| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on skew lines requiring the cross product formula for shortest distance and finding a plane equation. While it involves multiple steps and vector manipulation beyond standard A-level, it's a textbook application of well-defined formulas without requiring novel insight or particularly complex reasoning. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | (cid:167) 1(cid:183) (cid:167) 2(cid:183) (cid:167) 5(cid:183) |
| Answer | Marks |
|---|---|
| (cid:32) 6 | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | n |
| Answer | Marks |
|---|---|
| Using formula from the formula booklet | Accept any multiple |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | x(cid:16)2y(cid:16)z(cid:32)(cid:16)5 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Using their mutual perpendicular from |
| Answer | Marks |
|---|---|
| 7(i) | e |
| Answer | Marks |
|---|---|
| 7(i) | (continued) |
| Answer | Marks |
|---|---|
| 7(ii) | e |
Question 7:
7 | (i) | (cid:167) 1(cid:183) (cid:167) 2(cid:183) (cid:167) 5(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
n(cid:32) 2 (cid:117) (cid:16)1 (cid:32) (cid:16)10
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169)(cid:16)3(cid:185) (cid:169) 4(cid:185) (cid:169) (cid:16)5(cid:185)
(cid:167) 4(cid:183) (cid:167) 3(cid:183) (cid:167) 1(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
b(cid:16)a(cid:32) (cid:16)2 (cid:16) 5 (cid:32) (cid:16)7
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:168) (cid:184) (cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) 7(cid:185) (cid:169)(cid:16)2(cid:185) (cid:169) 9(cid:185)
Shortest distance is given by
(cid:167) 1(cid:183) (cid:167) 5(cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:16)7 (cid:16)10
(cid:168) (cid:184) (cid:168) (cid:184) p
(cid:11)b(cid:16)a(cid:12)(cid:152)n (cid:168) 9 (cid:184) (cid:168) (cid:16)5 (cid:184)
(cid:169) (cid:185) (cid:169) (cid:185)
D(cid:32) (cid:32)
n 150 S
(cid:32) 6 | M1
A1
B1
e
M1
A1
[5] | 1.1a
1.1
1.1
i
c
1.2
1.1 | n
e
Finding a mutual perpendicular
m
Any vector between the lines
Using formula from the formula booklet | Accept any multiple
Or any other valid method
7 | (ii) | x(cid:16)2y(cid:16)z(cid:32)(cid:16)5 | M1
A1
[2] | 3.1a
1.1 | Using their mutual perpendicular from
part (i)
oe
--- 7(i) ---
7(i) | e
p
S
(answer space continued on next page)
7(i) | (continued)
n
e
m
i
c
--- 7(ii) ---
7(ii) | e
p
S7 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\frac { x - 3 } { 1 } = \frac { y - 5 } { 2 } = \frac { z + 2 } { - 3 }$ and $\frac { x - 4 } { 2 } = \frac { y + 2 } { - 1 } = \frac { z - 7 } { 4 }$.\\
(i) Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.\\
(ii) Find a cartesian equation of the plane which contains $l _ { 1 }$ and is parallel to $l _ { 2 }$.
\hfill \mbox{\textit{OCR Further Pure Core 1 Q7 [7]}}