Q & A STPM 2022 - Mathematics (T)

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Term 1 2 3 Q & A STPM Q



Mathematics (T) [954] &

Q & A STPM fulfils the needs of students in mastering the
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FEATURES TITLES IN THIS SERIES: Mathematics
• Based on the latest syllabus ■ Pengajian Am
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• Common Errors to show commom ■ Ekonomi
mistakes frequently made by students ■ Pengajian Perniagaan STPM
while answering questions ■ Mathematics (T)
• Model Paper which follows the latest ■ Physics
STPM assessment format to prepare ■ Chemistry
students for the actual exam ■ Biology . .
Pengajian Am Ekonomi Physics Term 1 . 2 . 3 Term 1 2 3
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Lee Yoon Woh • Khoo Ee Sin • Tan Guan Hin
PELANGI



CVR_Q&A_STPM_2022.indd 2-3 7/26/21 1:58 PM

CONTENTS
CONTENTS


STPM Scheme of Assessment ........................................................................................... ii
Term 1
Chapter 1 Functions 1
Chapter 2 Sequences and Series 64

Chapter 3 Matrices 88
Chapter 4 Complex Numbers 125

Chapter 5 Analytic Geometry 144
Chapter 6 Vectors 163

STPM Model Paper 954/1 ................................................................................................ 191
Term 2
Chapter 1 Limits and Continuity 194

Chapter 2 Differentiation 212
Chapter 3 Integration 241

Chapter 4 Differential Equations 272
Chapter 5 Maclaurin Series 301

Chapter 6 Numerical Methods 318
STPM Model Paper 954/2 ................................................................................................ 344
Term 3

Chapter 1 Data Description 347
Chapter 2 Probability 382

Chapter 3 Probability Distributions 405
Chapter 4 Sampling and Estimation 429

Chapter 5 Hypothesis Testing 460
Chapter 6 Chi-Squared Tests 496

STPM Model Paper 954/3 ................................................................................................ 530
Answers .................................................................................................................................. 533

iv





00 Contents.indd 4 4/11/19 1:54 PM

Chapter Mathematics T Term 1 STPM Chapter 6 Vectors

6 Vectors







Question 1
1 –22 Term
1
Find the unit vector in the same direction as a = .
1
Answer:
1
 2
2
a = 1 –22 , |a| = 1 + (–2)

= 5
^
The required unit vector, a = a
|a|
1
1

=  1 –22
5
1
5
= 1  2 2
–

5
Note:
A unit vector is a vector with a magnitude of 1.
The 2-D and 3-D space can be easily built using the following unit
vectors.
0
1
2-D i = 1 0 2 , j = 1 1 2
0
0
1
1 2 1 2 1 2
3-D i = 0 , j = 1 , k = 0
0 0 1
Question 2
State the position vectors of the points with coordinates
(a) B(−2, 6) (b) D(−4,−9)

Answer:
(a) The position vector of point B is −2i + 6j.
(b) The position vector of point D is −4i − 9j.



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Mathematics T Term 1 STPM Chapter 6 Vectors

Question 3
Sketch the position vectors of:
r 1 = 2i + 3j, r 2 = −3i + 4j, r 3 = −2i − j, and r 4 = 4i − 4j.

Answer:
The vectors are shown below.
y
(–3, 4)
Term
4
1 (2, 3)
r 2
2
r 1
x
–4 –2 0 2 4
r 3
(–2, –1)
–2 r 4
–4
(4, –4)

Note:
The position vector is represented by a vector, joining origin to the
position of point object. Position vector in terms of components
(coordinates) is:
r = xi +yj (for two dimensions)
r = xi +yj + zk (for three dimensions)
where i, j and k are unit vectors in x, y and z directions.
a
a
0
1
0
1 b 2 = ai + bj = a 1 0 2 1 1 2 1 0 2 1 b 2
+ b
+
=
1
0
0
a
1 2 1 2 1 2 1 2
b = ai + bj + ck = a 0 + b 1 + c 0
c 0 0 1
0
0
a
1 2 1 2 1 2
= 0 + b + 0
0 0 c
Question 4
The coordinates of a point P are (3, 4, 5). Find the position vector of P and
represent this position vector on a three dimensional graph.
164





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Mathematics T Term 1 STPM Chapter 6 Vectors
Answer:
The position vector of P is,
p = 3i + 4j + 5k
The vector p is shown below

z z
5
k P(3, 4, 5) Term
P(3, 4, 5)
j
y 1
O
O y
i 4
3
x
x

Question 5
^
^
If a is a unit vector, express a vector in the same direction as a but having
modulus 5.
Answer:
^
The unit vector, a = a
|a|
^
The required vector, a = |a|a
^
= 5a

Question 6
A point A is given as (−2, 1, 2).
(a) Find the position vector of A.
(b) Find the unit vector that is in the same direction as the position vectors
of A.

Answer:
(a) Denoting the position vectors of A by a, then
a = −2i + j + 2k
(b) It has the magnitude, |a| = (–2) + 1 + 2
2
 2 2
= 3
^
The required unit vector, a = a = –2i + j + 2k
|a| 3
1
2
2
= – i + j + k
3 3 3
165




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Mathematics T Term 1 STPM Chapter 6 Vectors

Question 7
Find two unit vectors that are collinear with:
(a) a = 3i + 4j
(b) b = −2i – j + 3k


Answer:
(a) a = 3i + 4j
The magnitude, |a| = 3 + 4
 2
2
Term
1 = 5 a 3i + 4j
^
The required unit vectors, a = ± = ±
|a| 5
Thus, two unit vectors collinear with a are
4
3 i + j and – i – j
3
4
5 5 5 5
Note:
The answer can be plus or minus as collinear vectors need not be in the
same direction.
(b) b = −2i – j + 3k
The magnitude, |b| = (–2) + (–1) + 3
 2
2
2

= 14
^
The required unit vectors, b = ± b = ± –2i – j + 3k
|b| 
14
Thus, two unit vectors collinear with b are



14
14
– 14 i –  j + 3 14 k and  i +  j – 3 14 k
14
7 14 14 7 14 14
Question 8
The position vectors a, b and c of three points A, B and C respectively are given
by, a = 3i + 2j, b = 6i – 5j, and c = –4i + j. Find
(a) 3(2)b + 2(5)c
(b) 7(a – 2b)
Answer:
(a) 3(2)b + 2(5)c = 6(6i – 5j) + 10(–4i + j)
= (36 – 40)i + (–30 + 10)j
= –4i – 20j



166





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Mathematics T Term 1 STPM Chapter 6 Vectors

(b) 7(a – 2b) = 7a – 14b
= 7(3i + 2j) – 14(6i – 5j)
= (21 – 84)i + (14 + 70)j
= –63i + 84j

Note:
For any scalars k and l, and any vectors a and b
1. k(a + b) = ka + kb Term
2. (k + l)a = ka + la
3. k(l)a = (kl)a
1



Question 9
The position vectors p, q and r of three points P, Q and R respectively are given
by, p = −3i + 5k, q = 2i – 2j + 3k, and r = 4i + 2j – 6k. Find
1
(a) 2p + 3q + r (b) p – 2q + r
2
Answer:
(a) 2p + 3q + r = 2(−3i + 5k) + 3(2i – 2j + 3k) + (4i + 2j – 6k)
= (−6 + 6 + 4)i + (−6 + 2)j + (10 + 9 − 6)k
= 4i – 4j + 13k
1
1
(b) p – 2q + r = (−3i + 5k) − 2(2i – 2j + 3k) + (4i + 2j – 6k)
2 2
= (−3 – 4 + 2)i + (4 + 1)j + (5 − 6 − 3)k
= –5i + 5j – 4k


Question 10
→ →
ABC is a triangle, AB = a and BC = b. If M is the mid-point of AB, express the
→ →
vectors CA and MC in terms of a and b.

Answer:
→ → → Alternative method
CA = CB + BA C → →
→ → CA = –AC
= –BC – AB b → →
= –b – a ~ = –(AB + BC)
= –(b + a) B = –(a + b)
= –(a + b) = –a – b
a
A ~

167





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Mathematics T Term 1 STPM Chapter 6 Vectors


C
Tips
→ → → Tips
MC = MB + BC
= 1 AB + BC It is advisable to answer
2 the question with the
= 1 a + b B help of a graph.
2 M
A

Term
1 Question 11
EFGH is a parallelogram. State a single vector equal to each of the following.
→ → → →
(a) EF + EH (b) GH – GF
→ → →
(c) FH + GF + HG
H G





E F

Answer:
→ → → →
(a) Given that EFGH is a parallelogram, we have EF = HG and EH = FG.
→ → → →
Therefore EF + EH = EF + FG
→
= EG
from the properties of addition of vectors in the parallelogram EFGH.
→ →
(b) From the parallelogram, we have FG = −GF.
→ → → →
Thus, GH – GF = GH + (–GF)
→ →
= GH + FG
→
= FH

→ → →
(c) Since FH + HG = FG,
→ → → → →
we have FH + GF + HG = FG + GF
→ → → →
= –GF + GF FG = –GF
→
= 0 , which is a zero vector.


168





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Mathematics T Term 1 STPM Chapter 6 Vectors

Question 12
→ → →
If vectors p, q and r are represented by OP, OQ and OR in the tetrahedron as
→ → →
shown below. Find PQ, PR and RQ in terms of p, q and r.
R


Q Term

1
O
P

Answer:
→ →
From the tetrahedron, p + PQ = q, we have PQ = q – p.
→ →
Likewise, p + PR = r, or PR = r – p.
→ →
And, r + RQ = q, therefore RQ = q – r


Question 13
→
If AB = 3i + j + 4k and a point B is located at (2, –5, 3), determine the coordinates
of A.


Answer:
Let point A be located at (a 1 , a 2 , a 3 ), then Tips
→ → → Tips
AB = OB – OA Given two points
3i + j + 4k = (2, –5, 3) – (a 1 , a 2 , a 3 ) P 1 (x 1 , y 1 , z 1 ) and P 2 (x 1 , y 1 , z 1 )
3
= (2 – a 1 )i + (–5 – a 2 )j + (3 – a 3 )k in R , the displacement
vector from P 1 to P 2 is
Equating respective compenents of both rides: P 1 P 2 = (x 2 – x 1 , y 2 – y 1 , z 2 – z 1 )
2 – a 1 = 3 , –5 – a 2 = 1 , 3 – a 3 = 4 = (x 2 – x 1 )i + (y 2 – y 1 )j
+ (z 2 – z 1 )k
a 1 = –1 a 2 = –6 a 3 = –1
Thus, the coordinates of A is (–1, –6, –1).



Question 14
Show that the three points J(2, –1, –8), K(0, –5, –2) and L(–3, –11, 7) are collinear.



169





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Mathematics T Term 1 STPM Chapter 6 Vectors
Answer:
→
JK = (0 – 2)i + (–5 + 1)j + (–2 + 8)k = –2i –4j + 6k and
→
KL = (–3 – 0)i + (–11 + 5)j + (7 + 2)k = –3i –6j + 9k
→ 2 →
so, JK = 3 KL. Since the point K is common to both vectors, the three points
are collinear.

Term
1 Question 15
The position vectors p, q, r of the points P, Q, R, respectively, are given by
p = 2i + 2j + 3k,
q = –i + 5j + 2k,
r = 3i + 7j + 5k.
Find the position vector of the point S which forms a parallelogram PQRS
together with three other given points P, Q and R.


Answer:
Let point S be located at (s 1 , s 2 , s 3 ). Common error
Since PQRS is a parallelogram, we have
→
→
PQ = SR. There are students who finding
→
→
→
→
That is (–1 – 2)i + (5 – 2)j + (2 – 3)k PQ = RS instead of PQ = SR
= (3 – s 1 )i + (7 – s 2 )j + (5 – s 3 )k.
Or, s 1 = 6, s 2 = 4 and s 3 = 6.
The required position vector of the point S is 6i + 4j + 6k.
Question 16
The position vectors a and b of two points A and B respectively are given by,
a = 3i + 2j, b = 6i – 5j.
Find
(a) the scalar product, a · b
(b) the angle between the vectors a and b.


Answer:
(a) a = 3i + 2j, b = 6i – 5j
a · b = (3i + 2j) · (6i – 5j)
= (3)(6) + (2)(−5) Respective components are multiplied
= 8 together and the results are added.



170





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Mathematics T Term 1 STPM Chapter 6 Vectors

Note:
Dot Product (Scalar product)
1. If a = (a 1 , a 2 ) and b = (b 1 b 2 ), then the dot product of a and b is the
scalar
a · b = a 1 · a 2 2 = a 1 b 1 + a 2 b 2
1 b 1 2 1 b 2
2. If a = (a 1 , a 2 , a 3 ) and b = (b 1 , b 2 , b 3 ), then the dot product of a and b
is the scalar Term
1 2 1 2
a 1
b 1
a · b = a 2 · b 2 = a 1 b 1 + a 2 b 2 + a 3 b 3 1
a 3 b 3

(b) |a| = 3 + 2 , |b| = 6 + (–5)
2
 2
2
 2


= 13 = 61 Note:
cos q = a · b If the angle between
|a||b| two vector a and b is q,
= 8 From part (a) then a · b = |a||b| cos q.

13 61
≈ 0.2841
The angle between the vectors a and b is, q = cos (0.2841)
–1
= 73.5°
Question 17
The position vectors a and b are shown in the figure below. Vector a has
modulus 5 and vector b has modulus 7 and the angle between them is 60°.
Calculate a · b.

b
60°
a

Answer:
The angle between the two vectors is 60°.
a · b = |a||b| cos q
= 5 × 7 cos 60°
= 17.5






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Mathematics T Term 1 STPM Chapter 6 Vectors

Question 18
Determine whether each pair of the vectors are orthogonal:
(a) a = −i + 3j and b = 6i + 2j
(b) c = 2i + 4j + 4k and d = 5i − j + k

Answer:
(a) a · b = (−i + 3j) · (6i + 2j)
Term
= (−1)(6) + (3)(2)
1 = 0
Thus, a and b are orthogonal.

(b) c · d = (2i + 4j + 4k) · (5i − j + k)
= (2)(5) + (4)(−1) + (4)(1)
= 10
Thus, c and d are not orthogonal.



Question 19
The points P, Q and R have position vectors, relative to origin O, given
respectively by p = 2i + 3j + 4k, q = 5i − 2j + k and r = 3i + 5j + 2k.
Find the angle PQR, giving your answer to the nearest 0.1°.

Answer:
→ →
QP = p – q |QP| = (–3) + 5 + 3
2
 2 2
= (2i + 3j + 4k) – (5i − 2j + k) = 43

= –3i + 5j + 3k
→ →
QR = r – q |QR| = (–2) + 7 + 1
2
 2 2
= (3i + 5j + 2k) – (5i − 2j + k) = 54

= –2i + 7j + k
→ →
QP · QR = (–3i + 5j + 3k) · (–2i + 7j + k)
= (−3)(−2) + (5)(7) + (3)(1)
= 44
→ →
cos ∠PQR = QP · QR
→ →
|QP||QR|
44
=
 
43 54
≈ 0.9131

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Mathematics T Term 1 STPM Chapter 6 Vectors

∠PQR = cos (0.9131)
–1
= 24.1°


Question 20
The position vectors a and b of two points A and B respectively are given by,
a = 3i + 2j and b = 4i – 5j Term
Find the vector product, a × b.

1
Answer:
i j k
The vector product, a × b = 3 2 0
4 –5 0
2
j +
= u –5 0 0 u u 3 0 u u 3 2 u k
i –
4
0
4 –5
= 0i + 0j – 23k
= –23k
Note:
1. Definition of the cross product:
u × v is a vector perpendicular to both u and v, obeying the right-
hand rule as shown below:
u v

v

–n
u
n v


u v u

2. u × v = –(v × u)




Question 21
Find two unit vectors orthogonal to the following vectors:
a = 2j + k and b = i – k.



173





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Mathematics T Term 1 STPM Chapter 6 Vectors
Answer:
i j k
a × b = 0 2 1
1 0 –1
u 2 1 u u 0 1 u u 0 2 u
= 0 –1 i – 1 –1 j + 1 0 k

= (–2 – 0)i – (0 – 1)j + (0 – 2)k
= –2i + j – 2k
Term
1 |a × b| = (–2) + 1 + (–2)
 2 2
2
= 3
Thus, two unit vectors orthogonal to both a and b are
a × b –2i + j – 2k
± = ±
|a × b| 3
1
= ± (–2i + j – 2k)
3
Note:
Orthogonal Vector
a × b
a × b is orthogonal to both a and b. Thus, ± |a × b| are two unit vectors
orthogonal to both a and b.




Question 22
The position vectors a and b of two points A and B respectively are given by,
a = i + 3j − 5k and b = 2i – j + 3k
Find
(a) the vector product, a × b,
(b) the acute angle between the vectors a and b.


Answer:
i j k
(a) a × b = 1 3 –5
2 –1 3
3
i –
j +
= u –1 –5 u u 1 –5 u u 1 3 u k
2 –1
3
2
3
= (9 – 5)i – (3 + 10)j + (–1 – 6)k
= 4i – 13j – 7k
174




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Mathematics T Term 1 STPM Chapter 6 Vectors

(b) |a| = 1 + 3 + (–5) |b| = 2 + (–1) + 3
 2
2
2
2
 2 2


= 35 = 14
 2
|a × b| = 4 + (–13) + (–7 )
2
2

= 234
234
sin q = |a × b| = 
|a||b|   Term
35 14
≈ 0.6911
1
The acute angle between the vectors a and b is, q = sin (0.6911)
–1
= 43.7°
Note:
The same result can be obtained by using the scalar product method.
We can use the cross product to determine the angle between the given
two vectors as follows:
|a × b| = |a||b| sin q
sin q = a × b , where 0 < q < π is the angle between a and b.
|a||b|




Question 23
Given a = i − 3j + 2k, b = −i + 2j − k, and c = 2i − 3j − k, check by computation
that
(a) a × b = –(b × a)
(b) a × (b + c) = a × b + a × c


Answer:
i j k
(a) a × b = 1 –3 2
–1 2 –1
u –3 2 u u 1 2 u u 1 –3 u
= 2 –1 i – –1 –1 j + –1 2 k

= (3 – 4)i – (–1 + 2)j + (2 – 3)k
= −i − j − k




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Mathematics T Term 1 STPM Chapter 6 Vectors

i j k
b × a = –1 2 –1
1 –3 2

2
2
j +
= u –3 –1 u u –1 –1 u u –1 –3 u k
i –
1
1
2
2
= (4 – 3)i – (–2 + 1)j + (3 – 2)k
= i + j + k
Term
1 Thus, a × b = –(b × a).
Note:
Cross product of two vectors is a vector.

(b) (b + c) = (−i + 2j − k) + (2i − 3j − k)
= i − j − 2k

i j k
a × (b + c) = 1 –3 2
1 –1 –2

= u –3 2 u u 1 2 u u 1 –3 u k
j +
i –
1 –2
1 –1
–1 –2
= (6 + 2)i – (–2 – 2)j + (–1 + 3)k
= 8i + 4j + 2k
a × b = −i − j – k from part (b)
i j k
a × c = 1 –3 2
2 –3 –1

= u –3 2 u u 1 2 u u 1 –3 u k
j +
i –
–3 –1
2 –3
2 –1
= (3 + 6)i – (–1 – 4)j + (–3 + 6)k
= 9i + 5j + 3k
a × b + a × c = (−i − j – k) + (9i + 5j + 3k)
= 8i + 4j + 2k
Thus, a × (b + c) = a × b + a × c.





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Mathematics T Term 1 STPM Chapter 6 Vectors

Question 24

Given that a = 3i and b = 5i + 4j, find the area of the triangle OAB.

Answer:
1
Area of triangle OAB = |a × b|
2
= 1 |3i × (5i + 4j)| Term
2
= 1 |15i × i + 12i × j| 1
2
= 1 |15 × 0 + 12k|
2
= 6

Tips
Tips
1. i × i = j × j = k × k = 0
2. (a) i × j = k, j × k = i, k × i = j
(b) j × i = –k, k × j = –i, i × k = –j
3. Refer to the diagram on right, for the triangle AOB, let B
→ →
a = OA and b = OB, then
|a × b| = |a||b| sin q
= OA × (OB sin q) 
= OA × BP (= base × height) O P A
1
Thus, area of triangle OAB = |a × b|
2


Question 25
Show that the area of a parallelogram spanned by the vectors a and b is
|a × b|. Hence, evaluate the area of the parallelogram formed by the vectors
3a + b and a – b, if the angle between its sides is 30°, |a| = 1 and |b| = 3.


Answer:
Let the base of the parallelogram be a, and the angle between its sides a and
b be q, then its height is |b| sin q.
The area of the parallelogram is
= the length of its base × its height
= |a||b| sin q
= |a × b|

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Mathematics T Term 1 STPM Chapter 6 Vectors

The area of the parallelogram formed by the vectors 3a + b and a – b,
= |(3a + b) × (a – b)|
= |(3a) × a – (3a) × b + b × a – b × b|
= |0 – 3a × b – a × b – 0|
= |–4 a × b|
= 4|a||b| sin 30°

= 4 · 1 · 3 · 1
2
Term
= 6
1
Note:
We have shown from Question 23 that:
a × b = –b × a

The area of the parallelogram with two adjacent sides formed by a and b
is the magnitude of their vector product:
Area = |a × b| = |a||b| sin q,
where 0 < q < π is the angle between a and b.




Question 26
Write down the vector equation of the line which passes through the points
with position vectors.
(a) a = 3i + 2j and b = 4i + 5j.
(b) c = i – 2j – 3k and d = 3i + 2j + 4k.

Answer:
(a) The direction vector, b – a = (4i + 5j) – (3i + 2j)
= (4 – 3)i + (5 – 2)j
= i + 3j
The equation of the line is,
r = a + t(b − a)
r = (3i + 2j) + t(i + 3j)

(b) The direction vector, d – c = (3i + 2j + 4k) – (i – 2j – 3k)
= (3 – 1)i + (2 + 2)j + (4 + 3)k
= 2i + 4j + 7k
The equation of the line is,
r = c + t(d − c)
r = (i – 2j – 3k) + t(2i + 4j + 7k)



178





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Mathematics T Term 1 STPM Chapter 6 Vectors

Note:
The vector equation of the line through points A and B with position
vectors a and b is
r = a + t(b − a)
where each value of the parameter t ∈ R corresponds to a point on the
line,
r = xi + yj, for two dimensions, and r = xi + yj + zk, for three dimensions, Term
is a position vector to any unknown point on the line,
a = a 1 i + a 2 j, for two dimensions, and a = a 1 i + a 2 j + a 3 k, for three
dimensions, is a position vector to any known point on the line, and 1
b − a is a direction vector parallel to the line.




Question 27
→
(a) Write down the vector AB joining the points A and B with coordinates
(3, 2, 6) and (−1, 2, 1) respectively.
(b) Find the equation of the straight line through A and B.


Answer:
→ → →
(a) The direction vector, AB = OB – OA
= (−1, 2, 1) – (3, 2, 6)
= (−4, 0, –5)
(b) The equation of the straight line through A and B is
r = a + t(b − a)
r = (3i + 2j + 6k) + t(−4i – 5k)
Note:
→
If AB is a vector with initial point A(x 1 , y 1 , z 1 ) and terminal point
→ → →
B(x 2 , y 2 , z 2 ), then AB = OB – OA = (x 2 – x 1 , y 2 – y 1 , z 2 – z 1 ), as seen in
the following diagram.


B(x 2 , y 2 , z 2 )
,
M(x 2 – x 1 y 2 – y 1 , z 2 – z 1 )
A(x 1 , y 1 , z 1 ) v
O







179





06 Term 1 Chapter 6.indd 179 4/11/19 2:14 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Question 28
(a) Write down the Cartesian form of the equation of the straight line which
passes through the two points (7, 4,−2) and (3, 6,−1).
(b) State the equivalent vector equation.


Answer:
(a) In order to use the Cartesian equation, we find the vector,
→
Term
AB = (3 – 7)i + (6 – 4)j + [–1 – (–2)]k
1 = –4i + 2j + k
The Cartesian form of the equation is
x – 7 y – 4 z – (–2)
= =
3 – 7 6 – 4 –1 – (–2)
x – 7 = y – 4 = z + 2
–4 2 1

(b) The equivalent vector equation is
r = a + t(b − a)
r = (7i + 4j – 2k) + t[(3i + 6j – k) − (7i + 4j – 2k)]
= (7i + 4j – 2k) + t(–4i + 2j + k)
Note:
(a) In general, the Cartesian form of the equation of the straight line
which passes through the points with coordinates (a 1 , a 2 , a 3 ) and
(b 1 , b 2 , b 3 ) is
x – a 1 = y – a 2 = z – a 3
b 1 – a 1 b 2 – a 2 b 3 – a 3
(b) The equivalent vector equation is
r = a + t(b − a)
where each value of the parameter t corresponds to a point on the
line.




Question 29
The Cartesian equation of a straight line is given by
x – 3 = y + 1 = z + 5
2 (–6) 4
Rewrite it in vector form.



180





06 Term 1 Chapter 6.indd 180 4/11/19 2:14 PM

Mathematics T Term 1 STPM Chapter 6 Vectors
Answer:
Let x – 3 = y + 1 = z + 5 = t for some parameter t ∈ R, then the parametric
2 (–6) 4
equations of the line are:
x = 3 + 2t
y = –1 – 6t
z = –5 + 4t
The vector equation is, Term
(x, y, z) = (3, –1, –5) + t(2, –6, 4)
or r = (3i – j – 5k) + t(2i – 6j + 4k) 1


Note:
This equation is not unique (although the line is unique), the line can be
written as
r = (3i – j – 5k) + 2t(i – 3j + 2k)

By substituting s = 2t,
r = (3i – j – 5k) + s(i – 3j + 2k)




Question 30
Find the coordinates of the point where the line passing through P(3, 4, –5)
parallel to –2i – 3j + k crosses the xy-plane.


Answer:
The vector equation of the line is
r = p + ta
r = (3i + 4j – 5k) + t(–2i – 3j + k) for some parameter t ∈ R.
(xi + yj + zk) = (3 – 2t)i + (4 – 3t)j + (–5 + t)k
The parametric equations of the line are:
x = 3 – 2t
y = 4 – 3t
z = –5 + t
This line will cross the xy-plane when z = 0.
–5 + t = 0
t = 5
x = 3 – 2(5) = –7
y = 4 – 3(5) = –11
Thus, the coordinates of the point of intersection are (–7, –11, 0).



181





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Mathematics T Term 1 STPM Chapter 6 Vectors

Tips
Tips
Two vectors are parallel if one of them is a scalar multiple of the other. For
example:
The vector 2i – 3j + k is parallel to the vector 4i – 6j + 2k since
1
2i – 3j + k = (4i – 6j + 2k)
2

kv, k > 0
Term
1

v
O
kv, k < 0





Question 31
(a) Find the vector equation of the plane which passes through the point with
position vector i – 2j + 4k and which is perpendicular to 2i + 3k.
(b) Find the Cartesian equation of this plane.
(c) Show that the point –2i – 5j + 6k lies on the plane.


Answer:
(a) Using the formula r . n = a . n
The vector equation of the plane is
r · (2i + 3k) = (i – 2j + 4k) . (2i + 3k)
= 2 + 12
r · (2i + 3k) = 14

(b) The Cartesian equation of this plane is
(xi + yj + zk) · (2i + 3k) = 14
2x + 3z = 14

(c) For the point –2i – 5j + 6k, x = –2, y = –5, z = 6, satisfies the equation
2x + 3z = 14.
Thus, the point –2i – 5j + 6k lies on the plane.






182





06 Term 1 Chapter 6.indd 182 4/11/19 2:14 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Note:
A plane perpendicular to the vector n and passing through the point with
position vector a, has equation
(r – a) · n = 0
r · n = a · n

n Term

A
P
a 1
r
O



Question 32
(a) Find the vector equation of the plane through (3, 5,−2) for which
n = (1, 3, 2) is a vector normal to the plane.
(b) What is the distance of the plane from O?


Answer:
(a) Using the formula r · n = a · n
The vector equation of the plane is
r · (i + 3j + 2k) = (3i + 5j – 2k) . (i + 3j + 2k)
= 3 + 15 – 4
r · (i + 3j + 2k) = 14


(b) The magnitude of the vector normal to the plane is,
|n| = 1 + 3 + 2
2
2
 2

= 14
The distance from the origin is
^
a · n = a . n
|n|
14
= a . n = 14 from part (a)

14

= 14


183





06 Term 1 Chapter 6.indd 183 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Note:
^
A plane with unit normal n, which is a perpendicular distance d from O
is given by
^
r · n = d


Question 33
Term
Find the equation of a plane which is normal to 2i + j – 2k and which is a
1 distance 1 from the origin. Give your answer in Cartesian form.


Answer:
^
Using the formula r · n = d
2i + j – 2k
(xi + yj + zk) · = 1
2 + 1 + (–2)
 2
2
2
2x + y − 2z = 3
Thus, the required equation of the plane in Cartesian form is 2x + y − 2z = 3.
Note:
One of the following information is needed to construct a plane equation:
(a) 3 points lying on the plane.
(b) 2 points lying on the plane, and 1 directional vector.
(c) 2 lines lying on the plane.
(d) A point lying on the plane, and the normal vector of a plane.
A plane is simply just a flat surface in space. We first start by introducing
the vector equation of a plane,
r = a + sb + tc
where a is a position vector, b and c are 2 non-parallel vectors, s and t
being 2 arbitrary constants as shown in the following diagram.

z





b
c
a
0

x y


184





06 Term 1 Chapter 6.indd 184 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Question 34
Find the vector equation of the plane that passes through the points
a = 2i − j, b = −2j − k, and c = i + j − k.
Hence, deduce its Cartesian form.


Answer:
Using the formula r = a + s(b – a) + t(c – a) where s and t are 2 arbitrary Term
constants.
The required vector equation of the plane is 1
r = 2i – j + s[(–2j – k) – (2i – j)] + t[(i + j – k) – (2i – j)]
r = 2i – j + s(–2i – j – k) + t(–i + 2j – k)
r = (2 – 2s – t)i + (–1 – s + 2t)j + (–s – t)k
To find the Cartesian equation of the plane, compare with r = xi + yj + zk
x = 2 – 2s – t , y = –1 – s + 2t , z = –s – t



Question 35
Find the Cartesian equation of a plane that contains (3, −7, 2) and has normal
2i + 5j – 4k.


Answer:
The required Cartesian equation of the plane is
a(x – x 1 ) + b(y – y 1 ) + c(z – z 1 ) = 0
2(x – 3) + 5[y – (−7)] – 4(z – 2) = 0
2x + 5y – 4z = –37
Note:
The Cartesian equation of a plane that passes through a point P(x 1 , y 1 , z 1 )
and has a normal vector
n = ai + bj – ck is a(x – x 1 ) + b(y – y 1 ) + c(z – z 1 ) = 0
z

n = ai + bj – ck
M(x, y, z)


P(x 1 , y 1 , z 1 )
y

x


185





06 Term 1 Chapter 6.indd 185 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Question 36
Show that the line with equation r = 2i – j + 5k + t(3i + 2j – 2k) is parallel to
the plane whose equation is r · (2i – 2j + k) = –3.

Answer:
The direction vector of the line is, b = 3i + 2j – 2k
The normal vector to the plane is, n = 2i – 2j + k
Term
b · n = (3i + 2j – 2k) . (2i – 2j + k)
1 = (3)(2) + (2)(–2) + (–2)(1)
= 0
Thus, the given line is parallel to the plane.



Question 37
The equations of two lines are given as follows:
r 1 = i – 2j + 3k + s(4i + 5j – 6k)
r 2 = –2i + 4j + 2k + t(3i – 6j + k)
Calculate the acute angle between these lines, leave your answer correct to
one decimal place.


Answer:
The acute angle formed by these lines can be calculated as the acute angle
formed by the corresponding direction vectors, i.e.
b = 4i + 5j – 6k and d = 3i – 6j + k.
Let the angle be q, then by using the scalar product we have,
cos q = b · d
|b||d|
(4)(3) + (5)(–6) + (–6)(1)
=
4 + 5 + (–6) 3 + (–6) + 1
 2  2  2
2
2
2
–24
=

77 46
≈ –0.4033
The required acute angle is, q = cos (0.4033)
–1
= 66.2°



186





06 Term 1 Chapter 6.indd 186 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Tips
Tips
1. To find the angle between two lines it is sufficient to find the angle
between the two direction vectors. This means that it is also possible to
find the angle between skew lines.
b · d
2. To find the acute angle, q, we can use the formula cos q = .
|b||d|
3. The negative sign indicates that the angle between the two directions Term
is obtuse. However, the angle between two lines would normally be
considered to be acute.
1


Question 38
Find the point of intersection of the following two lines.
r 1 = 3i + 5j + s(i – 2j)
r 2 = 4i – 3j + t(–i + 4j),
where s and t ∈ R.


Answer:
If the lines intersect, then for some values of s and t,
r = (3 + s)i + (5 – 2s)j = (4 – t)i + (–3 + 4t)j
Equating coefficients of the i: 3 + s = 4 –t
s + t = 1 …… 1
Equating coefficients of the j: 5 – 2s = –3 + 4t
2s + 4t = 8
s + 2t = 4 …… 2
2 – 1, t = 3 …… 3
Substituting 3 into 1, s + 3 = 1
s = –2

Substituting s = –2, the point of intersection has position vector,
r = [3 + (–2)]i + [5 – 2(–2)]j
r = i + 9j
Thus, the point of intersection is (1, 9).







187





06 Term 1 Chapter 6.indd 187 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Note:
1. For two-dimensions, two lines are guaranteed to cross unless they
are parallel. At the point of intersection, the coordinates for both
lines are the same, and are usually found by solving simultaneous
equations.
2. For three-dimensions, two lines do not have to meet. If they do not
meet then the lines are called ‘skew’.
3. Finding where two lines meet is the same as for two-dimensions,
Term
except there will be three simultaneous equations (in two variables)
1 to solve. (A solution that does not work for all three equations means
the lines do not meet and are therefore skew.)




Question 39
Find the point at which the line
x – 1 = = z – 2
y
2 4 –3
intersects the plane: x + 2y – z + 1 = 0.


Answer:
Rewrite the Cartesian equation into parametric forms.

Let x – 1 = y = z – 2 = t, for some parameter t ∈ R, the parametric
2 4 –3
equations of the line are:
x = 1 + 2t
y = 4t
z = 2 – 3t
Substitute these parametric equations into the equation of the plane
x + 2y – z + 1 = 0
1 + 2t + 2(4t) – (2 – 3t) + 1 = 0
13t = 0
t = 0
Thus,
x = 1 + 2(0) = 1
y = 4(0) = 0
z = 2 –3(0) = 2
The required point is (1, 0, 2)





188





06 Term 1 Chapter 6.indd 188 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Question 40
Show that the line with equation
r = 5i + 5k + t(2i + j), where t ∈ R.
lies in the plane whose equation is r · (–i + 2j) = –5.


Answer: Term
r = 5i + 5k + t(2i + j), where t ∈ R.
r = (5 + 2t)i + tj + 5k ……… 1
Substituting 1 into left side of the plane, 1
r · (–i + 2j) = [(5 + 2t)i + tj + 5k] · (–i + 2j)
=(5 + 2t)(–1) + 2t + 0
= –5
= right side of the equation of the plane.
Thus, the line lies in the plane.



Question 41
Find an equation for the line of intersection of the following two planes
x + 2y – z = –5 and y + 2z = 5.


Answer:
The two normal vectors of the planes are u = i + 2j – k and v = j + 2k
respectively, so the direction vector of the line is parallel to u × v where,

i j k
u × v = 1 2 –1
0 1 2

i –
= u 2 –1 u u 1 –1 u u 1 2 u k
j +
0
1
2
2
0
1
= [4 – (–1)]i – (2 – 0)j + (1 – 0)k
= 5i – 2j + k
The planes intersect in an entire line, by substituting x = 0 into both equations
of planes (this is an arbitrary choice), we have
2y – z = –5 ……… 1
y + 2z = 5 ……… 2
2 × 1 + 2, 5y = –5
y = –1 ……… 3

189





06 Term 1 Chapter 6.indd 189 4/11/19 2:15 PM

Mathematics T Term 1 STPM Chapter 6 Vectors

Substituting 3 into 1, 2(–1) – z = –5
z = 3
Thus, one of the point of intersection for the planes is (0, –1, 3).
The vector equation for the line of intersection of the two planes is
0
5
1 2 1 2
r = –1 + t –2 , where t ∈ R
3 1
or r = –j + 3k + t(5i – 2j + k), where t ∈ R
Term
1
Note:
1. Two planes intersect in a line L.
2. n 1 × n 2 is the direction vector of the line of intersection, L, of the two
planes with normal vectors n 1 and n 2 respectively, as shown in the

following diagram.




n 1
n 2


L





























190





06 Term 1 Chapter 6.indd 190 4/11/19 2:15 PM

Answers





STPM Model Paper 954/1

1 – 2x if 1 – 2x > 0
1. y = |1 – 2x| =
–(1 – 2x) if 1 – 2x , 0
1 – 2x, x < 1
= 2
2x – 1, x . 1
2
1
x 0 — 1
2
y 1 0 1

y

y = |1 – 2x|
1 A 1
y = –
x
x
0
–2 –1 1 2
–1



From the graph, A is the point (1, 1).
The set of values of x is {x | x , 0, x . 1}

1 1
2
2
2. (x + y) – (x – y)
1 y 1 1 y 1
1
1
= x 1 + x 2 2 – x 1 – x 2 2
2
2
1 1 1 1 3
1 y 2 1 – 2 2 y 2 2 1 – 2 21 – 2 2 y 3
3
4
= x 1 + 2x + 2! 1 x 2 + 3! 1 x 2 + …
2
1 1 1 1 3
1 y 2 1 – 2 2 y 2 2 1 – 2 21 – 2 2 y 3
3
4
1
1
– x 1 – 2x + 2! – x 2 + 3! – x 2 + …
2
1 y y 2 y 3 1 y y 2 y 3
1
1
2
2
2
= x 1 + 2x – 8x 2 + 16x 3 + … – x 1 – 2x – 8x 2 – 16x 3 – …
2
1 y 3
= x 2 y + 8x 22
1 x
533
22 Answers.indd 533 4/11/19 2:58 PM

Mathematics T STPM Answers

Putting x = 4, y = 2,
1 1 1 8
3
3 4
(4 + 2) – (4 – 2) = 4 2 2 + 8(4 )
2
2
4
1
6 – 2 = 2
  1 1 + 64 2
2
= 33
32
1 1
2
(x + y) + (x – y) 2
1 y y 2 y 3 1 y y 2 y 3
1
2
1
= x 1 + 2x – 8x 2 + 16x 3 + … + x 1 – 2x – 8x 2 – 16x 3 – … 2
2
2
1 y 2
1
2
≈ x 2 – 4x 22
Putting x = 4, y = 2
1 1 1 4
3
2 4
2
2
(4 + 2) + (4 – 2) = 4 2 – 4(4 )
2
  31
6 + 2 =
8
3. First augment the matrix with the identity, forming (M | I), an n × 2n
matrix. Let R , R and R , represent row 1, row 2, and row 3 respectively
3
2
1
of the augmented matrix.
2
1 –1 1 3 1 0 0 2
0
1
0
0
4
0
1
0
1
3
0
–R ↔ R 1 1 0 –4 0 –1 0 2
3
1
2
0
1
0
2 1
3 1 0 0 0 1
R → R – 2R 1 1 0 –4 0 –1 0
2
2
R → R – 3R 1 1 0 1 11 1 2 0 2
3
3
0 1 12 0 3 1
R → R – R 1 0 –4 0 –1 0
3 3 2 1 0 1 11 1 2 0 2
0 0 1 –1 1 1
R → R + 4R 3 1 1 0 –4 3 4
1
1
R → R – 11R 3 1 0 0 0 12 –9 –11 2
2 2 0 1 1 –1 1 1

1 –4 3 4 2
Thus, M = 12 –9 –11
–1
1
1
–1
534
22 Answers.indd 534 4/11/19 2:58 PM

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