
| Answer | Marks |
|---|---|
| (d) | Substg. back p = 4, q = 5 into expression for z |
| Answer | Marks |
|---|---|
| z = 9 and Sh. Dist. = 3 (m) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 2.2a | cao |
| (e) | e.g. Because they are modelled as spheres, for any value of p and q the |
| Answer | Marks |
|---|---|
| the shortest distance is now 3 – 1 = 2 (m) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.5c |
| 1.1 | ft (d)’s answer Or statement distance is 1m less |
Question 8:
8 | (a) | 8q4p3
PQ = q – p = q p5
4q3p4
z = (PQ)2 = (8q – 4p +3)2 + (q – p + 5)2 + (4q – 3p + 4)2
= (64q2 + 16p2 + 9 – 64pq – 24p + 48q)
+ (q2 + p2 + 25 – 2pq – 10p + 10q)
+ (16q2 + 9p2 + 16 – 24pq – 24p + 32q)
= 81q2 + 26p2 + 50 – 90pq – 58p + 90q | M1
M1
A1
[3] | 3.3
1.2
1.1 | Attempted soi
Clear attempt to square at least two three-term brackets
AG from fully supported (visible) working
(b) | z z
= 52p – 90q – 58 = 162q – 90p + 90
p q
26p45q29
Setting both p.d.s to zero and solving simultaneously
45p81q45
p = 4, q = 5
3 | B1
B1
M1
A1
[4] | 1.1
1.1
3.1a
1.1 | BC
(c) | (Diagram may consist of two skew lines; P on one, Q on the other.)
Moving P, Q in “opposite” directions along their lines gives z indefinitely
large, hence stationary point is not a maximum
Symmetric properties of P, Q (i.e. p, q) gives both max or both min
so not a saddle-point | E1
E1 | 3.4
3.4 | Or z-p-q (-shaped) paraboloid drawn
Or z-p AND z-q (-shaped) parabola drawn
Alternative method I
z-p-q (-shaped) paraboloid OR z-p AND z-q (-shaped) parabola drawn | E1
Noting surface has a minimum for each section ( 2 shown) | E1
Alternative method II
Skew lines have a minimum distance, so z must have a minimum | E1
There is only one stationary point in (b), so it must be this minimum and not | E1
either a max. or a saddle point.
[2]
(d) | Substg. back p = 4, q = 5 into expression for z
3
z = 9 and Sh. Dist. = 3 (m) | M1
A1
[2] | 1.1a
2.2a | cao
(e) | e.g. Because they are modelled as spheres, for any value of p and q the
distance between them will simply be less than in the original model.
the shortest distance is now 3 – 1 = 2 (m) | M1
A1
[2] | 3.5c
1.1 | ft (d)’s answer Or statement distance is 1m less
PMT
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© OCR 20198 The motion of two remote controlled helicopters $P$ and $Q$ is modelled as two points moving along straight lines.
Helicopter $P$ moves on the line $\mathbf { r } = \left( \begin{array} { r } 2 + 4 p \\ - 3 + p \\ 1 + 3 p \end{array} \right)$ and helicopter $Q$ moves on the line $\mathbf { r } = \left( \begin{array} { l } 5 + 8 q \\ 2 + q \\ 5 + 4 q \end{array} \right)$.\\
The function $z$ denotes $( P Q ) ^ { 2 }$, the square of the distance between $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\item Show that $z = 26 p ^ { 2 } + 81 q ^ { 2 } - 90 p q - 58 p + 90 q + 50$.
\item Use partial differentiation to find the values of $p$ and $q$ for which $z$ has a stationary point.
\item With the aid of a diagram, explain why this stationary point must be a minimum point, rather than a maximum point or a saddle point.
\item Hence find the shortest possible distance between the two helicopters.
The model is now refined by modelling each helicopter as a sphere of radius 0.5 units.
\item Explain how this will change your answer to part (d).
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q8 [13]}}