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1. Vector quantity is any quantity that has both a magnitude and a direction.

Kuantiti vektor ialah kuantiti yang mempunyai kedua-dua magnitud dan arah.

2. Scalar quantity is a quantity that has magnitude but no direction.

Kuantiti skalar ialah kuantiti yang mempunyai magnitud tetapi tidak mempunyai arah.

3. A vector is usually represented by a directed line segment drawn as an arrow. The length of the line represents the magnitude or the size of the vector and the arrow indicates the direction of the vector.

Suatu vektor biasanya diwakili oleh tembereng garis berarah yang dilukis sebagai satu anak panah. Panjang garis mewakili magnitud atau saiz vektor itu dan anak panah menunjukkan arah vektor itu.

Magnitude Magnitud

4. A vector from an initial point A to a terminal point B can be written as →

AB, a~, AB or a.

Suatu vektor dari titik awal A ke titik terminal B boleh ditulis sebagai →

AB, a~, AB or a.

BTerminal point Titik terminal

Initial point Titik awal

A a~

5. Symbol used in representing the magnitude of a vector:

Simbol yang mewakili magnitud suatu vektor:

has the magnitude mempunyai magnitud Relating factor Faktor penghubung

AB→ AB

uABu→ a~ uABu a

ua~u uau

as as as

6. Vector –→

AB represents a vector in the opposite direction as →

AB, that is → BA = –→

AB.

Vektor –→

AB mewakili suatu vektor dalam arah yang bertentangan dengan →

AB, iaitu → BA = –→

AB.

7. Two vectors are equal if and only if both the vectors have the same magnitude and direction.

Dua vektor adalah sama jika dan hanya jika kedua-dua vektor mempunyai magnitud dan arah yang sama.

8. A zero vector 0~ has magnitude zero and its direction cannot be determined.

Vektor sifar 0~ mempunyai magnitud sifar dan arahnya tidak dapat ditentukan.

9. The vector a~ multiplied by the scalar k is also a vector and is written as ka~ where

Vektor a~ yang didarabkan dengan scalar juga merupakan suatu vector dan ditulis sebagai ka~ dengan keadaan

(i) |ka~| = k|a~|

(ii) if k . 0, then k a~ has the same direction as a~.

jika k . 0, maka arah k a̰ sama dengan a~.

(iii) if k , 0, then k _a~ has the opposite direction as a~.

jika k , 0, maka arah k a~ bertentangan dengan a~.

10. Vector a~ and b~ are parallel if and only if a~ = kb~ where k is a constant.

Vektor a~ dan b~ adalah selari jika dan hanya jika a~ = ka~

dengan keadaan k ialah pemalar.

8 ^Vectors Vektor

Vectors

Vektor

8.1

^Textbook

pg. 212 – 220

1. Identify whether each of the following quantity is a vector or a scalar by marking 7 and give your justification.

PL1

Kenal pasti sama ada setiap kuantiti yang berikut ialah kuantiti vektor atau skalar dengan menandakan 7 dan berikan justifikasi anda.

Quantity

Kuantiti

Scalar

Skalar

Vector

Vektor

Justification

Justifikasi

Example ^{30 km h}

^–1

⁷ No direction

(a) 10 kg 7 No direction

(b) 50 N 7 Possesses direction and magnitude

(c) 5% 7 No direction

(d) 1 m s

^–1

due north

ke utara

7 Possesses direction and magnitude

(e) 45°C 7 No direction

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2. Draw and label each of the following vectors.

^PL1

Lukis dan tandakan setiap vektor berikut.

Example

QR → represents a force of 70 N due east.

QR mewakili daya 70 N ke arah timur.→

70 N

Q R

(a) a~ m s

^–2

represents acceleration making an angle of 30° with the horizontal line.

a~ m s^–2 mewakili pecutan pada 30° dengan garis mengufuk.

30°

m s^–2 a~

(b) → HK represents a displacement of 56 m due north- east.

HK mewakili sesaran 56 m ke timur laut.→

N

56 m K

H

(c) v~ represents a velocity of 60 km h

^–1

due south.

v~ mewakili halaju 60 km h^–1 ke arah selatan.

60 km h^–1

3. (i) Determine the magnitude and direction of each vector in the diagram.

Tentukan magnitud dan arah bagi setiap vektor pada rajah di sebelah.

(ii) Hence draw the corresponding negative vector and state the magnitude and direction respectively.

Seterusnya lukis vektor negatif yang sepadan dan nyatakan magnitud serta arah masing-masing.

(i) Example e~

Magnitude = √2

²

+ 2

²

= √8

= 2√2 units

\ 2√2 units due north-west

Pythagoras' theorem Teorem Pythagoras c² = a² + b²

a c

b

(a) a~

Magnitude

= 5 units

\ 5 units due east

(b) b~

Magnitude = √2

²

+ 1

²

= √5 units

\ √5 units due south-east

(c) → AB

Magnitude

= 4 units

\ 4 units due west

(d) → MN Magnitude = √3

²

+ 3

²

= √18

= 3√2 units

\ 3√2 units due south-east

(ii)

– (a)

(b) M

N (c)B A

(d)

–a ~ –b ~

N/U e~

– e~: 2√2 units due south-east

(a) –a~ : 5 units due west

(b) –b~ : √5 units due north-west (c) – → AB : 4 units due east

(d) – → MN : 3√2 units due north-west

PL1 (a)

(b) M

N (c)B A

(d)

a~

b~

N/U e~

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4. State the pairs of vectors which are equal.

^PL2

Nyatakan pasangan vektor yang sama.

P

Q

U

F

E t~ V

w~

r~ s~

Answer/

Jawapan

: UV and → t~

6. Prove that the following pairs of vectors are parallel. Then, determine whether the points involved are collinear.

^PL3

Buktikan bahawa pasangan vektor berikut adalah selari. Kemudian, tentukan sama ada titik-titik yang terlibat adalah segaris atau tidak

Example

AB → = 3k~, CD → = –1.2k~

AB → = 3k~ ⇒ k~ = 1 3

AB → CD → = –1.2 1 ¹ ₃ ^{AB →} 2

= – 2 5

AB →

Hence, → AB and → CD are parallel but points A, B, C and D are not collinear as there is no common point.

Vector a~ and b~ are parallel if and only if

~ = kba ~ where k is a constant.

Vektor a~ ^{dan b}~ adalah selari jika dan hanya jika a~ ^{= kb}~ dengan keadaan k ialah pemalar.

(a) → EF = 3

2 m ~ , FG → = 5m ~ EF → = 3

2 m ~ , ⇒ m ~ = 2 3

EF → FG → = 5 1 ² ₃ ^{EF →} 2

= 10 3

EF →

Hence, EF → and → FG are parallel. Points E, F and G are collinear as F is the common point.

(b) → HI = –2n ~, → IK = –4n ~ HI = –2n → ~, ⇒ n ~ = – 1

2 HI →

→ IK = –4 1 ^{– 1} ₂ ^{→ HI} 2

= 2→ HI

Hence, HI and → → IK are parallel. Points H, I and K are collinear as I is the common point.

(c) → RS = –3q ~ , → TU = –5q ~ RS = –3q → ~ , ⇒ q ~ = – 1

3 RS → TU = –5 → 1 ^{– 1} 3

RS → 2

= 5

3 RS →

Hence, RS and → → TU are parallel but points R, S, T and U are not collinear as there is no common point.

(d) → VW = 25p ~ , → XY = 8p ~ VW = 25p → ~ , ⇒ p ~ = 1

25 VW → XY = 8 → 1 25 ¹

VW → 2

= 8

25 VW →

Hence, VW and → → XY are parallel but points V, W, X and Y are not collinear as there is no common point.

5. Express vectors below in terms of r~.

^PL2

Ungkapkan vektor berikut dalam sebutan r~.

Example

(b) (c)

(a) r~

Example _– ¹ 2 r~

(a) – 3 2 r~

(b) 3 r~

(c) 2 r~

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7. U, V and W are three collinear points. Point V divides UW in the ratio of p : q and → UW = 6 r~. Determine the vectors → UV and → VW in terms of r~ when

^PL3

U, V dan W ialah tiga titik segaris. Titik V membahagikan UW dengan nisbah p : q dan →UW = 6 r~. Tentukan vektor UV dan →→ VW dalam sebutan r~ apabila

U, V and W are collinear points means these points are lying on a single straight line. The vectors connecting these points are parallel.

U, V dan W ialah titik-titik yang segaris bermaksud titik-titik tersebut terletak pada satu garis lurus. Vektor-vektor yang menyambungkan titik- titik tersebut adalah selari.

Example

p = 2 and /

dan

q = 3 UV = → 1 2 + 3 ² 2 ⁶

^r~

= 12 5

^r~

VW = → 1 _{2 + 3} ³ 2 ⁶

^r~

= 18 5

^r~

2 V

U

3 W

(a) p = 4 and /

dan

q = 2 UV = → 1 ⁴ ₆ 2 ⁶

^r~

= 4

_r~

VW = → 1 ² ₆ 2 ⁶

^r~

= 2

_r~

4 V

U

2 W

(b) p = 5 and /

dan

q = 4 UV → = 1 ⁵ ₉ 2 ⁶

^r~

= 10 3

^r~

VW = → 1 ⁴ ₉ 2 ⁶

^r~

= 8

3

^r~

5 V

U

4 W

(c) p = 1 and /

dan

q = 3 UV = → 1 ¹ ₄ 2 ⁶

^r~

= 3

2

^r~

VW = → 1 ³ ₄ 2 ⁶

^r~

= 9

2

^r~

1V U

3 W

8. In the following equations, m and n are constants. Vectors a~ and b~ are not parallel and non-zero. Determine the values of m and n in each of the following cases.

^PL3

Dalam persamaan berikut, m dan n ialah pemalar. Vektor-vektor a~ dan b~ tidak selari dan bukan sifar. Tentukan nilai m dan n dalam setiap kes berikut.

Example

(3m – 2)a~ = (n + 5)b~

3m – 2 = 0 3m = 2

m = 2 3 n + 5 = 0 n = –5

(a) (m + 3)b~ = (2 – n)a~

m + 3 = 0 m = –3

2 – n = 0 n = 2

(b) (4 – 2m)a~ = (n + 1)b~

4 – 2m = 0 2m = 4 m = 2 n + 1 = 0 n = –1

(c) (5m + 6)a~ – (2n + 4)b~ = 0 (5m + 6)a~ = (2n + 4)b~

5m + 6 = 0 2n + 4 = 0

5m = –6 2n = –4

m = – 6

5 n = –2

UV = →

1

2 + 3²

2

^VW→

VW = →

1

2 + 3³

2

^UW→

If a~ and b~ are not parallel and non-zero, and ha~ = kb~, then h = k = 0.

Jika a~ ^{dan b}~ tidak selari dan bukan sifar, dan ha~ ^{= kb}~^, maka h = k = 0.

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Notes SMART

Addition and Subtraction of Vectors

Penambahan dan Penolakan Vektor

8.2

pg. 221 – 226^Textbook

1. A resultant vector is the combination of two or more single vectors.

Suatu vektor paduan ialah kombinasi dua atau lebih vektor.

2. Parallel vectors/ Vektor selari

(a) The addition of two parallel vectors a~ and b~ can be combined to produce a resultant vector represented as a~ + b~. For example,

Penambahan bagi dua vektor selari a~ dan b~ boleh digabungkan untuk menghasilkan vektor paduan yang diwakili sebagai a~ + b~. Misalnya,

= 2 m s^–1

a~ b~= 1 m s^–1 a~+ b~= 3 m s^–1

+ =

(b) The subtraction of vector b~ from vector a~ is the sum of vector a~ and negative vector b~, that is a~ – b~ = a~ + (–b~). For example,

Penolakan vektor b~daripada vektor a~ialah hasil tambah vektor a~ dan vektor negatif b~, iaitu a~ – b~ = a~ + (–b~).

Misalnya,

= 2 m s^–1 a~

= 2 m s^–1

a~ –b~= –1 m s^–1 a~– b~= 1 m s^–1

= 1 m s^–1

– b~

= + =

3. Non-parallel vectors/ Vektor tidak selari The resultant vector can be obtained using Vektor paduannya boleh diperoleh menggunakan

• triangle law hukum segi tiga

a~

a~

b~

b~

+

• parallelogram law hukum segi empat selari

a~ a~

b~

b~

+

• polygon law hukum poligon

a~

a~ b~

b~ c~

c~

+ +

9. It is given that → CD = 2u ~, EF = 1 →

3 u ~ and FG = u → ~.

^PL3

Diberi bahawa →CD = 2u~, →EF = 13u~ dan →

FG = u~.

(i) Express each of the following expressions in term of u ~.

Ungkapkan setiap ungkapan berikut dalam sebutan u~.

(ii) If uu~u = 4 units, find the magnitude of each of the following expressions.

Jika uu~u = 4 unit, cari magnitud bagi setiap ungkapan berikut.

Example 4→ CD + 9→ EF

(i) 4→ CD + 9→ EF = 4(2u ~) + 9 1 ¹ 3 ~ u 2

= 11u ~

(ii) u4→ CD + 9→ EF u = 11uu~u

uka~u = kua~u

= 11(4)

= 44 units

(a) → CD – 3→ EF + 2→ FG

(i) → CD – 3→ EF + 2→ FG = 2u ~ – 3 1 ¹ ₃ ~ ^u 2 ^{+ 2(u} ~)

= 3u ~ (ii) u→ CD – 3→ EF + 2→ FG u = 3uu~u

= 3(4)

= 12 units

(b) 1 4

FG + 3 → 2

EF – 1 → 8

DC →

(i) 1 4

FG + 3 → 2

EF – 1 → 8

DC = 1 →

4 (u ~) + 3

2 1 ¹ ₃ ^u ~ 2 ^{– 1} ₈ ^(–2u ~)

= 1

4 ~ + u 1 2 u ~ + 1

4 u ~

= u ~

(ii) u ¹ ₄ ^{FG + 3 →} ₂ ^{EF – 1 →} ₈ ^{DC →} u ⁼ uu~u

= 4 units

(c) → GF + 3→ CD – 6→ EF

(i) → GF + 3→ CD – 6→ EF = –u ~ + 3(2u ~) – 6 1 ¹ 3 ~ u 2

= 3u ~ (ii) u→ GF + 3→ CD – 6→ EF u = 3uu~u

= 3(4)

= 12 units

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10.

p~

q~

The diagram shows vectors p ~ and q ~ . Draw the resultant vectors for the following.

^PL3

Rajah di atas menunjukkan vektor-vektor p~ dan q~. Lukis vektor paduan bagi berikut.

Example p ~ + 3q ~

p~ p~

q~

q~

+ 3 3

(a) 2p

~ – q

~

p~

2 p~

2 –q~

q~

–

(b) – 3 2 p

~ + 2q ~

q~ – 3_p~

2 p~

– 3_2 2

q~

+ 2

(c) 2p ~ + 2q ~

q~

2

p~

2 q~

p~+ 2 2

(d) –3p ~ – 4q ~

q~

–4

q~

– 4 p~

–3

p~

–3

(e) –2p ~ + 3q ~

q~

3

q~

+ 3

p~

–2 p~

–2

11. Use the given vectors to draw the resultant vectors in the space provided using parallelogram law.

^PL3

Gunakan vektor yang diberi untuk melukis vektor paduan di ruang yang disediakan dengan menggunakan hukum segi empat selari.

Example r~ + s~

_r~

r~

r~

s~

s~

s~

+

(a) c~ + 1 3 b ~

c~

c~

c~

b~

b~ 1–b~

3 1–3

+

(b) 2a~ + d~

a~

d~

a~

d~

d~

+ 2

a~

2

12. The diagram shows an octagon ABCDEFGH. Find the resultant vector for each of the following.

^PL3

Rajah di bawah menunjukkan sebuah oktagon ABCDEFGH. Cari vektor paduan bagi setiap yang berikut.

A

B C

D

E F G

H

Example

BC → + → CD + → DE = → BE

(a) → DE + → EF + → FG + → GH = DH →

(b) → DC + → ED + → FE = FC →

(c) → CD – → ED + → EA = CA →

(d) → DE + → EF – → GF – → AG = DA →

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13. Solve the following problems.

^PL4

Selesaikan masalah berikut.

Example

Two forces F

₁

= 40 N and F

₂

= 20 N are acting on a moving object. Draw vector diagrams to represent the following situations.

Daily Application

Suatu objek yang sedang bergerak dikenakan dua daya F₁ = 40 N dan F₂ = 20 N. Lukis gambar rajah vektor untuk mewakili situasi berikut.

(i) F

₁

and F

₂

are in the same direction

F₁ dan F₂ dalam arah yang sama

(ii) F

₁

and F

₂

are in the opposite direction

F₁ dan F₂ dalam arah yang bertentangan

Hence, determine the magnitude of the force and the direction of the movement of the object.

Seterusnya, tentukan magnitud daya dan arah pergerakan objek tersebut.

(i)

F₁ F₂

Magnitude =

F₁

+

F₁

= 40 + 20

= 60 N

The magnitude of the force acted on the object is 60 N in the same direction as

F₁

and

F₂

.

(ii)

F₁ F₂

Magnitude =

F₁

+ (–

F₂

)

= 40 + (–20)

= 20 N

The magnitude of the force acted on the object is 20 N in the same direction as

F₁

.

(a) Displacement is a vector quantity that shows the change in position of an object. An aeroplane flies 200 km north from airport A to airport B.

Then, it flies to airport C which is 400 km north- west of the airport B. Draw a vector diagram to represent the path taken by the aeroplane and the displacement of the aeroplane.

Daily Application

Sesaran ialah kuantiti vektor yang menunjukkan perubahan kedudukan suatu objek. Sebuah kapal terbang bergerak 200 km ke utara dari lapangan terbang A ke lapangan terbang B. Kemudian, kapal terbang itu bergerak ke lapangan terbang C yang berada di 400 km barat laut lapangan terbang B. Lukis satu gambar rajah vektor untuk mewakili perjalanan kapal terbang itu dan sesaran kapal terbang itu.

A B C

Displacement 45°

1 cm represents 200 km

(b)

a~ b~

c~

T W

V Q P

U R S

The diagram shows a cube PQRSTUVW.

According to the diagram, express in terms of a~,

~ and c~, vector b

Rajah di atas menunjukkan sebuah kubus PQRSTUVW.

Berdasarkan rajah tersebut, ungkapkan dalam sebutan a~, b~ dan c~, vektor

(i) → PV (ii) → PW (iii) → UQ

(i) → PV = a~ + b~

(ii) → PW = a~ + b~ + c~

(iii) → UQ = –b~ + a~

b~

a~

a~

+ b~

V

Q P

c~

b~+ c~

a~+

b~

a~+

W

P V

b~

– –b~ a~

a~

+ U

Q P

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Notes SMART

(c)

A

D

X B

s~ C r~

In the diagram, ABCD is a rectangle and X is the midpoint of AB. Given that → AD = r~ and DC → = s~, express in terms of r~ and s~,

Dalam rajah di atas, ABCD ialah sebuah segi empat tepat dan X ialah titik tengah AB. Diberi bahawa →→ AD = r~ dan DC = s~, ungkapkan dalam sebutan r~ dan s~,

(i) → AC (ii) → DB (iii) → CX (iv) → XD

(i) → AC = → AD + → DC = r~ + s~

(ii) → DB = → DA + → AB

= –r~ + s~

(iii) → CX = → CB + 1 2

BA →

= –r~ – 1 2 s~

(iv) → XD = → XA + → AD

= – 1 2 s~ + r~

(d)

B

A M

O C

The diagram shows a quadrilateral OABC with AO → = a~, OB → = b~, OC → = c~ and 2OM = MB. The lines AC and OB intersect at point M which is the midpoint of the line AC. Find the vector → OM in terms of a~ and c~.

Rajah di atas menunjukkan sebuah sisi empat OABC dengan AO = a~, → →

OB = b~, →

OC = c~ dan 2OM = MB. Garis AC dan garis OB menyilang pada titik M yang merupakan titik tengah bagi garis AC. Cari vector →OM dalam sebutan a~

dan c~.

OB → = b~ and 2OM = MB, hence OM = 1 → 3

OB = 1 → 3 ~ b AM = → → MC

AO + → → OM = → MO + → OC – → OA + → OM = – → OM + → OC –a~ + 1

3 b ~ = – 1 3 ~ + c~ b

2

3 b ~ = a ~ + c~

b ~ = 3

2 (a~ + c~) OM = 1 →

3 ~ b

= 1

3 [ 3

2 (a~ + c~)]

= 1

2 (a~ + c~) Vectors in a Cartesian Plane

Vektor dalam Satah Cartes

8.3

pg. 227 – 235^Textbook

1. Vectors in a Cartesian plane can be written as Vektor dalam satah Cartes boleh ditulis sebagai (a) x i~ + y j~

(b)

1

^x_y

2

^{, where/} dengan keadaan i~ =

1

¹₀

2

^{, j}_~ ⁼

1

⁰₁

2

i~ and j~ are vectors of magnitude 1 unit that parallel to x-axis and y-axis respectively.

i~ dan j~ ialah vektor dengan magnitud 1 unit yang masing-masing selari dengan paksi-x dan paksi-y.

2. If A(x1, y₁) is a point on a Cartesian plane, the vector formed from the origin O to point A is →

OA = x₁i~ + y¹j~.

OA is known as a position vector.→

Jika A(x₁, y₁) ialah suatu titik pada satah Cartes, vektor yang terbentuk dari asalan O ke titik A ialah →

OA = x₁i~ + y¹j~. → OA dikenali sebagai vektor kedudukan.

3. The magnitude for a vector r~ = x i~+ y j~ is Magnitud bagi suatu vektor r~ = x ḭ + y j̰ ialah

u r~u = √x ² + y ² 4. The unit vector in the direction of r~ is

Vektor unit dalam arah r~ ialah

~ = ^r r~

u r~u = x i~ + y j~

√x ² + y ²

5. The magnitude of the unit vector in the direction of a vector is 1 unit.

Magnitud vektor unit dalam arah suatu vektor ialah 1 unit.

6. (a) Addition/ ^Penambahan

(a i~ + b j~) + (c i~ + d j~) = (a + c) i~ + (b + d) j~

(b) Subtraction/ Penolakan

(a i~ + b j~) – (c i~ + d j~) = (a – c) i~ + (b – d) j~

(c) Pendaraban dengan skalar/ Multiplication by scalar k(a i~ + b j~) = ka i~ + kb j~

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14. The diagram shows some vectors drawn on the Cartesian plane.

^PL3

Rajah di sebelah menunjukkan beberapa vektor dilukis pada satah Cartes.

(i) Express each of the following vectors in the form of

Ungkapkan setiap vektor berikut dalam bentuk

(a) x i~ + y j~

(b) 1 ^x _y 2

(ii) Hence, find the magnitude of each of the vectors.

Seterusnya, cari magnitud bagi setiap vector tersebut.

Example q ~

(i) (a) 2 i~ + j~

(b) 1 ² ₁ 2

(ii) uq ~u = √2

²

+ 1

² √x² + y²

= √5 units

(a) r~

(i) (a) –3 i~ – 2 j~

(b) 1 ^–3 _–2 2

(ii) ur~u

= √(–3)

²

+ (–2)

²

= √13 units

(b) s~

(i) (a) –3 i~ + 2 j~

(b) 1 ^–3 ₂ 2

(ii) u s~u

= √(–3)

²

+ 2

²

= √13 units

(c) t~

(i) (a) –4 i~ + 3 j~

(b) 1 ^–4 3 2

(ii) u t~u

= √(–4)

²

+ 3

²

= √25 units

= 5 units

(d) u ~

(i) (a) 3 i~ – j~

(b) 1 –1 ³ 2

(ii) uu~u

= √3

²

+ (–1)

²

= √10 units

(e) v~

(i) (a) 3 j ~ (b) 1 ⁰ 3 2

(ii) uv~u = √3

²

= 3 units

(f) w ~

(i) (a) –3 i~ – 3 j~

(b) 1 ^–3 –3 2

(ii) uw ~u

= √(–3)

²

+ (–3)

²

= √18

= 3√2

15. Determine unit vector for each of the following.

^PL3

Tentukan vektor unit bagi setiap berikut.

Example

(i) → AC = 3 i~ + 4 j~

Unit vector in the direction of → AC = AC →

u → AC u = 3 i~ + 4 j~

√3

²

+ 4

²

x i~ + y j~

√x² + y²

= 3

5 i~ + 4 5 j

~

(a) a~ = 12 i~ – 5 j~

~ =

^{^}

a 12 i~ – 5 j~

√12

²

+ (–5)

²

= 12

13 i~ – 5 13 j

~

(b) → PQ = –3 i~ – 3 j~

Unit vector in the direction of → PQ = –3 i~ – 3 j~

√(–3)

²

+ (–3)

²

= –3 i~ – 3 j~

3√2

= – 1

√2 i~ – 1

√2 j

~

(ii) s~ = 1 _–12 ⁵ 2

^

s~ = 1

√5

²

+ (–12)

²

1 _–12 ⁵ 2

= 1 ^{– 12 13 5} ₁₃ 2

(c) → RS = 1 ⁴ 6 2

Unit vector in the direction of → RS = 1 √4

²

+ 6

²

1 ⁴ ₆ 2

= 1

2√13 1 ⁴ ₆ 2 ⁼ 1 ^√13 _√13 ^{2 3} 2

(d) r~ = 1 15 ^–8 2

^{^}

r~ = 1

√(–8)

²

+ 15

²

1 ₁₅ ^–8 2

= 1 ^{– 8 15} ₁₇ ¹⁷ 2

q~

w~ v~

u~

s~ t~

r~ y

O x

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16. Given the vectors p = –8 i~ + 2j~, q = 3i~ + 5j~, r = –2i~ – 6j~, s = 1 _–6 ³ 2 ^{, t =} 1 ^–10 _–4 2 ^{and u =} 1 ⁷ ₇ 2 , find in the form of x i~+ y j~ and 1 ^x _y 2 for each of the following.

^PL3

Diberi vektor-vektor p =–8 i~+ 2 j~, q = 3 i~+ 5 j~, r = –2 i~ – 6 j~, s =

1

–6³

2

^{, t =}

1

^–10–4

2

^{dan u =}

1

⁷7

2

, cari dalam bentuk x i~+ y j~ dan

1

^xy

2

bagi setiap berikut.

Example p + q

= (–8 i~ + 2 j~) + (3i~ + 5j~)

= –8 i~ + 3 i~ + 2 j~ + 5j~

= –5 i~ + 7 j~

= 1 ^–5 ₇ 2

• Add or subtract the corresponding components in the vectors involved.

Tambah atau tolak komponen yang sepadan dalam vektor-vektor yang terlibat.

• Multiply each component with the scalar involved.

Darab setiap komponen dengan skalar yang terlibat.

(a) 2u – 3p + 4r

= 2(7 i~ + 7 j~) – 3(–8i~ + 2j~) + 4(–2 i~ – 6 j~)

= 14 i~ + 14 j~ + 24i~ – 6j~

– 8 i~ – 24 j~

= 30 i~ – 16 j~

= 1 –16 ³⁰ 2

(b) 5s + 2t – u + p

= 5(3 i~ – 6 j~) + 2(–10i~ – 4j~) – (7 i~ + 7 j~) + (–8i~ + 2j~)

= 15 i~ – 30 j~ – 20i~ – 8j~

– 7 i~ – 7 j~ – 8i~ + 2j~

= –20 i~ – 43 j~

= 1 ^–20 _–43 2

(c) 4r + 2q + 2 3 s

= 4(–2 i~ – 6 j~) + 2(3i~ + 5j~) + 2

3 (3 i~ – 6 j~)

= –8 i~ – 24 j~ + 6i~ + 10j~

+ 2 i~ – 4 j~

= –18 j ~ = 1 –18 ⁰ 2

(d) – 1

2 t – u – q = – 1

2 (–10 i~ – 4 j~) – (7i~ + 7j~) – (3 i~ + 5 j~)

= 5 i~ + 2 j~ – 7i~ – 7j~ – 3i~ – 5j~

= –5 i~ – 10 j~

= 1 _–10 ^–5 2

17. It is given that the position vectors of P, Q and R are i~ + j~, 3i~ – 2j~ and 6i~ + kj~ respectively.

^PL4

Diberi bahawa vektor kedudukan P, Q dan R masing-masing ialah i~ + j~, 3 i~ – 2 j~ dan 6 i~ + k j~.

Example

Find the vector of → RQ in term of k.

Cari vektor →RQ dalam sebutan k.

RQ = → → RO + → OQ

–OR + → OQ→

= –(6 i~ + k j~) + (3i~ – 2j~)

= –6 i~ – k j~ + 3i~ – 2j~

= –3 i~ – (k + 2) j~

(a) Find the vector of → PQ.

Cari vektor →PQ .

PQ = → → PO + → OQ

= –( i~ + j~) + (3i~ – 2j~)

= – i~ – j~ + 3i~ – 2j~

= 2 i~ – 3 j~

(b) Find the vector of → PR in term of k.

Cari vektor →PR dalam sebutan k.

PR = → → PO + → OR

= –( i~ + j~) + (6i~ + kj~)

= – i~ – j~ + 6i~ + kj~

= 5 i~ + (k – 1) j~

(c) If → PQ = µ → PR, find the values of µ and k.

Jika →PQ = µ →PR , cari nilai µ dan k.

→ PQ = µ → PR

2 i~ – 3 j~ = µ[5i~ + (k – 1)j~]

By comparing the like terms, 2 = 5µ –3 = (k – 1)µ µ = 2

5 –3 = (k – 1) 1 ² ₅ 2

– 15

2 = k –1

k = – 13

2

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bitan P

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18. Solve each of the following.

^PL5

Selesaikan setiap yang berikut.

Example

The current of a river is flowing parallel to its bank with a velocity of 1.25 km h

^–1

. A swimmer is swimming at 2.5 km h

^–1

perpendicularly to the river bank across the river. Calculate

Daily Application

^{HOTS Applying}

Arus sungai mengalir selari dengan tebing sungai dengan halaju 1.25 km h^–1. Seorang perenang berenang secara berserenjang kepada tebing sungai itu dengan halaju 2.5 km h^–1. Hitung

(i) the magnitude of the resultant velocity of the swimmer in km h

^–1

,

magnitud halaju paduan perenang itu dalam km h^–1,

(ii) the time taken, in hour, if the width of the river is 200 m.

masa yang diambil, dalam jam, jika lebar sungai itu ialah 200 m.

Assuming the direction of the water current is along the positive x-axis and the direction of the swimmer is along the positive y-axis.

The magnitude of the water current = 1.25 km h

^–1

The magnitude of the swimmer = 2.5 km h

^–1

Let v km h

^–1

represents the resultant velocity of the swimmer.

v~ = 1.25 i~ + 2.5 j~

(i) |v~| = √1.25

²

+ 2.5

²

= 2.795 km h

^–1

(ii) Time taken = Displacement Velocity

= 0.2

2.795

200 m = 0.2 km

= 0.07156 hour

(a) In this question, i~ and j~ are the unit vectors, in metre, due east and due north respectively. The initial positions of particles P and Q relative to the origin O are 2 i~ + 4 j~ and 8i~ + 12j~ respectively. Particles P dan Q move from their initial position at the same time with constant velocities in m s

^–1

. Velocities of particles P and Q are i~ + j~ and –2i~ – 3j~ respectively. Show that after 2 seconds, both the particles P and Q will meet.

Daily Application ^HOTS Applying

Dalam soalan ini, i~ dan j~ masing-masing ialah vektor unit, dalam meter, ke timur dan ke utara. Kedudukan awal zarah P dan zarah Q relatif kepada titik asalan O masing-masing ialah 2 i~ + 4 j~ dan 8 i~ + 12 j~. Zarah P and zarah Q bergerak pada masa yang sama dari kedudukan awal dengan halaju tetap dalam m s^–1. Halaju zarah P dan halaju zarah Q masing-masing ialah

i~ + j~ dan –2 i~ – 3 j~. Tunjukkan selepas 2 saat, kedua-dua zarah P dan Q akan bertemu.

Initial position:

OP = → 1 ² ₄ 2

OQ = → 1 12 ⁸ 2

Velocities of P and Q:

v~

^P

= 1 ¹ 1 2

v~

^Q

= 1 ^–2 _–3 2

After 2 seconds,

P: displacement = 2 1 ¹ ₁ 2 ⁼ 1 ² ₂ 2

Q: displacement = 2 1 ^–2 –3 2 ⁼ 1 ^–4 –6 2

OP →

_{t = 2}

= 1 ² ₄ 2 ⁺ 1 ² ₂ 2 ⁼ 1 ⁴ ₆ 2

OQ →

_{t = 2}

= 1 ₁₂ ⁸ 2 ⁺ 1 ^–4 _–6 2 ⁼ 1 ⁴ ₆ 2

At t = 2 s, the position vectors of both the particles P and Q are the same. Hence, P meets Q after 2 s.

1.25 km h^–1

2.5 km h^–1 v km h^–1

displacement

= time taken × velocity

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(b) Given R, S and T are three collinear points. If RS : ST = 3 : 2 and → RT = 10

_r~

.

Diberi R, S dan T adalah tiga titik segaris. Jika RS : ST = 3 : 2 dan →RT = 10 r~.

(i) Find RS and → → ST in terms of

_r~

.

Cari →RS dan →ST dalam sebutan r~.

(ii) If u

_r~

u = 4 units, find u → RSu and u → ST u.

Jika

u

_r~

u

= 4 unit, cari

u

^→RS

u

dan

u

^→ST

u

.

(i) → RS = 3 5 × 10

_r~

= 6

_r~

ST → = 2 5 × 10

_r~

= 4

_r~

(ii) u → RSu = 6u

_r~

u

= 6 × 4

= 24 units u → STu = 4u

_r~

u

= 4 × 4

= 16 units

(c) It is given that u ~ = 1 _{p + 3} ⁸ 2 , v~ = 1 ² ₁ 2 ^{and w} ~ = 1 ^–5 ₄ 2 ^.

^{HOTS Applying}

Diberi bahawa u~ =

1

p + 3⁸

2

^{, v~ =}

1

²1

2

^{dan w~ =}

1

^–54

2

^.

(i) If u ~ is parallel with v~, find the value of p.

Jika u~ selari dengan v~, cari nilai p.

(ii) With the value of p obtained, determine

Dengan nilai p yang diperolehi, tentukan

(a) the value of h and of k such that hu ~ – v~ – kw ~ = 0.

nilai h dan nilai k dengan keadaan hu~ – v~ – kw~ = 0.

(b) the vector in the same direction and parallel to 3v~ + w~ and has a magnitude of 3√2.

vektor dalam arah yang sama dan selari dengan 3v~ + w~ dan mempunyai magnitud 3√2.

(i) u ~ = lv~ where l is a constant 1 _{p + 3} ⁸ 2 ^{= l} 1 ² ₁ 2

= 1 ^2l _l 2

By comparing, 8 = 2l p + 3 = l l = 4 p + 3 = 4

p = 1

(ii) (a) u ~ = 1 ⁸ ₄ 2

h 1 ⁸ ₄ 2 ^– 1 ² ₁ 2 ^{– k} 1 ^–5 ₄ 2 ^{= 0}

1 8h – 2 + 5k 4h – 1 – 4k 2 ^{= 0}

8h – 2 + 5k = 0 ……… 1 4h – 1 – 4k = 0 ……… 2 2 × 2: 8h – 2 – 8k = 0 ……… 3 3 – 1: –13k = 0

k = 0

Substitute k = 0 into 2,

4h – 1 = 0

h = 1

4

(ii) (b) 3v~ + w~ = 3 1 ² ₁ 2 ⁺ 1 ^–5 ₄ 2

= 1 ⁶ ₃ 2 ⁺ 1 ^–5 ₄ 2

= 1 ¹ 7 2

u3v~ + w~u = √1

²

+ 7

²

= √50

= 5√2

Unit vector in the direction of 3v~ + w~

= 1

5√2 1 ¹ ₇ 2

The required vector is 3√2 × 1

5√2 1 ¹ ₇ 2 ^{= 3} ₅ 1 ¹ ₇ 2

= 1 ^{21 3 5} ₅ 2

3 S

R

2 T

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(d) In the diagram, M and R are the midpoints of PQ and OM respectively. The lines OM and PN intersect at R and ON = 1

3 OQ. If OP = p → ~ and → OQ = q ~ ,

^{HOTS Applying}

Dalam rajah di sebelah, M dan R masing-masing ialah titik tengah bagi PQ dan OM. Garis OM dan garis PN bersilang di R dan ON = 13OQ. Jika →OP = p~ dan →OQ = q~,

(i) find, in terms of p ~ and q ~ , in its simplest form,

cari, dalam sebutan p~ dan q~, dalam bentuk termudah,

(a) → OM (b) → OR (c) → PR

(ii) find the ratio PR : PN in its simplest form.

cari nisbah PR : PN dalam bentuk termudah.

(i) (a) → OM = → OQ + → QM = q ~ + 1 2

QP → = q ~ + 1

2 ( → QO + → OP ) = q ~ + 1

2 (–q ~ + p ~ ) = 1 2 p

~ + 1 2 q

~ (b) OR = 1 →

2

OM = 1 →

2 1 ¹ ₂ ~ ^p + 1 2 q

~ 2 ^{= 1} ₄ ~ ^p + 1 4 q

~ (c) PR = → → PO + → OR = –p ~ + 1 ¹ ₄ ~ ^p + 1

4 q ~ 2 ^{= – 3} ₄ ^p ~ + 1 4 ~ q (ii) → PN = → PO + → ON = –p ~ + 1

3

OQ = –p → ~ + 1 3 q ~ PR : PN = 1 ^{– 3} ₄ ^p ~ + 1

4 q ~ 2 ^: 1 ^–p ~ + 1

3 ~ q 2 ^{= 1} ₄ 1 ^–3p ~ + q ~ 2 ^{: 1} ₃ 1 ^–3p ~ + q ~ 2 ^{= 1} ₄ ^{: 1} ₃ ^{= 3 : 4}

(e) Points P and Q have the position vectors p ~ and q ~ respectively. A point R lies on the line OP such that OR = 1

2 OP and a point S lies on line OQ such that OS = 1

3 OQ. Express QR → and → SP in terms of p ~ and q ~ . The lines QR and SP intersect at T.

^{HOTS Analysing}

Titik P dan titik Q mempunyai vektor kedudukan p~ dan q~ masing-masing. Suatu titik R berada pada garis OP dengan keadaan OR = 12OP dan suatu titik S berada pada garis OQ dengan keadaan OS = 13OQ. Ungkapkan →OQ dan →SP dalam sebutan p~ dan

~q. Garis QR dan garis SP bersilang di T.

(i) Given that → PT = l SP, express → → OT in terms of l, p ~ and q ~ .

Diberi bahawa →PT = l →SP , ungkapkan →OT dalam sebutan l, p~ dan q~.

(ii) Given that → QT = µ → QR, express → OT in terms of µ, p ~ and q ~ .

Diberi bahawa →QT = µ →QR, ungkapkan →OT dalam sebutan µ, p~ dan q~.

Hence, calculate the value of l and of µ. With the values obtained, find the position vector of T.

Seterusnya, hitung nilai bagi l dan µ. Dengan nilai-nilai yang diperolehi, cari vektor kedudukan bagi T.

OP → = p

~ , → OQ = q

~ , → OR = 1 2 p

~ and → OS = 1 3 q

→ ~

QR = → QO + → OR = –q

~ + 1 2 p

~ = 1 2 p

~ – q

→ ~

SP = → SO + → OP = – 1 3 q

~ + p

~ = p

~ – 1 3 q

~ (i) → OT = → OP + → PT

= p

~ + l SP →

= p

~ + l 1 ~ ^p – 1 3 q

~ 2

= (1 + l)p ~ – 1 3 lq ~ (ii) → OT = → OQ + → QT = ~ q + µ → QR

= q

~ + µ( 1 2 p

~ – q

~ )

= 1

2 µp

~ + (1 – µ)q ~

Hence, (1 + l)p ~ – 1 3 lq ~ = 1

2 µp

~ + (1 – µ)q

~ 1 + l = 1

2 µ ——————— 1

– 1

3 l = 1 – µ —————— 2 From 1, l = 1

2 µ – 1 ——— 3 Substitute 3 into 2,

– 1

3 1 ¹ 2 µ – 1 2 ^{= 1 – µ}

– 1

2 µ + 1 = 3 – 3µ

5

2 µ = 2

µ = 4

5 l = 1

2 1 ⁴ ₃ 2 ^{– 1}

= – 3 5

\ OT → = 1 ^{1 – 3} 5 2 ~ ^p – 1

3 1 ^{– 3} 5 2 ~ ^q = 2 5 p

~ + 1 5 q

~

P M

Q

O N

R

Q

T

P O

S R

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bitan P

elang

i S dn. Bhd .

Paper 1

1. It is given →OP =

1

^m5

2

^{, →OQ =}

1

²3

2

^{and →OR =}

1

–7ⁿ

2

^{, where m}

and n are constants. Express m in terms of n, if points P, Q and R lie on a straight line.

Diberi →

OP = 1 ^m ₅ 2 ^{, OQ = →} 1 ² ₃ 2 ^{dan OR = →} 1 _–7 ⁿ 2 , dengan keadaan m dan n ialah pemalar. Ungkapkan m dalam sebutan n, jika titik-titik P, Q dan R terletak pada satu garis lurus.

[3]

Ans: m = 15(12 – n)

2. The diagram shows a regular hexagon with centre O.

Rajah di bawah menunjukkan sebuah heksagon sekata dengan pusat O.

F E

D A

B O

C

(a) Express →BF + →FD + →FA as a single vector.

Ungkapkan → BF + →

FD + →

FA sebagai satu vektor tunggal.

(b) Given →OA = a~, →

OB = b~ and the length of each side of the hexagon is 2 units, find the unit vector in the direction of →AB, in terms of a~ and b~.

Diberi →

OA = a~, →

OB = b~ dan panjang setiap sisi heksagon itu ialah 2 unit, cari vektor unit dalam arah →

AB, dalam sebutan a~ dan b~.

[3]

Ans: (a) →BC (b) – 12a~ + 1

2b~

3. Given a~ = –2 i~ + 3 j~, b~ = 2 i~ – 7 j~ and c~ = – i~ + 4 j~.

If la~ + µb~ = 2c~, find the values of l and µ.

Diberi bahawa a~ = –2 i~ + 3 j~ , b~ = 2 i~ – 7 j~ dan c~ = – i~ + 4 j~.

Jika λa~ + µb~ = 2c~, cari nilai λ dan µ.

[3]

Ans: l = – 1

4 , µ = – 54

4. The diagram shows a trapezium ABCD.

Rajah di bawah menunjukkan trapezium ABCD.

A

B

D

C q~

p~

SPM 2015

SPM 2016

SPM 2017

Given p~ =

1

⁴3

2

^{and q}~ =

1

k – 1²

2

where k is a constant, find the value of k.

Diberi p~ = 1 ⁴ ₃ 2 _{dan q~ =} 1 _{k – 1} ² 2 , dengan keadaan k ialah pemalar, cari nilai k.

[3]

Ans: k = 5 2

5. The diagram shows the vectors →OP , →OQ and →OR drawn on a square grid.

Rajah di bawah menunjukkan vektor-vektor → OP, →

OQ dan → OR dilukis pada grid segi empat sama.

O P Q

R S

p~ q~

(a) Express →OR in the form mp~ + nq~, where m and n are constants.

Ungkapkan →

OR dalam bentuk mp~ + nq~, dengan keadaan m dan n ialah pemalar.

(b) On the same diagram, mark and label the point S such that →RS + →OQ = 2→OP.

Pada rajah yang sama, tanda and label titik S dengan keadaan →

RS + → OQ = 2 →

OP.

[3]

Ans: (a) p~ + 2q~

(b) Refer to the diagram above

6. A(–2, 3) and B(2, 5) lie on a Cartesian plane. It is given that 3 →OA = 2 →OB + →OC. Find

A(–2, 3) dan B(2, 5) terletak pada satah Cartes. Diberi bahawa 3 →

OA = 2 → OB + →

OC. Cari

(a) the coordinates of C,

koordinat C.

(b) u →ACu.

[4]

Ans: (a) (–10, –1) (b) 4√5 units

SPM 2018

SPM 2018

SPM ^Practice 8

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bitan P

elang

i S dn. Bhd .

7. (a) Calculate the magnitude of the vector

1

^–35

2

^.

Hitung magnitud bagi vektor 1 ^–3 ₅ 2 ^.

(b) Hence, state the unit vector of

1

^–35

2

^.

Seterusnya, nyatakan vektor unit bagi 1 ^–3 ₅ 2 ^.

[3]

Ans: (a) √34 units

(b)

1 ^–

_√34^√3453

2

8. Given the points P(5, 3), Q(2, –4) and R(–1, –2),

Diberi titik-titik P(5, 3), Q(2, – 4) dan R(–1, –2),

(a) write →RP in the form of

1

^xy

2

^,

tulis →

RP dalam bentuk 1 ^x _y 2 ^,

(b) find →RP – →RQ in the form of a i~ + b j~.

cari → RP – →

RQ dalam bentuk a i~ + b j~.

_[3]

Ans: (a)

1

⁶5

2

(b) 3 i~ + 7 j~

9. Given →OP = 5p~, →OQ = 15q~ and →OR = 24q~ – 3p~, show that Q lies on the line PR.

Diberi →

OP = 5p~, →

OQ = 15q~ dan →

OR = 24q~– 3p~, tunjukkan bahawa Q berada pada garis PR.

[3]

Ans: Refer to Answer Section

10. Given r~ = c~ + p(c~ + 2d~) and s~ = 2c~ + d~ + qc~, where c~ and d~ are non-parallel and non-zero. If r~ = s~, find the values of p and q, where p and q are constants.

Diberi r~ = c~ + p(c~ + 2d~) dan s~ = 2c~ + d~ + qc~, dengan keadaan c~ dan d~ tidak selari dan bukan sifar. Jika r~ = s~, cari nilai p dan q, dengan keadaan p dan q ialah pemalar.

[3]

Ans: p = 12, q = – 12

Paper 2

1. The diagram shows a triangle ABC.

Rajah di bawah menunjukkan sebuah segi tiga ABC.

C T S

A R

B

SPM 2018

SPM 2015

It is given that AR : RB = 1 : 2, BT : TC = 2 : 1, →→ AR = 2x~ and AC = 6y~.

Diberi AR : RB = 1 : 2, BT : TC = 2 : 1, →

AR = 2x~ dan → AC = 6y~.

(a) Express in terms of x~ and y~,

Ungkapkan dalam sebutan x~ dan y~,

(i) →CR , (ii) →CT .

[3]

(b) Given x~ = 2 i~ and y~ = – i~ + 4 j~, find u→CTu.

Diberi x~ = 2 i~ dan y~ = –i~ + 4j~, cari

^u

→ CT

^u

.

[2]

(c) Given →CS = p→CR and →ST = q→AT, where p and q are constants, find the values of p and q.

Diberi → CS = p →

CR dan → ST = q →

AT, dengan keadaan p dan q ialah pemalar, cari nilai p dan q.

[5]

Ans: (a) (i) 2x~ – 6y~ (ii) 2x~ – 2y~

(b) 10 units (c) p = 35, q = 25

2. The diagram shows the position and the direction of boats P, Q and R on a river.

Rajah di bawah menunjukkan kedudukan dan arah bot P, Q dan R di suatu sungai.

Riverbank Tebing sungai

R Q

P

Both boats P and Q move in the direction of the water current which flow with a velocity of w~ = (2 i~ + 1

2~j) m s^–1. The velocity of boat P is p~ = ( i~ + j~) m s^–1 and the velocity of boat Q is q~ = (3 i~ + 2 j~) m s^–1.

Kedua-dua bot P dan bot Q bergerak mengikut arah arus air yang mengalir dengan halaju w~ = (2 i~ + 1

2 j~ ) m s

^–1

. Halaju bot P ialah p~ = (i~ + j~) m s

^–1

dan halaju bot Q ialah q~ = (3i~ + 2j~) m s

^–1

.

(a) Determine how many times the resultant velocity of boat Q compare to the resultant velocity of boat P.

Tentukan berapa kali ganda halaju paduan bot Q berbanding halaju paduan bot P.

[4]

(b) At the half way, boat R changes its velocity to r~ = ( i~ – 3

2~j) m s^–1. Find

Pada separuh perjalanan, bot R menukar halajunya ke r~ = ( i~ – 3

2 j~ ) m s

^–1

. Cari

(i) the resultant velocity of boat R,

halaju paduan bot R,

(ii) the unit vector in the direction of boat R.

vektor unit dalam arah bot R.

[3]

Ans: (a) 53 times

(b) (i) (3 i~ – j~) m s^–1 (ii) 3

√10i~ – 1

√10~j

SPM 2016