e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity Chapter 8: Further calculus
8.3 Integration of k f ′(x)
f(x)
Since integration is the reverse process of differentiation and ln x only exists for x . 0, we
can say that:
KEY POINT 8.3 REWIND
∫ f ′(x) dx = ln f(x) + c In Chapter 4, we
f(x) learnt that
This can be extended to:
d [ ln f(x) ] = f ′(x) .
∫ kf ′(x) dx = kln f(x) + c
f(x) dx f(x)
Review WORKED EXAMPLE 8.5
∫Find 2x 3 dx.
x2 +
Answer
If f(x) = x2 + 3 then f ′(x) = 2x.
∫∴ 2x dx = ln x2 +3 +c
x2 + 3
189
Review WORKED EXAMPLE 8.6
Review ∫Find tan 3x dx.
Answer
∫ ∫tan 3x dx = sin 3x dx
cos 3x
If f(x) = cos 3x then f ′(x) = −3sin 3x.
∫ ∫tan 3x dx = − 1 −3sin 3x dx
3 cos 3x
= − 1 ln cos 3x + c
3
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WORKED EXAMPLE 8.7
∫Find 3x2 + 5x dx.
2x3 + 5x2
Answer
If f(x) = 2x3 + 5x2 then f ′(x) = 6x2 + 10x.
∫ ∫3x2 + 5x dx = 1 6x2 + 10x dx
2x3 + 5x2 2 2x3 + 5x2
Review = 1 ln 2x3 + 5x2 + c
2
WORKED EXAMPLE 8.8
∫Find1 2ex dx
0 1 + ex
Answer
If f(x) = 1 + ex then f ′(x) = ex.
∫ ∫1 2ex 1 ex
0 1 + ex 0 1 + ex
dx = 2 dx
190 = 2 ln 1 + ex 10 Substitute limits.
Simplify the logarithms.
Review = 2 ln(1 + e) − 2 ln 2
= 2 ln 1 + e
2
EXERCISE 8C
1 Find the following integrals.
∫a 6x2 1 dx ∫b 1 cos x dx ∫c 4x − 10 1 dx
x3 − + sin x x2 − 5x +
∫d cot x dx ∫e 2 x dx ∫f sec2 x dx
− x2 1 + tan x
2 Find the exact value of each of these integrals.
∫a 1 π ∫b 2 3x2 ∫c ln 2 e2x
4 0 x3 + 2 0 1 + e2x
tan x dx dx dx
1 π
Review 6
∫d 2 x +1 ∫e 1 π ∫f 1 2x + 3
1 + 2x − 4 0 + 1)(x +
dx cot x dx dx
x2 1 1 π (x 2)
6
∫3 Show that ex + e−x dx = ln(1 − e2x ) − x + c.
ex − e−x
∫4 1π sin x cos x = − 1 2.
Show that 4 1 − 2 cos 2x dx 8 ln
1 π
3
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PS 5 y
O px
The diagram shows part of the curve y = 4x . Find the exact value of p for which the shaded region has
x2 + 1
an area of 4.
Review 8.4 Integration by substitution REWIND
Integration by substitution can be considered as the reverse process of differentiation by the In the Pure
chain rule. Mathematics 1
Coursebook Chapter 7,
This method is used when a simple substitution can be applied that will transform a we learnt the chain rule
difficult integral into an easier integral. The integral must be completely rewritten in terms for differentiation:
of the new variable.
dy = dy × du
dx du dx
WORKED EXAMPLE 8.9
∫Use the substitution u = 2x + 1 to find 4x 2x + 1 dx. 191
Review Answer Rewrite the integral in terms of u and simplify.
Integrate with respect to u.
u = 2x + 1 ⇒ x = u − 1 Rewrite in terms of x.
2
u = 2x + 1 ⇒ du = 2 ⇒ dx = du
dx 2
∫ ∫4x 2x + 1 dx = 4 u − 1 u du
2 2
∫= 3 − 1 du
u 2
u2
= 2 5 − 2 3 +c
u2 u2
53
= 2 (2x 5 − 2 (2x 3 + c
+ 1)2 + 1)2
53
Review
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WORKED EXAMPLE 8.10
∫Use the substitution x = 2 tan u to find 8 4 dx.
x2 +
Answer
x = 2 tan u ⇒ x2 = 4 tan2 u
x = 2 tan u ⇒ dx = 2 sec2 u ⇒ dx = 2 sec2 u du
du
∫ ∫8 = 8 2 sec2 Rewrite the integral in terms of u and simplify.
x2 + 4 tan2 u
dx + 4 u du Use tan2 u + 1 ≡ sec2 u.
4 Simplify.
Review
∫= 16 sec2 u Integrate with respect to u.
4 tan2 u + 4 du Rewrite in terms of x.
∫= 16 sec2 u du
4 sec2 u
∫= 4du
= 4u + c
= 4 tan−1 x + c
2
192 WORKED EXAMPLE 8.11
Review ∫Use the substitution u = sin x to find sin2 2x cos x dx.
Review Answer Replace dx by du and simplify. TIP
cos x
u = sin x ⇒ sin2 2x = (2 sin x cos x)2 When using
Replace sin2 2x by 4u2 − 4u4. substitution in definite
= 4 sin2 x cos2 x Integrate with respect to u. integrals you must also
= 4u2 (1 − u)2 Rewrite in terms of x. remember to convert
u = sin x ⇒ du = cos x ⇒ dx = du the limits to limits for u.
dx cos x
∫ ∫sin2 2x cos x dx = sin2 2x cos x du
cos x
∫= sin2 2x du
∫= (4u2 − 4u4 ) du
= 4 u3 − 4 u5 + c
35
= 4 sin3 x − 4 sin5 x + c
35
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WORKED EXAMPLE 8.12
∫Use the substitution u = x + 1 to find 3 x 1 dx.
0 x+
Answer
u = x + 1 ⇒ x = u − 1
u = x + 1 ⇒ du = 1 ⇒ dx = du
dx
Find the new limits for u:
Review x = 3⇒ u = 3+1= 4
x = 0⇒ u = 0+1=1
∫ ∫x=3 x 1 dx = u=4 u − 1 du Write the integral in terms of u and simplify.
x=0 x+ u=1 u Integrate with respect to u.
Evaluate the function at the limits.
∫= 4 1 − −1 du
1 u 2
u2
= 2 3 − 1 4
3
u2 2u 2 1
= 16 − 4 − 2 − 2
3 3
=8 193
3
Review EXERCISE 8D
∫1 Given that I = ∫x dx, use the substitution u = x2 − 3 to show that I = 1 1 du.
x2 − 3 2 u
Hence find I .
2 Find these integrals, using the given substitution.
∫a (x x dx, u = x + 2 ∫b x 1 − 2x2 dx, u = 1 − 2x2
+ 2)4
∫c cos x sin5 x dx, u = sin x ∫d ex ex + 2 dx, u = ex + 2
∫e 5x 1 dx, u = 5x + 1 ∫f x 3x − 1 dx, u = 3x − 1
5x +
1 x2 1 π
2
∫ ∫3 Given that I =
0 1 − x2 dx, use the substitution x = sin θ to show that I = sin2 θ dθ .
Review 0
Hence find the exact value of I .
4 Find the exact value of each of these integrals.
∫a 1 π cos3 = ∫b4 x = −
2 0 −
x dx, u sin x dx, u 3 x
3 x
0
23 ∫d π 2x cos x2 x2
1 x 2− 4 4
∫ ( )c dx, u = x dx , u = cos
x
0
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∫e 1 π sec2 tan3 dx , ∫ ( )f 2 5 = 2 tan θ
3 −2 +
x x u = tan x dx, x
1 π x2 4 2
6
∫g 1 x2 , x = 2 cos θ ∫2
0 4 − x2
h x(2 − x) dx, x = 2 sin2 θ
0
∫5 2 4
Use the substitution u = x − 2 to find the exact value of 1 1+ (x − 2)2 dx.
6 The diagram shows part of the curve y = 2 sin2 x cos3 x. Using the y
substitution u = sin x, find the exact area of the shaded region.
Review O πx
y 2
7 The diagram shows part of the curve y = ln x . Using the substitution
x
u = ln x, find the exact value of the volume of the solid formed when
the shaded region is rotated completely about the x-axis.
O ex
8 The diagram shows part of the curve y = 3 ecos x sin x. The curve y
has rotational symmetry of order 2 about the origin. Using
the substitution u = cos x, find the exact value of the total area
194 bounded by the curve and the x-axis between x = −π and x = π. −π O πx
Review8.5 The use of partial fractions in integration
EXPLORE 8.1
∫Consider 2 1 dx.
x2 −
Try using some of the integration techniques that you have learnt so far to find this
integral and discuss what happens with your classmates.
∫In the integral 2 1 dx, the denominator x2 −1 can be factorised to (x − 1)(x + 1).
x2 −
2
Hence the fraction x2 − 1 can be split into partial fractions.
Review 2 ≡ x A 1 + B Multiply throughout by (x − 1)(x + 1).
x2 − 1 − x +1 (1)
2 ≡ A(x + 1) + B(x − 1)
Let x = 1 in equation (1):
2 = 2A
A=1
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Let x = −1 in equation (1):
2 = −2B
B = −1
∴ 2 ≡ 1 − 1
x2 − 1 x −1 x +1
∫ ∫Hence 2 1 dx = 1 − 1 dx
x2 − x −1 x +1
= ln x − 1 − ln x + 1 + c
= ln x −1 +c
x +1
Review WORKED EXAMPLE 8.13
∫Find (x + 3−x + 3) dx.
1)(x2
Answer
First split into partial fractions.
(x 3−x + 3) ≡ A + Bx +C Multiply throughout by (x + 1)(x2 + 3).
+ 1)(x2 x +1 x2 +3
3 − x ≡ A(x2 + 3) + Bx(x + 1) + C(x + 1)
Let x = −1 4 = 4A ⇒ A=1 195
Let x = 0 3 = 3A + C ⇒ C = 0
Review Let x = 1 2 = 4A + 2B + 2C ⇒ B = −1
∫ ∫∴ + 3−x + dx = 1 − x dx
(x 1)(x2 3) x + 1 x2 + 3
∫ ∫= x 1 1 dx − 1 2x 3 dx
+ 2 x2 +
= ln x +1 − 1 ln x2 +3 +c
2
WORKED EXAMPLE 8.14
∫Find (x + 9 − 1)2 dx.
2)(x
Review Answer
First split into partial fractions.
9 AB C Multiply throughout by (x + 2)(x − 1)2.
(x + 2)(x − 1)2 ≡ x + 2 + x − 1 + (x − 1)2
9 ≡ A(x − 1)2 + B(x + 2)(x − 1) + C(x + 2)
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Let x = 1 9 = 3C ⇒ C = 3
Let x = −2 9 = 9A ⇒ A = 1
Let x = 0 9 = A − 2B + 2C ⇒ B = −1
9 1 1 3
+ 2)(x + − − 1)2
∫ ∫∴ dx = − + dx
(x − 1)2 x 2 x 1 (x
∫= x 1 2 − x 1 1 + 3(x − 1)−2 dx
+ −
= ln x + 2 − ln x − 1 − 3(x − 1)−1 + c
= ln x+2 − 3 + c
x −1 x −1
Review
EXERCISE 8E
1 Find each of these integrals.
∫a 1 x) dx ∫b 2x − 5 dx ∫c 2(4x − 9) dx
x(2 − (x + 2)(x − 1) (2x + 1)(x − 5)
∫d x2 + 2x − 5 dx ∫e 6x2 − 19x + 20 dx ∫f 2x2 + 4x − 21 dx
(x − 3)(x2 + 1) (x + 1)(x − 2)2 (x − 1)(x + 4)
2 Evaluate each of these integrals.
∫196 a 2 (x 5x + 13 3) dx ∫b 2 (2x 4x + 5 2) dx ∫c 2 4 − 3x dx
1 + 2)(x + 0 + 1)(x + 1 (2x − 1)(x + 2)
∫d 1 (x 3x2 + x 1) dx ∫e 1 x2 +x + 2 ) dx ∫f0 (2 − 8 + x2 ) dx
−1 − 2)(x2 + 0 (x + 1)(1 + x2 −2 x)(4
Review
∫g 3 1 − x − 2x2 dx ∫h 1 (2x 1 − 2x 1)2 dx ∫i 2 8x2 − 3x − 2 dx
2 x2 (1 − x) 0 + 1)(x + 1 4x3 − 3x + 1
3 By first dividing the numerator by the denominator, evaluate each of these integrals.
∫a 3 x2 + 2x + 3 dx ∫b 3 2x2 + 5x + 1 dx ∫c 2 4x2 + 2x − 5 dx
2 (x − 1)(x + 2) 1 (x + 1)(x + 2) 1 (x + 1)(2x − 1)
∫4 3 2 x+ 3 2
Show that 2 2 x2 − 2 dx = ln 3.
∫5 1π cos x = 1 ln 2.
Show that 9 − sin2 x dx 6
2
0
∫6 ln 2 e2x dx = 1 ln 27 .
Show that 0 (1 + ex )(2ex + 1) 2 20
Review
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8.6 Integration by parts REWIND
In Chapter 4, we learnt
∫Consider x e2x dx. The function to be integrated is clearly the product of two simpler the product rule for
differentiation:
functions x and e2x. If we try using a substitution to simplify the integral we will find that d (uv) = u dv + v du .
it does not help. To integrate this function we require a new technique. dx dx dx
Integrating both sides of the equation for the product rule for differentiation with respect to 197
x and using the fact that integration is the reverse process of differentiation we can say that:
∫ ∫uv =udv dx + v du dx
dx dx
Review Rearranging gives the formula for integration by parts:
KEY POINT 8.4
∫ ∫u dv dx = uv − v du dx
dx dx
The following Worked examples show how to use this formula.
WORKED EXAMPLE 8.15
∫Find x e2x dx.
Review Answer
u = x ⇒ du = 1
dx
dv = e2x ⇒ v = 1 e2x
dx 2
∫ ∫Substitute into u dv dx = uv − v du dx:
dx dx
∫ ∫x e2x dx = 1 xe2x − 1 e2x dx
22
= 1 x e2x − 1 e2x + c
24
It is important to choose our values for u and dv carefully.
dx
∫Consider trying to find x ln x dx. If we let u = x and dv = ln x , then finding du is
dx dx
Review
easy but we do not know how to integrate ln x to find v. Hence we must let u = ln x and
dv = x. This is demonstrated in Worked example 8.16.
dx
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WORKED EXAMPLE 8.16
∫Find x ln x dx.
Answer
u = ln x ⇒ du = 1
dx x
dv = x ⇒ v = 1 x2
dx 2
Review ∫ ∫Substitute into u dv dx = uv − v du dx:
dx dx
∫ ∫x ln x dx = 1 x2 ln x − 1 x2 ⋅ 1 dx
2 2x
∫= 1 x2 ln x − 1 x dx
22
= 1 x2 ln x − 1 x2 + c
24
WORKED EXAMPLE 8.17
∫198 Find ln x dx.
Review Answer
∫ ∫First write ln x dx as 1 ⋅ ln x dx.
u = ln x ⇒ du = 1
dx x
dv = 1 ⇒ v=x
dx
∫ ∫Substitute into u dv dx = uv − v du dx:
dx dx
∫ ∫ln x dx = x ln x − 1 ⋅ x dx
x
∫= x ln x − 1 dx
= x ln x − x + c
Review
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Sometimes we might need to use integration by parts more than once to integrate a function.
WORKED EXAMPLE 8.18
∫Find x2 sin x dx .
Answer
u = x2 ⇒ du = 2x
dx
Review dv = sin x ⇒ v = − cos x
dx
∫ ∫Substitute into u dv dx = uv − v du dx
dx dx
∫ ∫x2 sin x dx = −x2 cos x − − 2x cos x dx
∫= −x2 cos x + 2x cos x dx (1)
∫Use integration by parts a second time to find 2x cos x dx
u = 2x ⇒ du = 2 199
dv = cos x dx
dx
⇒ v = sin x
Review ∫ ∫Substitute intou dv dx = uv − v du dx
dx dx
∫ ∫2x cos x dx = 2x sin x − 2 sin x dx
= 2x sin x + 2 cos x + c
Substituting in (1) gives:
∫ x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + c
WORKED EXAMPLE 8.19
∫Given that I = ln x dx, use integration by parts to show that I = 1 (ln x)2 + c.
x2
Review Answer ⇒ du = 1
u = ln x dx x
dv = 1 ⇒ v = ln x
dx x
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∫ ∫Substitute into u dv dx = uv − v du dx
dx dx
∫I = (ln x)2 − ln x dx Rearrange.
x Divide both sides by 2.
2I = (ln x)2 + k
I = 1 (ln x)2 + k
2 2
∴ I = 1 (ln x)2 +c
2
Review WORKED EXAMPLE 8.20
∫Find the exact value of 1 π
2
4x sin 2x dx.
0
Answer ⇒ du = 4
u = 4x dx
dv = sin 2x ⇒ v = − 1 cos 2x
dx 2
200 Substitute into u dv dx = uv − v du dx
∫ ∫dx dx
∫ ∫4x sin 2x dx = −2x cos 2x + 2 cos 2x dx
= −2x cos 2x + sin 2x + c
Review
∫∴ 1 π 1 π
2
4x sin 2x dx = −2x cos 2x + sin 2x 02
0
= (−π cos π + sin π) − (0 + sin 0)
=π
EXERCISE 8F
1 Find each of these integrals. ∫b x cos x dx ∫c x ln 2x dx
∫e x3 ln x dx ∫f ln x dx
∫a 3xex dx
∫2 x
Review ∫d x sin 2x dx
b x ln 2x dx
2 Evaluate each of these integrals.
1
∫a 1 π ∫1
6 ∫1
x cos 3x dx c xe2x dx
e (2 − x)e−3x dx
0 0
0
∫3 ∫0f (1 − 3xe−5x ) dx
− 1
d ln x dx 5
1
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3 Use integration by parts twice to find the exact value of each of these integrals.
∫2 ∫b 1 π ∫π
4 x2 sin x dx
a (ln x)2 dx x2 cos 2x dx c 0
1 0 ∫π
ex sin x dx
∫e ∫0 f 0
d x(ln x)2 dx e x2ex dx
0 −∞
4 Find the exact value of the area for each of these shaded regions.
ay by
y = x2 ln x y = x sin x
Review O e O πx
y = x2 e–x
cy ln x dy
x
y=
O ex O 4x
5 y 201
Review Ox
The diagram shows the curve y = e−x x + 2. Find the exact value of the volume of the solid formed when the
shaded region is rotated completely about the x-axis.
8.7 Further integration
Some integrals can be found using different integration techniques.
The activities in Explore 8.2 look at why our answers to an indefinite integral might ‘appear’
to be different when using different integration techniques when they are actually equivalent.
EXPLORE 8.2
Review ∫1 Nadia and Richard are asked to find x(x − 2)7 dx.
Nadia uses the substitution u = x − 2 and her answer is: 1 (x − 2)9 + 1 (x − 2)8 + c
94
Richard uses integration by parts and his answer is: 1 x(x − 2)8 − 1 (x − 2)9 + c
8 72
Use each of their methods to show that they are both correct and then use algebra to explain why their answers
are equivalent.
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∫2 a Find 4x +1 1 dx using:
2x2 + x−
i the substitution u = 2x2 + x − 1
ii partial fractions.
∫b Suggest another method for finding 4x + 1 1 dx.
2x2 + x −
c Use algebra to show that your two answers to part a are equivalent.
ReviewExplore 8.3 is designed to make you think about the different integration techniques that
might be available to you for a particular integral.
This activity expects you to be able to recognise the appropriate substitution to use when
integrating by substitution.
EXPLORE 8.3
Discuss with your classmates which integration methods can be used to find each of these integrals.
∫ x2 2x + 3 dx ∫ 3x2 dx ∫ x ex2 dx ∫ 2x dx
+ 3x − 10 x3 + 1 x+5
∫ ln 5x dx ∫ x 2x + 1 dx ∫3 1)5 dx ∫ 3x − 2 dx
(2x +
202
∫ x2 2x + 3 10 dx ∫ cos5 x sin x dx ∫ x sin 4x dx ∫3 sin x x dx
+ 3x − + cos
Review Find each integral.
Checklist of learning and understanding
Integrating kf ′(x)
f(x)
∫● kf ′(x) dx = kln f(x) + c
f(x)
Integration by substitution
Review ● Substitutions can sometimes be used to simplify the form of a function so that its integral can
be easily recognised. When using a substitution, we must ensure that the integral is completely
rewritten in terms of the new variable before integrating.
Integrating rational functions
● Some rational functions can be split into partial fractions that can then be integrated.
Integration by parts
∫ ∫● u dv dx = uv − v du dx
dx dx
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END-OF-CHAPTER REVIEW EXERCISE 8
∫1 [5]
1 Find the exact value of 2 xe−2x dx.
0
Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2016
∫2 Find the exact value of 4 ln x dx. [5]
1x
Cambridge International A Level Mathematics 9709 Paper 31 Q3 November 2013
∫3 1 π (1 + 3 tan x)
Use the substitution u = 1 + 3tan x to find the exact value of 4 cos2 x dx
[5]
0
Review Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2014
4 i Prove that cot θ + tan θ ≡ 2 cosec 2θ. [3]
∫ii 1 π = 1
3
Hence show that cosec 2θ dθ ln 3. [4]
1 π 2
6
Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2013
∫5 1 x2
Let I = 0 (4 − x2 ) dx.
∫i 1 π
Using the substitution x = 2 sin θ , show that I = 6 sin2
4 θ dθ. [3]
0
ii Hence find the exact value of I . [4]
Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2010 203
Review ∫2 [3]
6 The integral I is defined by I = 4t3 ln(t2 + 1) dt.
0
∫5
i Use the substitution x = t2 + 1 to show that I = (2x − 2) ln x dx.
1
ii Hence find the exact value of I. [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2011
7 i Express 2 in partial fractions. [2]
(x + 1)(x + 3) [2]
ii Using your answer to part i, show that (x 2 + 2 ≡ (x 1 − 1 + 1 3 + (x 1 3)2 .
+ 1)(x 3) + 1)2 x +1 x+ +
∫iii 1 4 dx = 7 − ln 3.
Hence show that 0 (x + 1)2 (x + 3)2 12 2 [5]
Review ∫8 a Find (4 + tan2 2x) dx. Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2010
[3]
∫ ( )b Find the exact value of 1 π sin x + 1 π
2 6
dx. [5]
1 sin x
4 π
Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2015
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9 i Prove the identity tan 2θ − tanθ ≡ tanθ sec 2θ. [4]
∫ii 1 π = 1 3.
6
Hence show that tan θ sec 2θ dθ ln [4]
0 22
Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2016
10 y
M
Review O 2x
The diagram shows part of the curve y = (2x − x2 )e 1 x and its maximum point M .
2
i Find the exact x-coordinate of M . [4]
ii Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis. [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2016
11 By first expressing 4x2 + 5x + 3 in partial fractions, show that 4 4x2 + 5x + 3 dx = 8 − ln 9. [10]
0 2x2 + 5x + 2
∫2x2 + 5x + 2
Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2012
204 y
12
Review O 2x
M
The diagram shows the curve y = x3 ln x and its minimum point M .
i Find the exact coordinates of M. [5]
ii Find the exact area of the shaded region bounded by the curve, the x-axis and the line x = 2. [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2010
13 y
M
Review R
O1 px
The diagram shows the curve y = 1 x2 for x ù 0, and its maximum point M . The shaded region R is
+ x3
enclosed by the curve, the x-axis and the lines x = 1 and x = p.
i Find the exact value of the x-coordinate of M . [4]
[6]
ii Calculate the value of p for which the area of R is equal to 1. Give your answer correct
to 3 significant figures.
Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2015
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CROSS-TOPIC REVIEW EXERCISE 3
P3 This exercise is for Pure Mathematics 3 students only.
1 Expand (1 − 3x)−5 in ascending powers of x, up to and including the term in x3. [3]
[3]
∫2 1 1 [4]
Evaluate 0 3x2 + 1 dx.
3 Expand 32 in ascending powers of x, up to and including the term in x3.
(x + 2)3
Review 4 Expand 1 in ascending powers of x, up to and including the term in x2. [4]
4 − 2x [5]
[5]
∫ 1π
5 Evaluate 2 x2 sin x dx.
0
6 Expand 1 + 2x in ascending powers of x, up to and including the term in x3.
1− x
∫7 1 + 2)e−2x dx = 5 − 7 .
Show that 4 4e2 [5]
(x
0
8 i Expand 1 in ascending powers of x, up to and including the term in x2, simplifying
(1 − 4x )
the coefficients. [3] 205
[2]
ii Hence find the coefficient of x2 in the expansion of 1 + 2x ) .
− 16x
(4
Review
Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2012
9 When (1 + ax)−2, where a is a positive constant, is expanded in ascending powers of x,
the coefficients of x and x3 are equal.
i Find the exact value of a. [4]
ii When a has this value, obtain the expansion up to and including the term in x2, [3]
simplifying the coefficients.
Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2012
10 y
Review O αx
pTohiendt iiasgαraamndshtohwe sshtahdeecdurrvegeioyn=is8esnicnlo21sexd−bytatnhe21cxurvfoeran0døthxe,linπe.sTxhe=xα-coanodrdiyna=te0o. f the maximum
i Show that α = 2 π. [3]
3
ii Find the exact value of the area of the shaded region. [4]
Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2012
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
∫11 5 5 − 3x dx 4 ln 4.
Show that 0 (x + 1)(3x + 1) = 3 [7]
∫12 1 4x2 [3]
Let I = 2 dx. [4]
0 1 − x2 π [5]
∫a Using the substitution x = sin θ, show that I = 6 4 sin2 θ dθ . [3]
b Hence show that I = π − 3 . 0
32
Review ∫a
13 It is given that x ln x dx = 30, where a . 1.
1
a Show that a = 2 119 .
ln a − 1
b Use the iterative formula an+1 = 119 to determine the value of a correct to
2 ln an − 1
2 decimal places. Give the result of each iteration correct to 4 decimal places.
14 y
M
206
Ox
The diagram shows the curve y = x2e2−x and its maximum point M.
Review
i Show that the x-coordinate of M is 2. [3]
∫ii Find the exact value of 2 x2e2−x dx. [6]
0
Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2015
15 Let f(x) = 12 + 8x − x2 ) .
(2 − x)(4 + x2
i Express f(x) in the form A + Bx + C . [4]
2−x 4 + x2
∫ii 1 f(x) dx = ln 25 .
Show that 0 2 [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2011
Review ∫a
16 It is given that ln(2x) dx = 1, where a . 1.
1
1 +
i Show that a = 1 exp ln 2 , where exp(x) denotes ex. [6]
2 a [3]
ln 2
ii Use the iterative formula an+1 = 1 exp 1 + an to determine the value of a correct to
2
2 decimal places. Give the result of each iteration to 4 decimal places.
Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2014
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17 i Express 4 + 12x + x2 in partial fractions. [5]
(3 − x)(1 + 2x)2 [5]
ii Hence obtain the expansion of 4 + 12x + x2 in ascending powers of x, up to and
(3 − x)(1 + 2x)2
including the term in x2.
Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2014
Review 18 Let f(x) = x2 − 8x + 9 .
(1 − x)(2 − x)2
i Express f(x) in partial fractions. [5]
ii Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x2. [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2014
19 Let f(x) = (x 5x − 2 − 1) .
− 1)(2x2
a Express f(x) in partial fractions. [5]
b Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x3. [5]
[5]
20 a Given that x3 − 2 ≡ A+ B + C + D , find the values of the constants A, B, C and D. 207
x2 (2x − 1) x x2 2x − 1 [5]
∫b 2 x3 − 2 = 3 − 9 + 1
Review Hence show that 1 x 2 (2x − 1) dx 2 2 ln 4 4 ln 3.
21 y
OM e x
The diagram shows the curve y = x2 ln x and its minimum point M.
i Find the exact values of the coordinates of M. [5]
ii Find the exact value of the area of the shaded region bounded by the curve, the x-axis [5]
and the line x = e.
Review Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2011
∫4 [5]
22 i Show that 4x ln x dx = 56 ln 2 − 12.
2
∫ 1π [5]
ii Use the substitution u = sin 4x to find the exact value of 24 cos3 4x dx.
0
Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2013
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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
23 i Express 4 cos θ + 3 sin θ in the form R cos(θ − α ), where R.0 and 0 , α , 1 π. Give
2
the value of α correct to 4 decimal places.
[3]
ii Hence
a solve the equation 4 cos θ + 3 sin θ = 2 for 0 , θ , 2π, [4]
[3]
∫b find (4 cos θ 50 sin θ )2 dθ .
+3
Review Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2013
208
Review
Review
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209
Review Chapter 9
Vectors
Review
P3 This chapter is for Pure Mathematics 3 students only.
In this chapter you will learn how to:
use standard notations for vectors, in two dimensions and three dimensions
■■ add and subtract vectors, multiply a vector by a scalar and interpret these operations
geometrically
calculate the magnitude of a vector and find and use unit vectors
■ use displacement vectors and position vectors
■ find the vector equation of a line
■ find whether two lines are parallel, intersect or are skew
■ find the common point of two intersecting lines
■■ find and use the scalar product of two vectors.
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University
Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3
Univer5csimtCyoPprye-ssC-aRmebrviidegwePREREQUISITE KNOWLEDGE Copy Copy Copy Copy
Where it comes from What you should be able to do Check your skills
Recall and apply: 1 The diagram shows a cuboid with CD = 5cm,
IGCSE / O Level
Mathematics ● Pythagoras’ theorem DH = 6 cm and EH = 3 cm.
● simple trigonometry
BF
to solve problems in two and C G
three dimensions. A
E
Review D 6 cm 3 cm
H
Find the angle between AG and AH.
UniversitCyoPprye-ssC-aRmebrviidegweIGCSE / O LevelShow familiarity with the2 Three points A, B and C are such that A(1, 7),
Mathematics following areas of coordinate B(7, 3) and C(0, −6).
Pure Mathematics 1 geometry in two dimensions:
Coursebook, Chapter 3 a Find the equation of the perpendicular
● Use and apply the general bisector of the line AB. Give your answer in
equation of a straight line the form y = mx + c.
in gradient/intercept form. b Write down the equation of the line parallel
● Calculate the coordinates to AB that passes through the point C .
210 of the midpoint of a line.
● Find and interpret the
Review gradients of parallel and
perpendicular lines.
IGCSE / O Level Divide a quantity in a given 3 The diagram shows a cuboid with CD = 5 cm,
Mathematics ratio. DH = 6 cm and EH = 3 cm.
Recall and apply properties B F
of prisms, other solids C
and plane shapes, such as G
trapezia and parallelograms. A E
UniversitCyoPprye-ssC-a5cRmmebrviidegwe
D 6 cm 3 cm
H
The point M lies on DH such that
DM :MH = 1:2.
The point N is the midpoint of EF .
Review a Write down the length of MH .
b Find the exact length of NH .
IGCSE / O Level Translate points in 4 The point P has coordinates (3, 9). The vector
Mathematics 2-dimensional space using
column vectors.
- Cambridge PQ = −1 .
ess - Review 3
Write down the coordinates of the point Q.
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Review Why are vectors important?
Vectors are used to model many situations in the world around us. They are used by WEB LINK
engineers to ensure that buildings are designed safely. They are used in navigation to
ensure that planes and boats, for example, arrive at their destinations in differing winds or Explore the Vector
water currents. There are countless examples of the use of vectors to model the movement geometry station on
of objects in mathematics and physics. Displacement, velocity and acceleration are all the Underground
examples of vector quantities. Vectors are essential to describing the position and location Mathematics website.
of objects. A boat lost at sea can be found if its bearing and distance from each of two
fixed points is known. These are just a few examples. Our world is 3-dimensional, as far as 211
we know, and in this chapter we will be studying vectors in two and three dimensions.
Review As the mathematics of vectors can be applied to all vectors, the study of vector geometry
and algebra in general is important.
It is helpful if you have a good understanding of simple coordinate geometry before you
start working with vectors because you will be able to make comparisons between the
coordinate geometry that you have previously studied and the vector geometry introduced
in this chapter.
Recall that a vector has both magnitude (size) and direction.
You should have some knowledge of how to work with vectors in 2-dimensional space
from your previous studies. You will be reminded of this here, before being introduced to
vectors in 3-dimensional space.
9.1 Displacement or translation vectors
A movement through a certain distance and in a given direction is called a displacement or
a translation. The magnitude or length of a displacement represents the distance moved.
Examples of displacement or translation vectors
● Sami walks 100 m in a straight line on a bearing of 045° from place A to place B. Jan is
also standing at place A. Jan walks directly to Sami. Jan therefore also walks for 100 m on
a bearing of 045° from place A. Jan follows the same displacement vector as Sami.
● The diagram shows two arrows.
y
3 Q
M
2
1
–2 –1 O 1 2 3x
–1
Review P
–2
L
–3
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ReviewThe arrow from L to M is the same length and has been drawn in the same direction as the TIP
arrow from P to Q.
In some books LM is
This means that the points L and P have been displaced or translated by the same vector to used for LM.
M and Q, respectively.
TIP
The arrow LM and the arrow PQ represent the same displacement.
When you write your
Notation using upper case vectors using the
correct notation, you
A displacement from L to M is written as LM. are less likely to make
errors.
The order of the letters is important as ML would represent a displacement from M to L.
The arrow above the upper case letters always goes from left to right. Reverse the letters to
show the opposite direction.
In the previous diagram, LM = PQ. Even though the arrows are drawn on different parts
of the page, the displacement shown by each is the same.
Components of a displacement in two dimensions
Any displacement can be represented by the displacement in the x-direction followed by the
displacement in the y-direction. These are called the components of the displacement.
In the following example, the displacement vector AB = 3 .
1
y 3 is the x-component and 1 is the y-component of the
3 displacement. As 3 and 1 are both positive, the movement is
212 2B
to the right and up, respectively.
Review 1
A
–3 –2 –1 O 1 2x
Components of a displacement in three dimensions
In a 3-dimensional system of coordinates, a point P is located using three coordinates,
(x, y, z). The axis for the third coordinate, the z-coordinate, intersects with the x- and
y-axes at the origin and the three axes are perpendicular to each other. The axes may be
rotated, as shown.
yz
z
P(x, y, z)
P(x, y, z) P(x, y, z) y
x
Review xy
zx
A displacement vector in a 3-dimensional system needs to have three components.
For example, the point P(1, 3, 7) translates to the point Q(5, 3, 8) by the displacement
4
vector PQ = 0 .
1
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Notation using lower case
Sometimes lower case letters are used for vectors. In text, these are in bold type.
For example, a or b.
When written with a pen or pencil, they should be underlined with a bar or a short, wavy line
instead. For example, a or b. Sometimes the bar or wavy line is written over the top of the
ɶ
lower case letter.
This diagram shows three vectors, a, b and c .
Review abc
The vectors a and b have the same direction but not the same magnitude or length. Vectors
a and b are parallel but not equal. In fact, a is half the length of b. When the length of a
is doubled, the vectors are equal. In other words b = 2a . When a is scaled so that it is the
same length as b, then the vectors are equal.
a
b
a
Vector c is 3 of the length of vector b, but it has opposite direction. TIP 213
In other 4 This may also
3 be written, for
words c = − 4 b. example, as
4c = −3b.
Review A quantity (usually a number) used to scale a vector is called a scalar. So in ‘2a’ the vector a Can you write an
expression for b in
is being multiplied by the scalar 2. terms of c ?
Parallel vectors C
Two vectors are therefore parallel provided that one is a scalar multiple of the other. B
For example, when a = λb, where λ is some scalar, then a and b are parallel.
Combining displacements (adding and subtracting)
Starting at point A, make the displacement AB = 4 followed by the displacement
2
3
BC = 7 .
What is the result?
The overall displacement shown by AC, is 7 . This is usually called the
resultant vector. 9
Review
● The x-component of the resultant vector is easily found by adding the
x-components of AB and BC. A
● The y-component of the resultant vector is easily found by adding the
y-components of AB and BC.
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This operation is called vector addition.
KEY POINT 9.1
The formula for vector addition:
AB + BC = AC
The displacement AC followed by the displacement CA must return you to point A.
In other words AC + CA = 0 , as there is no overall displacement.
0
Review 7 + −7 = 0 so CA = −7
9 −9 0 , −9 .
As −7 and −9 are both negative, the movement is to the left and down, respectively.
Alternatively, we may write 7 − 7 = 0 . In other words AC − AC 0
9 9 0 = 0 .
It is reasonable, therefore, to state that CA = −AC.
Review KEY POINT 9.2
CA = −AC is always true.
214
Taking the negative of a vector changes the direction of the vector and does not change
its length.
Vectors AC and CA have opposite direction.
As a − b is the same as a + −b, vector subtraction is defined as the addition of the negative
of a vector. The displacement a − b is therefore the displacement a followed by the
displacement −b.
EXPLORE 9.1
a, b and c are displacement vectors such that c = a + b.
Using diagrams to help you, show that a = c − b and that a + b = b + a.
Review KEY POINT 9.3
a + b = b + a means that the addition of vectors is commutative.
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WORKED EXAMPLE 9.1
QR
aM
Review P bS
PQRS is a quadrilateral. M is the midpoint of RS.
PQ = a, PS = b and PM = 1 (2a + 5b).
4
a Find an expression in terms of a and b for SM.
Give your answer in its simplest form.
b Find an expression for QR.
c Describe quadrilateral PQRS, justifying your answer.
Answer
a SM = SP + PM Substitute SP = −PS .
Substitute given vectors in a and b.
SM = −PS + PM or PM − PS Expand and simplify.
SM = 1 (2a + 5b) − b Use SR = 2SM and QP = −PQ. 215
4 Substitute known vectors in a and b.
Expand and simplify.
SM = 1 a + 5 b − b = 1 a + 1 b
24 24
Review b QR = QP + PS + SR
QR = −PQ + PS + 2SM
QR = −a + b + 2 1 a + 1 b
2 4
QR = 3 b
2
c QR = 3 PS so the lines QR and PS are parallel and
2
PQRS is a trapezium.
Review Magnitude of a vector TIP
The magnitude of a vector is the length of the line segment representing the vector. This is A diagram can often
found using the components of the displacement and Pythagoras’ theorem. be useful in vector
questions.
Notation: The magnitude of a vector p is written as p . This is also known as the modulus
of the vector.
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WORKED EXAMPLE 9.2
a Find the magnitude of each of the vectors a = 4 , −3 and a – b.
3 b = 4
5 −1
b p = 0 and q = 8 . Find the magnitude of the vector p + q.
8 −3
Answer
Review a Draw a diagram showing the x- and
y -components.
a3
4 Apply Pythagoras’ theorem.
| a | = 42 + 32 = 25 = 5 Again, draw a diagram to show the x- and
y-components.
b
4
216
Review 3 Apply Pythagoras’ theorem.
b = (−3)2 + 42 = 25 = 5 The diagram has been given to help you to see
the components of the vector a − b.
a 3 –b
–4
4
3
a−b = 7 = 72 + (−1)2 = 49 + 1 = 50 = 5 2 Apply Pythagoras’ theorem. (Notice that
−1 a − b is not the same as a − b .)
Review 5 −1 4
b p+q = 0 + 8 = 8 and It is not a good idea to try to draw a diagram
8 −3 5 here.
4 Apply Pythagoras’ theorem in 3 dimensions.
8 = 42 + 82 + 52 = 105
5
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Unit vectors
A unit vector is any vector of magnitude or length 1.
Notation: aˆ (read as a hat) is the unit vector in the direction of a.
To find a unit displacement vector in a particular direction:
● find the magnitude of the displacement
● then divide the components of the vector by the magnitude. p1
p
p2
p1 p .
This means that, for a vectorp p = p2 , the corresponding unit vector is pˆ =
p3 p3
Review p
WORKED EXAMPLE 9.3
5
Find the unit vector in the direction 2 .
Answer
3 The diagram shows that, geometrically, we 217
2 are finding lengths of similar triangles when
finding components of unit vectors.
Review 12
3 Apply Pythagoras’ theorem to find the
magnitude of the given vector.
√5 √5
3 Divide each component by the magnitude.
5 = ( 5 )2 + 22 = 9 =3
2
5
3
5 is .
Unit vector in the direction 2 2
3
WORKED EXAMPLE 9.4
Review 3 11
Points A, B and C are such that AB = 4 and BC = −2 .
1 −1
Find the unit displacement vector in the direction AC.
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Answer C AC is the resultant vector formed from
AB + BC.
B
A
AC = AB + BC
3 11 14
Substitute column vectors and sum components.
Review AC = 4 + −2 = 2
1 −1 0 Now find the magnitude by applying
Pythagoras’ theorem.
AC = 142 + 22 + 02 = 200 = 10 2 Divide each component by the magnitude of
AC .
Unit displacement vector in the direction of AC is
1 14
10 2 2
0 .
EXERCISE 9A A
218
1 The diagram shows the vectors
AB and BC on 1 centimetre
squared paper.
Review C
Using 1cm as 1 unit of displacement:
a write AB and BC as column B
vectors
b write down the column vector AC .
2 The diagram shows the vectors DE and DF on 1 centimetre squared paper.
F
Review E
D
Using 1cm as 1 unit of displacement:
a find the column vector EF
b show that DF − DE = EF .
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3Q
R
P
Justify the statement QR = PR − PQ.
Review P 4 In triangle ABC , X and Y are the midpoints of AB and AC, X A
respectively. B Y
AX = a and AY = b. C
G
a Use a vector method to prove that BC is parallel to XY . 219
b XY = k BC H
C
Write down the value of the constant k.
M
PS 5 The diagram shows a cuboid, ABCDEFGH. N B
F
z
yE
D
A
Review O x
M is the midpoint of BC and the point N is on FG such that FN : NG is 1:3.
12
Given that AG = 4 , find the displacement vector:
2
a AM b AN. A pB
q s
6 ABCDEFGH is a regular octagon. AB = p, AH = q, HG = r and HC = s. H
r
a Using the vectors p, q, r and s, write down expressions for: C
G D
i BC ii EB.
b Explain why s = kp and find the exact value of k.
Review FE
P 7 Use a vector method to show that, when the diagonals of a quadrilateral bisect one another, then the opposite
sides are parallel and equal in length.
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9.2 Position vectors
A position vector is simply a means of locating a point in space relative to an origin,
usually O. This means a position vector is fixed in space. Position vectors are sometimes
called location vectors.
Up to this point, the displacement vectors we have been working with have been free REWIND
vectors. This means they are not fixed to any particular starting point or origin. Many free
vectors can represent the same displacement. Look back at vectors
LM and PQ at the
Position vectors are different because they start at a given point or origin. This fixed start of this chapter.
starting point means there is only one arrow that represents each vector.
ReviewSince we can apply the same mathematics to all vectors, the results derived for
displacement vectors (free vectors) also apply to position vectors.
The fixed starting point is often called O, as in the origin of a system of Cartesian
coordinates.
1
So a point with coordinates A(1, 2, 3) can be represented by the position vector OA = 2 .
3
Position vector of any point in a plane
Consider this case in 2-dimensional space.
P
220
a
Review O
b
Given that a and b are position vectors relative to the origin O and that P is any other point
in the same plane as a and b, then the position vector of P can be written using a and b.
This is because OP is some scalar multiple of a plus some scalar multiple of b:
OP = ma + nb, where m and n are scalars.
All the other vectors in this plane can be written using combinations of the vectors a and b.
ReviewCartesian components
When the vectors a and b are
● perpendicular to each other
and
● unit vectors
the plane looks like the usual set of x- and y-axes (called the Oxy plane) we use for
Cartesian coordinates.
In this case, the unit vector in the x-direction is called i and the unit vector in the
y-direction is called j.
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y
P(x, y)
j
Oi x
OP = xi + yj, where x and y are scalars.
Review In this case, OP is some scalar multiple of i plus some scalar multiple of j:
The point P with coordinates (4, 5) has position vector 4 which can also be written as
4i + 5j. 5 ,
This can be extended to three dimensions.
The unit vector in the z-direction is called k.
z
Q(x, y, z)
kj y 221
Oi x
Review In this case, OQ is some scalar multiple of i plus some scalar multiple of j plus some scalar
multiple of k:
OQ = xi + yj + zk, where x, y and z are scalars.
4
The point Q with coordinates (4, 2, 3) has position vector 2 , which can also be written
as 4i + 2j + 3k. 3
WORKED EXAMPLE 9.5
The position vectors of the points A, B and C are given by a, b and c , respectively, where
2 0 −4
Review a = 5 , b = 4 and c= 2 .
−3 6 24
a Write a, b and c in terms of i, j and k.
b Giving your answers in terms of i, j and k, find the vectors representing:
i AB ii AC
c Describe the relationship between the points A, B and C , justifying your answer.
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Answer B
a a = 2i + 5j − 3k, b = 4j + 6k and c = −4i + 2j + 24k. A
b i A simple sketch should help you to calculate the
answer correctly.
AB = AO + OB = OB − OA O
Substitute the Cartesian form.
Review AB = b − a = 4j + 6k − (2i + 5j − 3k) Expand and simplify.
AB = −2i − j + 9k
ii AC = AO + OC = OC − OA Substitute the Cartesian form.
AC = c − a = −4i + 2j + 24k − (2i + 5j − 3k) Expand and simplify.
AC = −6i − 3j + 27k
c AC = 3AB and so these vectors are parallel. Since the vectors also have point A in common, then A, B and
C must be collinear (so they lie on the same straight line).
WORKED EXAMPLE 9.6
222 The diagram shows a cube OABCDEFG with sides of length 4 cm. Unit vectors i, j
and k are parallel to OC , OA and OD, respectively. The point M lies on FB such that
FM = 1 cm, N is the midpoint of OC and P is the centre of the square face DEFG .
Review
Express each of the vectors NP and MP in terms of i, j and k.
E F The cube is of side 4 cm and P′ is the centre of
GM the square AOCB. NP is parallel to the
P y-direction and so the first part of the vector NP
D is 2j.
k jA B P is directly above this point P′ and so the
Oi N C second component P′P is parallel to the
z-direction and is +4k.
Answer
Using the displacement of P from N and M :
NP = 2j + 4k
Review E F
P M
G
D
k jA B
P′
Oi N C
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MP = −2i − 2j + k
E G F Starting with the x-direction, move 2i left. This
P M is the opposite to the positive direction indicated
by the red arrow, so the first component is −2i.
D
In the y-direction, again we are moving in the
A B negative direction, so the second component is −2j.
kj
In the z-direction, we must move a distance
Review Oi C equal to MF . This is 1 unit and so the third
component is +k.
N
Or we can use position vectors:
NP = OP − ON OP is 2 units in x-direction, 2 units in
y-direction and 4 units in the z-direction and
OP = 2i + 2j + 4k and ON = 2i ON is 2 units in x-direction. 223
NP = OP − ON = 2i + 2j + 4k − 2i
Review NP = 2j + 4k Subtract vectors.
MP = OP − OM Simplify.
OP = 2i + 2j + 4k and OM = 4i + 4j + 3k OM is 4 units in x-direction, 4 units in
y-direction and 3 units in z-direction.
MP = OP − OM = 2i + 2j + 4k − (4i + 4j + 3k)
MP = −2i − 2j + k Subtract vectors.
Expand and simplify.
EXERCISE 9B
1 Relative to an origin O, the position vectors of the points A and B are given by OA = 5i + 3j and
OB = 5i + λ j + µk.
a In the case where λ = 0 and µ = 7, find the unit vector in the direction of AB.
b Given that triangle OAB is right-angled with area 5 34 and µ = 10, find the value of λ.
Review 2 The diagram shows a cuboid OABCDEFG with a horizontal base G 6 cm NF
OABC . The cuboid has a length OA of 8 cm, a width AB of 4 cm
and a height BF of d cm. The point M is the midpoint of CE and D M d cm
the point N is the point on GF such that GN = 6 cm. The unit C 8 cm E
vectors i, j and k are parallel to OA, OC and OD, respectively.
It is given that OM = 4i + 2j + k. kj B
Oi
a Write down the height of the cuboid. 4 cm
A
b Find ON.
PS c Find the unit vector in the direction MN.
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3 The position vectors of the points A, B, C and D are given by a, b, c and d, respectively, where
2 13 5 −6
a = 2 , b = 5 , c = −3 and d = −6 .
0 4 4 0
Review a i Find AD , BC , AB and DC .
ii Deduce that ABCD is a parallelogram.
b Find the coordinates of
i the point M , the midpoint of the line AB
ii the point P such that BP = 1 BD .
3
4 Relative to an origin O, the position vector of A is i − 2j + 5k and the position vector of B is 3i + 4j + k.
a Find the magnitude of AB.
b Use Pythagoras’ theorem to show that OAB is a right-angled triangle.
c Find the exact area of triangle OAB.
PS 5 The position vectors of the points A, B and C are given by
a = 5i + j + 2k, b = 7i + 4j − k and c = i + j + qk.
Given that the length of AB is equal to the length of AC, find the exact possible values of the constant q.
224 6 The diagram shows a triangular prism, OABCDE . The D
Review uniform cross-section of the prism, OAE, is a right-angled E 7 cm N B
triangle with base 24 cm and height 7 cm. The length, AB, of k C
the prism is 20 cm. The unit vectors i, j and k are parallel O 20 cm
to OA, OC and OE, respectively. The point N divides the j A
length of the line DB in the ratio 2:3. i 24 cm
a Find the magnitude of DB.
b Find ON.
PS 7 Three points O, P and Q are such that OP = 2i + 5j + ak and OQ = i + (1 + a)j − 3k.
Given that the magnitude of OP is equal to the magnitude of OQ, find the value of the constant a.
PS 8 Relative to an origin O, the position vectors of the points P and Q are given by
−6k 2k + 13
OP = −2 and OQ = −8 .
8(1 + k) −32k
Review Given that OPQ is a straight line:
a find the value of the constant k
b write each of OP and OQ in the form xi + yj + zk
c find the magnitude of the vector PQ.
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PS 9 An ant has an in-built vector navigation system! It is able to go on complex journeys to find food and then
return directly home (to its starting point) by the shortest route.
An ant leaves home and its journey is represented by the following list of displacements:
1 −3
i + 3j + 6k 6 10i −1 4i − k
0 −5
The ant finds food at this point.
What single displacement takes the ant home?
One unit of displacement is 10 cm.
Review What is the shortest distance home from the point where the ant finds food?
EXPLORE 9.2 225
A system of objects with the properties:
• can be represented by a column of numbers
• can be combined using addition
• can be multiplied by a number to give another object in the same system
is called a vector space.
Do quadratic expressions of the form ax2 + bx + c, where a, b and c are real constants, form a vector space?
Explain your answer.
Review 9.3 The scalar product
We have seen how to add and subtract vectors and also how to multiply a vector by
a scalar. It is also possible to multiply two vectors. It is possible to find two different
products by multiplying a pair of vectors. The first product gives an answer that is a vector.
This is called the vector product or cross product (for example, the cross product of
the vectors a and b is a × b). The cross product is not required for Pure Mathematics 3.
The second product gives an answer that is a scalar. This is called the scalar product or dot
product (the dot product of the vectors a and b, for example, is a ⋅b).
Introductory ideas
The angle between two vectors is defined to be the angle made
● either when the direction of each vector is away from a point
● or when the direction of each vector is towards a point.
Review θθ
Directions both away from a point. Directions both towards a point.
As vectors have magnitude and direction we can think of the scalar product as being
multiplication in relation to a particular direction.
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A useful way to understand this is to think about the work done by a person when they
pull an object along a flat surface using a rope:
work done by the person = force needed to pull the object × distance the object moves
Direction of pull a a
Displacement of object b
θ
x
Review The diagram shows vector a representing the pull and vector b representing the
displacement of the object.
The direction of the pull is not in the same as the direction of the displacement. This means
that only some of the force the person uses to pull the object makes it move. In fact, only
the x-component of the pull is in the direction of the displacement.
a is the magnitude of the pull. Using the angle made between the two directions, we can
see that the magnitude of the x-component of the pull is x = a cos θ .
b is the magnitude of the displacement, which is a measure of the distance moved.
Applying this information to the formula, we can see that the
( )work done by the person = a cosθ b .
Finding the scalar (number) that is work done by the person is one application of the scalar
product.
226
KEY POINT 9.4
In general, the geometric definition of the scalar product of two vectors is:
a ⋅ b = a b cosθ
Review
This formula allows us to calculate the value that represents the impact of one vector on
the other, taking account of direction.
WORKED EXAMPLE 9.7
8 4
p = −6 , q = −3 and the angle, θ, between the two vectors is 45°. Find the value of the scalar product, p ⋅ q.
−10 0
Review Answer
p ⋅q = p q cosθ
p = 82 + (−6)2 + (−10)2 = 10 2 and q = 42 + (−3)2 + 02 = 5
p ⋅q = (10 2 ) (5) cos 45° = (10 2 ) (5) 2 = 50
2
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Special case 1: Parallel vectors a
b
When two parallel vectors, a and b, have the same direction (not opposite directions),
θ = 0° or 0 radians. b
Therefore a⋅b = a b cos 0
This also means that, for example: a⋅b = a
a⋅a = a b
a or a 2
a
Review Special case 2: Perpendicular vectors
When two vectors, a and b, are perpendicular, θ = 90° or π radians.
2
a ⋅b = a b cos 90°
a⋅b = 0
KEY POINT 9.5
This is very important. When two vectors are perpendicular, their scalar product is zero. It is also
true that when the scalar product is zero, the two vectors are perpendicular.
EXPLORE 9.3 227
Review Show that the scalar product has the following properties:
a⋅b = b⋅a
a ⋅ mb = ma ⋅b = m(a ⋅b), where m is a scalar
a⋅(b + c) = a⋅b + a⋅c
(a + b)⋅(c + d) = a ⋅c + a ⋅d + b ⋅c + b ⋅d
Review Scalar product for component form
When we write a = a1i + a2j + a3k and b = b1i + b2j + b3k the properties of the scalar
product give an algebraic method of finding the same value that we found using
a b cos θ.
a ⋅ b = (a1i + a2j + a3k)(b1i + b2j + b3k)
= a1i(b1i + b2j + b3k) + a2j(b1i + b2j + b3k) + a3k(b1i + b2j + b3k)
= a1b1i ⋅ i + a1b2i ⋅ j + a1b3i ⋅ k
+ a2b1j ⋅ i + a2b2j ⋅ j + a2b3j ⋅ k
+ a3b1k ⋅ i + a3b2k ⋅ j + a3b3k ⋅ k
However, as i is parallel to i, j is parallel to j and k is parallel to k then:
i⋅i = j⋅j = k⋅k = 1
Also, as i, j and k are perpendicular to each other, scalar products such as i ⋅ j, j⋅ k, etc. are
all 0.
a ⋅ b = a1b1i ⋅ i + a1b2i ⋅ j + a1b3i ⋅ k
+ a2b1j ⋅ i + a2b2j ⋅ j + a2b3j ⋅ k
+ a3b1k ⋅ i + a3b2k ⋅ j + a3b3k ⋅ k
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KEY POINT 9.6
In general, the algebraic definition of the scalar product of two vectors is:
a ⋅ b = a1b1 + a2b2 + a3b3
a1 b1
a2 b2
When written in column form, a = and b = , so it is perfectly valid to
a3 b3
Reviewuse the components of column vectors in the algebraic form of the scalar product
a ⋅ b = a1b1 + a2b2 + a3b3.
We now have a way to find the dot product when we do not know the angle between the
two vectors.
Sometimes the two versions of the scalar product are written together like this:
a ⋅ b = a1b1 + a2b2 + a3b3 = a b cosθ
This can be useful as it allows us to find the angle between the two vectors when we know
the components of the vectors.
WORKED EXAMPLE 9.8
228 p = 8i − 6j − 10k and q = 4i − 3j
Show that the angle, θ , between the two vectors is 45°.
Review Answer Find the value of the scalar product p ⋅q.
p ⋅ q = p1q1 + p2q2 + p3q3 Find the magnitude of p and of q using
p ⋅q = 8(4) + (−6)(−3) + (−10)(0) = 32 + 18 = 50 Pythagoras’ theorem.
p = 82 + (−6)2 + (−10)2 = 10 2 Use p ⋅q = p q cosθ.
Make cosθ the subject.
q = 42 + (−3)2 + 02 = 5 Solve for θ .
50 = (10 2 ) (5) cosθ
50 = cosθ
50 2
θ = cos−1 1 = 45°
2
Review
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WORKED EXAMPLE 9.9
Relative to the origin O, the position vectors of the points A and B are given by
5 −1
OA = −2 and OB = 3 .
4 −1
Find angle OBA. TIP
A sketch will help
Review Answer you to make sure
you choose a correct
A pair of directions
for your scalar
OB and AB are the two directions product.
towards the point B, so these are the
vectors we need to use. 229
O First of all, find the column vector AB .
B
Find OB ⋅ AB.
−1 5 −6 −1
= − = − =
AB OB OA 3 −2 5 Using the given vector OB = 3 ,
−1 4 −5 −1
find the value of OB ⋅ AB.
OB ⋅ AB = (−1)(−6) + (3)(5) + (−1)(−5) = 26 Find the magnitude of OB and of
AB using Pythagoras’ theorem.
Review
−6
(−6)2 + 52 + (−5)2 =
AB = 5 = 86
−5
−1
= = (−1)2 + 32 + (−1)2 =
OB 3 11 Use OB ⋅ AB = OB AB cosOBA
−1
26 = 86 11 cosOBA Make cosOBA the subject.
26 = cosOBA Solve for angle OBA.
86 11
Review Angle OBA = 32.3°
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EXERCISE 9C
1 Find which of the following pairs of position vectors are perpendicular to one another.
For any position vectors that are not perpendicular, find the acute angle between them.
2 7 5 −3 4 −4
a a = 8 , b = −1 b c = 12 , d = −1 c e = −9 , f = −2
−2 3 13 3 −2 1
−5 1
2 Relative to the origin O, the position vectors of the points A and B are given by OA = 0 and OB = 7 .
3 2
Review
Find angle BOA.
3 Given that the vectors 6i − 2j + 5k and ai + 4j − 2k are perpendicular, find the value of the constant a.
4 Relative to the origin O, the position vectors of the points P and Q are given by
5k k
OP = −3 , OQ = k+2 where k is a constant.
7k + −1
9
a Find the values of k for which OP and OQ are perpendicular to one another.
b Given that k = 2, find the angle, θ , between OP and OQ.
230 5 The diagram shows a cuboid OABCDEFG in which OC = 6 cm, E F
OA = 4 cm and OD = 3 cm. Unit vectors i, j and k are parallel to OC, OA P M
and OD respectively. The point M lies on FB such that FM = 1cm, N is
Review
the midpoint of OC and P is the centre of the rectangular face DEFG. D G
Using a scalar product, find angle NPM. A B
kj
O iN C
G M
6 The position vector a is such that a = 4i − 8j + k. Find a ⋅ j and hence
find the angle that a makes with the positive y-axis. F
E
7 Explain why a ⋅b ⋅c has no meaning. D
B
8 The diagram shows a cube OABCDEFG with side length 4 units. The unit N
A
vectors i, j and k are parallel to OA, OC and OD, respectively. The point M is
the midpoint of GF and AN : NB is 1:3. C
a Find each of the vectors OM and NG. kj
Review b Find the angle between the directions OM and NG. Oi
PS 9 The diagram shows a triangular prism, OABCDE . D
The uniform cross-section of the prism, OAE, is a right- M
angled triangle with base 77 cm and height 36 cm. The E
length, AB, of the prism is 60 cm. The unit vectors i, j and
k are parallel to OA, OC and OE, respectively. The point 36 cm
N divides the length of the line DB in the ratio 4:1 and M
is the midpoint of DE . k C N B
j 60 cm
Use a scalar product to find angle MAN .
O i 77 cm A
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PS 10 The diagram shows a pyramid VOABC . The base of the pyramid, OABC , V
is a rectangle. The unit vectors i, j and k are parallel to OA, OC and OV ,
respectively. The position vectors of the points A, B, C and V are given by N
OA = 6i, OC = 3j and OV = 9k.
The points M and N are the midpoints of AB and BV , respectively. k jC B
a Find the angle between the directions of AN and OC. O i M
b Find the vector MN. A
c The point P lies on OV and is such that PNM is a right angle. Find the
231
position vector of the point P.
Review 9.4 The vector equation of a line
We already know that, in 2-dimensional space, a line can be represented by an equation
of the form y = mx + c or ax + by = c. In 3-dimensional space, we need a new approach
because introducing the third dimension gives an equation of the form ax + by + cz = d,
which is actually the Cartesian equation of a plane. One way around this problem is to
write the equation of the line using vectors.
The equation of a line in vector form is valid in both 2- and 3-dimensional space.
So what information do we need to be able to do this?
When a line passes through a fixed given point and continues in a known direction then it
is uniquely positioned in space.
Therefore, to form the vector equation of a straight line, we need:
Review ● a point on the line, A(a1, a2 , a3 ), and a direction vector, b = b1 , that is parallel to
b2
the line. b3
A (a1, a2, a3) R (x, y, z) b1
b = bb23
Review O (0, 0, 0) TIP
R is any point on the line and so OR represents the position vector of any point on the Sometimes Greek
line. letters such as λ and µ
To reach R, starting at O, take the path OR = OA + AR. are used as parameters.
OA is a vector we know. To move from A to R, we need to travel in the direction of b.
Multiplying b by different scalars changes the position of R on the line, so simply gives
us different points on the line. Using the parameter t to represent the variable scalar
OR = OA + tb
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KEY POINT 9.7
Using lower case notation for vectors, the vector equation of the line is:
r = a + tb
● Or two points on the line, A(a1, a2, a3 ) and C(c1, c2, c3 ). One can be used as the fixed
given point and the direction can be found from the displacement between the two points.
C (c1, c2, c3) R (x, y, z)
A (a1, a2, a3)
Review
O (0, 0, 0)
TIP
The situation is almost the same. In this case, however, we do not know the direction of
the line. We can find the direction using the position vectors of the points A and C . Any known point can
be used for the position
Using the parameter t as before, vector, so you could
OR = OA + tAC. also use
OR = OC + tAC
KEY POINT 9.8
232 Using lower case notation for vectors, the vector equation of a line is:
r = a + t(c − a)
ReviewIn each case, the structure is:
general point on line = position of known point + some way along a known direction
Parametric form
R(x, y, z) represents any point on the line. x
The position vector of this point can be written as r = y or r = xi + yj + zk.
z
In the equation r = a + tb, using the position vectors in component form:
x a1 b1
a2 b2
y = a3 + t b3 or xi + yj + zk = a1i + a2j + a3k + t(b1i + b2j + b3k)
z
ReviewSimplifying the right-hand side of this:
x a1 + tb1 xi + yj + zk = a1i + tb1i + a2j + tb2j + a3k + tb3k
a2 + tb2 xi + yj + zk = (a1 + tb1)i + (a2 + tb2 )j + (a3 + tb3 )k
y = a3 + tb3 or
z
Comparing the components:
x = a1 + tb1
y = a2 + tb2
z = a3 + tb3
These three equations are called the parametric equations of the line.
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WORKED EXAMPLE 9.10
A line passes through the point with position vector 3i – 2j + 4k and is parallel to the vector i – j + k.
Find:
a the vector equation of the line
b the parametric equations of the line.
Review Answer General point on line =
Using t as the parameter:
a General point = r
Known point = 3i – 2j + 4k position of known point
+ some way along a known direction.
Known direction = i – j + k
Write r as a column vector and collect
3 1 together the x-, y- and z-components of the
right-hand side.
r = −2 + t −1
4 1 Compare components.
3 1
b r = −2 + t −1 or r = 3i – 2j + 4k + t(i − j + k) 233
4 1
Review x 3+t
y = −2 − t or
z 4 + t
xi + yj + zk = (3 + t)i + (–2 − t)j + (4 + t)k
x = 3 + t y = −2 − t z = 4 + t
WORKED EXAMPLE 9.11
Review The position vectors of the points A and B are given by
a = 3i + 2j − 7k and b = i − 4j + 8k,
respectively.
a Write down the vector AB.
b Find the vector equation of the line AB.
Answer Substitute for a and b.
Expand and simplify.
a AB = b − a
AB = i − 4j + 8k − (3i + 2j − 7k)
AB = −2i − 6j + 15k
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lines is the angle between
the two direction vectors
Cambridge International AoSf t&heAlinLees.vel Mathematics: Pure Mathematics 2 & 3
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b Using a = 3i + 4j − 7k as the known
point, t as the parameter and AB from
part a as the known direction:
General point = r General point on line =
Known point = 3i + 4j − 7k position of known point REWIND
+ some way along a known direction.
Known direction = −2i − 6j + 15k Recall, you could also
use b = i − 4j + 8k as
r = 3i + 2j − 7k + tAB the known point.
r = 3i + 2j − 7k + t(−2i − 6j + 15k)
Review KEY POINT 9.9
The angle between two lines is the angle between the two direction vectors of the lines.
WORKED EXAMPLE 9.12
The vector equation of the line L1 is r = 3i + j − k + λ(i − 8j + 5k).
a The line L2 is parallel to the line L1 and passes through the point A(0, 2, 5).
Write down the vector equation of the line L2.
234 b Find the angle between the line L1 and the line r = 4i + 3j − 2k + µ(5i − j + 2k).
Review Answer General point on line =
a Using µ as the parameter, position of known point
General point is r + some way along a known direction point.
Known point is 2j + 5k Find the scalar product d1 ⋅d2
Now find the magnitude of d1 and d2 using
The direction of the line L1 is i − 8j + 5k Pythagoras’ theorem
The line L2 is therefore
r = 2j + 5k + µ(i − 8j + 5k)
b Find the angle between
d1 = i − 8j + 5k and d2 = 5i − j + 2k.
d1 ⋅d2 = (1)(5) + (−8)(−1) + 5(2) = 23
Review d1 = 12 + (−8)2 + (5)2 = 3 10 Use d1 ⋅ d2 = d1 d2 cosθ .
d2 = 52 + (−1)2 + 22 = 30
23 = 3 10 30 cosθ Make cos θ the subject.
cosθ = 23 Solve for θ .
3 10 30
cos−1 3 23 30 = 63.727 … ° = 63.7°
10
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EXERCISE 9D
1 Write down the vector equation for the line through the point A that is parallel to vector b when:
a A(0, −1, 5) and b = 2i + 6j – k
b A( 0, 0, 0 ) and b = 7i – j – k
c A(7, 2, −3) and b = 3i – 4k.
Review 2 Write each of your answers to question 1 in parametric form.
3 Show that the line through the points with position vectors 9i + 2j – 5k and i + 7j + k is parallel to the line
r = λ(16i − 10j − 12k).
4 Find the parametric equations of each of the lines.
a r = 2i + 13j + k + t(i + j − k)
b r = 10j + t(2i + 5j)
c r = i − 3j + t(2i + 3j + 4k)
5 a Find the parametric equations of the line L through the points A(0, 4, −2) and B(1, 1, 6).
b Hence find the coordinates of the point where the line L crosses the Oxy plane.
PS 6 The line L1 has vector equation r = 2i – 3j + k + λ(6i + j + 3k) and the line L2 has parametric equations 235
x = µ + 4, y = µ − 7, z = 3µ.
Review
a Show that L1 and L2 are not parallel.
b Find the angle between the line L1 and the line L2.
PS 7 The vector equation of the line L1 is given by
x −3 4
y 0
= 8 + t −1 .
z −3
a Find the vector equation of the line L2 that is parallel to L1 and which passes through the point A(5, −3, 2).
b Show that A(5, −3, 2) is the foot of the perpendicular from the point B(4, −7, 2) to the line L2.
PS 8 The points A, B and C are position vectors relative to the origin O, given by
3 0 1
OA = 1 , OB = −1 and OC = 2 .
5 2 3
Review a Find a vector equation for the line L passing through A and B.
b The line through C , perpendicular to L, meets L at the point N. Find the exact coordinates of N .
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Review9.5 Intersection of two lines
In 2-dimensional space, two lines are either parallel or they must intersect at a point.
In 3-dimensional space it might be that:
● they are parallel
● or they are not parallel and intersect
● or they are not parallel and do not intersect.
Lines that are not parallel and do not intersect are called skew.
Parallel lines have the same direction vectors and so if a pair of lines is parallel, this can
easily be seen by examining the vector equations of the lines.
Non-parallel intersecting lines
In 2-dimensional space, to find the point of intersection of a pair of non-parallel lines
using their Cartesian equations ( ax + by = c ), we solve the equations of the lines
simultaneously. In other words, we put the two equations equal to each other.
The process is the same in 3-dimensional space and using the vector equations of the lines.
WORKED EXAMPLE 9.13
Given that the lines with vector equations
r = i − 2j − 4k + λ(−i − j + k)
236
and r = 2i − 2j + k + µ(i + 2j − 7k)
intersect at the point A, find the coordinates of A.
Review
Answer
The lines are equal at the point of intersection, so write the vector equation in
parametric form so that the components can be equated.
r = i − 2j − 4k + λ(−i − j + k) r = 2i − 2j + k + µ(i + 2j − 7k)
x =1−λ x =2+µ
y = −2 − λ y = −2 + 2µ
z = −4 + λ z = 1 − 7µ
Now equate the components.
1− λ = 2 + µ (1)
−2 − λ = −2 + 2µ (2)
−4 + λ = 1 − 7µ (3)
Review Solve these equations to find the value of λ and of µ:
From (1): λ = −1 − µ
Substitute into (3): −4 − 1 − µ = 1 − 7µ
6µ = 6 µ = 1
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When µ = 1: λ = µ = 1: λ = −1 − 1 = −2
Check:
λ == −2 µ =1 TIP
x = 1 − (−2) = 3 x = 2+1= 3 It is good to check that
y = −2 − (−2) = 0 y = −2 + 2 = 0 the value of λ and of
z = −4 − 2 = −6 z = 1 − 7 = −6 µ that we have found
give us the same point.
The lines intersect at the point A(3, 0, −6).
Review Skew lines
Skew lines have no point of intersection.
When two lines are skew, they are not parallel and they do not intersect.
WORKED EXAMPLE 9.14
Show that the lines with vector equations 237
r = 4i + 7j + 5k + λ(−2i + j + 3k)
Review r = i + j + 10k + µ(−6i + j + 4k)
do not intersect.
Answer
As −2i + j + 3k is not a scalar multiple of −6i + j + 4k the lines are not parallel.
Writing the equations in parametric form: r = i + j + 10k + µ(−6i + j + 4k)
x = 1 − 6µ
r = 4i + 7j + 5k + λ(−2i + j + 3k) y = 1+µ
x = 4 − 2λ z = 10 + 4µ
y =7+λ
z = 5 + 3λ
Now equate the components.
4 − 2λ = 1 − 6µ (1)
7 + λ = 1+µ (2)
5 + 3λ = 10 + 4µ (3)
Review Attempt to solve these equations to find the value of λ and of µ that should give a consistent point:
From (2): λ = µ − 6
Substitute into (3): 5 + 3( µ − 6) = 10 + 4µ
When µ = −23: µ = −23
λ = −23 − 6 = −29
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Check:
λ == −29 µ == −23
x = 4 − 2(−29) = 62 x = 1 − 6(−23) = 139
y = 7 − 29 = −22 y = 1 + (−23) = −22
z = 5 + 3(−29) = −82 z = 10 + 4(−23) = −82
The values of λ and µ do not give a consistent point, so there is no point of intersection and the lines are skew.
Review EXERCISE 9E TIP
1 Find whether the following pairs of lines are parallel, intersecting or skew. • Always carry out the
check with both λ
State the position vector of the common point of any lines that intersect. and µ as it is easy to
make a mistake.
a x = 3λ + 4 x = µ+1
• Re-read the question
y = 5λ + 1 y= µ+4 to check whether you
were asked to find the
z = 4λ − 3 z = 2µ + 5 position vector or the
coordinates of the
b r = i − 5j + 4k + λ(3i − 4j + 9k) r = i − 5j + 4k + λ(−6i + 8j − 18k) point of intersection
r = 4j + k + µ(8i + j + 14k) and give your answer
c r = 7i + 2j + 6k + λ(i + 3j + 9k) r = 4i − 9j + 2k + µ(−i − 8j − k) in the correct form.
d r = 2i − 3j + 4k + λ(6i + 4j −2k)
238
PS 2 Given that the lines with vector equations
Review r = 2i + 9j + k + λ(i – 4j + 5k)
and r = 11i + 9j + pk + µ(−i − 2j + 16k)
intersect at the point P, find the value of p and the position vector of the point P.
3 Three lines, L1, L2 and L3, have vector equations
r = 16i − 4j − 6k + λ(−12i + 4j + 3k)
r = 16i + 28j + 15k + µ(8i + 8j + 5k)
r = i + 9j + 3k + ν(4i − 12j − 8k)
The 3 points of intersection of these lines form an acute-angled triangle.
For this triangle, find:
a the position vector of each of the three vertices
b the size of each of the interior angles
c the length of each side.
Review 4 The line L1 passes through the points (3, 7, 9) and (−1, 3, 4).
a Find a vector equation of the line L1.
b The line L2 has vector equation
r = i + 2j + k + µ(3j + 2k).
Show that L1 and L2 do not intersect.
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