CAIE FP1 2018 November — Question 10 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicular distance point to line
DifficultyStandard +0.8 This is a comprehensive 3D vectors question requiring multiple techniques: showing lines are parallel via direction vectors, finding shortest distance between skew lines using the scalar triple product formula, and finding the angle between planes using normal vectors from cross products. While each technique is standard for Further Maths, the question requires sustained multi-step reasoning across three parts with 12 total marks, placing it moderately above average difficulty.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane4.04h Shortest distances: between parallel lines and between skew lines
The position vectors of the points \(A, B, C, D\) are $$\mathbf{i} + \mathbf{j} + 3\mathbf{k}, \quad 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}, \quad -\mathbf{i} + 3\mathbf{k}, \quad m\mathbf{j} + 4\mathbf{k},$$ respectively, where \(m\) is a constant.
  1. Show that the lines \(AB\) and \(CD\) are parallel when \(m = \frac{3}{2}\). [1]
  2. Given that \(m \neq \frac{3}{2}\), find the shortest distance between the lines \(AB\) and \(CD\). [5]
  3. When \(m = 2\), find the acute angle between the planes \(ABC\) and \(ABD\), giving your answer in degrees. [6]
Question 10:
AnswerMarks
10(ii)2 1 −2  −1 −4
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71)
         
AB= 3 , CD= m and AC = −1 or AD = m−1 or BC= −4
         
         
AnswerMarks
2 1  0   1  −2B1
i j k 3−2m  1 
   
n = 2 3 2 = 0 ( so parallel to 0 )
   
   
AnswerMarks Guidance
1 m 1 2m−3 −1M1A1 Finds common perpendicular using cross
product.
( )+0+0
AC.n −2 3−2m
= o.e.
n ( )2 +( )2
AnswerMarks Guidance
3−2m 2m−3M1 Uses formula for shortest distance.
2
= = 2
AnswerMarks
2A1
5
AnswerMarks
10(iii)i j k  2   1 
   
2 3 2 = −4 ~ −2 o.e.
   
   
AnswerMarks Guidance
−2 −1 0  4   2 M1A1 Finds normal to plane ABC (AEF).
i j k  1 
 
2 3 2 = −4 o.e.
 
 
AnswerMarks Guidance
−1 1 1  5 A1 Finds normal to ABD(AEF).
1+8+10  19 
cosθ= =
 
AnswerMarks Guidance
12 +22 +22 12 +42 +52  378M1A1FT Uses formula for angle between two lines.
⇒θ =12.2°A1 CAO.
6
AnswerMarks Guidance
QuestionAnswer Marks
11E(i)1 1
dy dy dt 12t2 6t2
= × = =
AnswerMarks Guidance
dx dt dx 18−2t 9−tB1 AEF.
d2y d dy dt
= ×
 
AnswerMarks Guidance
dx2 dtdx dxM1 Uses chain rule again to find second
derivative.
1 1
− ( )+6t2
3t 2 9−t
=
( )2( )
AnswerMarks Guidance
9−t 18−2tdM1 Uses quotient (or product rule)
1
3t − 2 ( 9−t+2t ) 3 ( 9+t )
= =
( )3 1
2 9−t ( )3
AnswerMarks Guidance
2t2 9−tA1 AG.
4
AnswerMarks
11E(ii)1 56 d2y
∫ dx
56 dx2
AnswerMarks Guidance
0B1 Uses correct formula for mean value
t=4
 1 
1 6t2 
=
 
56 9−t
 
 
AnswerMarks Guidance
t=0M1 Finding limits correctly
M1Using expression
1  6 4  3
=   = .
 
56 9−4 70
AnswerMarks Guidance
 A1 Inserts correct values of t and obtains
answer (AG.)
4
AnswerMarks Guidance
QuestionAnswer Marks
11E(iii)2 2
 dx +  dy =( 18−2t )2 +144t=4 ( t+9 )2
   
AnswerMarks Guidance
dt  dt M1A1 2 2
dx dy
Simplifies   +   .
dt  dt 
2π∫ 4 8t 3 2   ( 2 ( t+9 )) dt =32π∫ 4 t 5 2 +9t2 3 dt
 
 
AnswerMarks Guidance
0 0M1A1 ds
Uses 2π∫y dt.
dt
4
2 7 18 5 
=32π t2 + t2 
 7 5 
AnswerMarks Guidance
0M1 Integrates term by term.
211×83 169984
= π= π=15300
AnswerMarks Guidance
35 35A1 Accept exact answer or decimal rounding
to 15300.
6
AnswerMarks
11O(i)( )n 2 2 ( )n−1
I = x x2 −1 −2n∫ x2 x2 −1 dx
n  
1
AnswerMarks Guidance
1M1A1 Integrates by parts.
2( )( )n−1
= 2−2n∫ x2 −1+1 x2 −1 dx
AnswerMarks Guidance
1M1 Uses x2 = x2 −1+1.
= 2−2nI −2nI
AnswerMarks
n n−1A1
⇒ ( 2n+1 ) I = 2−2nI
AnswerMarks Guidance
n n−1A1 AG.
5
AnswerMarks Guidance
QuestionAnswer Marks
11O(ii)dx
=tanθsecθ
AnswerMarks Guidance
dθM1A1 Differentiates secθ.
π
secθ= 2⇒θ= secθ=1⇒θ=0
AnswerMarks Guidance
4B1 Changes limits.
π π
4( )n 4
I =∫ sec2θ−1 tanθsecθ dθ=∫ tan2n+1θsecθ dθ
n
AnswerMarks Guidance
0 0B1 Uses sec2θ−1=tan2θ. (AG.)
4
AnswerMarks
11O(iii)π
4 sin7θ
∫ dθ=I
cos8θ 3
AnswerMarks Guidance
0B1 Deduces that integral is I .
3
2− 2
I = 2−1 (or I = )
AnswerMarks Guidance
0 1 3B1 Calculates I or I
0 1
2− 2
3I = 2−2I ⇒I = .
AnswerMarks Guidance
1 0 1 3M1A1 Uses reduction formula to find I or I
1 2
………
2 4 7 2−8
I = − I =
AnswerMarks
2 5 5 1 15Finds I .
2
2 6 16−9 2
I = − I =
AnswerMarks Guidance
3 7 7 2 35A1 Finds I .
3
5
Question 10:
--- 10(ii) ---
10(ii) | 2 1 −2  −1 −4
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71)
         
AB= 3 , CD= m and AC = −1 or AD = m−1 or BC= −4
         
         
2 1  0   1  −2 | B1
i j k 3−2m  1 
   
n = 2 3 2 = 0 ( so parallel to 0 )
   
   
1 m 1 2m−3 −1 | M1A1 | Finds common perpendicular using cross
product.
( )+0+0
AC.n −2 3−2m
= o.e.
n ( )2 +( )2
3−2m 2m−3 | M1 | Uses formula for shortest distance.
2
= = 2
2 | A1
5
--- 10(iii) ---
10(iii) | i j k  2   1 
   
2 3 2 = −4 ~ −2 o.e.
   
   
−2 −1 0  4   2  | M1A1 | Finds normal to plane ABC (AEF).
i j k  1 
 
2 3 2 = −4 o.e.
 
 
−1 1 1  5  | A1 | Finds normal to ABD(AEF).
1+8+10  19 
cosθ= =
 
12 +22 +22 12 +42 +52  378 | M1A1FT | Uses formula for angle between two lines.
⇒θ =12.2° | A1 | CAO.
6
Question | Answer | Marks | Guidance
11E(i) | 1 1
dy dy dt 12t2 6t2
= × = =
dx dt dx 18−2t 9−t | B1 | AEF.
d2y d dy dt
= ×
 
dx2 dtdx dx | M1 | Uses chain rule again to find second
derivative.
1 1
− ( )+6t2
3t 2 9−t
=
( )2( )
9−t 18−2t | dM1 | Uses quotient (or product rule)
1
3t − 2 ( 9−t+2t ) 3 ( 9+t )
= =
( )3 1
2 9−t ( )3
2t2 9−t | A1 | AG.
4
11E(ii) | 1 56 d2y
∫ dx
56 dx2
0 | B1 | Uses correct formula for mean value
t=4
 1 
1 6t2 
=
 
56 9−t
 
 
t=0 | M1 | Finding limits correctly
M1 | Using expression
1  6 4  3
=   = .
 
56 9−4 70
  | A1 | Inserts correct values of t and obtains
answer (AG.)
4
Question | Answer | Marks | Guidance
11E(iii) | 2 2
 dx +  dy =( 18−2t )2 +144t=4 ( t+9 )2
   
dt  dt  | M1A1 | 2 2
dx dy
Simplifies   +   .
dt  dt 
2π∫ 4 8t 3 2   ( 2 ( t+9 )) dt =32π∫ 4 t 5 2 +9t2 3 dt
 
 
0 0 | M1A1 | ds
Uses 2π∫y dt.
dt
4
2 7 18 5 
=32π t2 + t2 
 7 5 
0 | M1 | Integrates term by term.
211×83 169984
= π= π=15300
35 35 | A1 | Accept exact answer or decimal rounding
to 15300.
6
11O(i) | ( )n 2 2 ( )n−1
I = x x2 −1 −2n∫ x2 x2 −1 dx
n  
1
1 | M1A1 | Integrates by parts.
2( )( )n−1
= 2−2n∫ x2 −1+1 x2 −1 dx
1 | M1 | Uses x2 = x2 −1+1.
= 2−2nI −2nI
n n−1 | A1
⇒ ( 2n+1 ) I = 2−2nI
n n−1 | A1 | AG.
5
Question | Answer | Marks | Guidance
11O(ii) | dx
=tanθsecθ
dθ | M1A1 | Differentiates secθ.
π
secθ= 2⇒θ= secθ=1⇒θ=0
4 | B1 | Changes limits.
π π
4( )n 4
I =∫ sec2θ−1 tanθsecθ dθ=∫ tan2n+1θsecθ dθ
n
0 0 | B1 | Uses sec2θ−1=tan2θ. (AG.)
4
11O(iii) | π
4 sin7θ
∫ dθ=I
cos8θ 3
0 | B1 | Deduces that integral is I .
3
2− 2
I = 2−1 (or I = )
0 1 3 | B1 | Calculates I or I
0 1
2− 2
3I = 2−2I ⇒I = .
1 0 1 3 | M1A1 | Uses reduction formula to find I or I
1 2
………
2 4 7 2−8
I = − I =
2 5 5 1 15 | Finds I .
2
2 6 16−9 2
I = − I =
3 7 7 2 35 | A1 | Finds I .
3
5
The position vectors of the points $A, B, C, D$ are
$$\mathbf{i} + \mathbf{j} + 3\mathbf{k}, \quad 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}, \quad -\mathbf{i} + 3\mathbf{k}, \quad m\mathbf{j} + 4\mathbf{k},$$
respectively, where $m$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Show that the lines $AB$ and $CD$ are parallel when $m = \frac{3}{2}$. [1]

\item Given that $m \neq \frac{3}{2}$, find the shortest distance between the lines $AB$ and $CD$. [5]

\item When $m = 2$, find the acute angle between the planes $ABC$ and $ABD$, giving your answer in degrees. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q10 [12]}}
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