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Ch
apter 5: Vector &
Sca
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Speed Maths
aa
a
B
A
a
TOPIC 5: VECTOR AND SCALAR 5.1 Vector
Vector is a quantity with both magnitude and direction, represented by an arrow which
indicates the direction of the quantity and the length of which is proportional to the magnitude. In example, velocity, displacement, acceleration, weight and force.
Scalar is a quantity possessing only magnitude. In example, speed, length, distance, power and work. a) Vector Notation
It usually written in the form of OB , a
a~ or ~a .
b) Vector Representation c) Equality of Vectors
Two vectors are said to be equal if they have the same magnitude and
direction. In the figure below, three equal vectors, a
, b
and c
have
been represented. The equality of vectors is expressed in the following
way cba
.
The three vectors are also same in magnitude.
cba
d) Negative Vector
Negative of a vector means the vector having the same magnitude but
opposite direction. For example, the negative vector for AB is
AB or BA .
O
B
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cba ~~~
y-axis
x-axis
x
y
O
B (x, y)
a
A
5.2 Vector in Cartesian Coordinate a) Position Vector is a vector which expresses the position of a point with respect to the origin. Point B (x, y) has a position vector of:
ABOAOB (Triangle Law)
yjxiOB (engineering notation in terms of i and j)
y
xOB (in the form of column vector)
b) Unit Vector in OB direction is,
22 yx
yjxi
=
y
x
yx 22
1in terms of i and j or column vector.
c) The magnitude of OB is given by the formula,
OB = 22 yx or
a
= 22 yx .
5.3 Operations of Vectors Resultant Vectors (addition and subtraction of vector) can be solved using Polygon Methods
a) Triangle Construction Method
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cba ~~~
b) Parallelogram Method
Example 1:
Draw the diagram for the following vector if given aOA ~ .
a) aOB ~
2
1
b) aOC ~2
Answer:
a) aOB ~
2
1
O
A
O
B
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b) aOC ~2
Example 2: The following diagram shows a polygon OABCD. Find the resultant vectors for each of the following.
a) CDBCAB
b) ODAOBACB
Answer:
a) ADCDBCAB
O
C
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b) CDODAOBACB
Resultant Vectors can also be solved using by algebraic operations. Example 3:
Given baOA~
3~ , baOB~~2 , baOC
~~ and baOD~
2~3 .
Express the following vectors in terms of a~ and b~
.
a) AB
b) BC2
c) CD
d) CA2
1
Answers:
a) AB OBAO
OBOA
)~~2()
~3~( baba
bbaa~~
3~2~
b) BC2 OCBO 2 OCOB 2 )~~()~~2(2 baba baba ~~~~22 ba ~2~32 ba
~4~6
c) CD ODCO ODOC )~2~3()~~( baba
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baba ~2~3~~ ba ~3~4 ba
~3~4
d) CA2
1 OACO
2
1
OAOC 2
1
)~3~()~~(2
1baba
baba ~3~~~2
1
ba ~2~22
1
ba~~
Example 4:
Given jia 2~ and jib 43~
, find the following.
a) ba~~
b) ba~
2~
c) )~~(2 ba
d) ba~
4~3
Answers:
a) ba~~ jiji 432
ji 35
b) ba~
2~ jiji 4322 jiji 862
ji 94
c) )~~
(2 ab jiji 2432 ji 52 ji 102
d) ba~
4~3 jiji 43423 jiji 161236
ji 196
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Example 5:
If kjip 22 and kjiq 43 , express the following in terms of i, j and k.
a) qp
b) pq 23
Answers:
a) qp kjikji 4322
kji 3
b) pq 23 kjikji 222433 kjikji 4243129
kji 71413
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1. Find the resultant vectors for each of the following.
a)
i. BDAB
ii. ODCO
iii. BCCA
iv. DOOB
b)
i. AB + BC + CD
ii. AC + DE + CD
iii. AB + CD + EF + BC + DE + FA
2. Find the following vector in terms of ~a ,
~b ,
~c and
~d .
a)
i. DA
ii. CE
iii. EB
cbabadcb ~~~,
~~,~~~
b)
i. EA
ii. CE
iii. AD
cbadcdcba ~~~,
~~,~~~~
A B
C
D E
F
A
B C
D
E F
EXERCISE
A B
C D
O
A
B
C
D E
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apter 5: Vector &
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3. Given baOA~
3~ , baOB~~2 , baOC
~~ and baOD~
2~3 . Express the
following vectors in terms of a~ and b~
.
a) AB
b) BC
c) CD
d) DA2
e) CA3
f) BA2
g) ACAD 2
h) CBDB 2
i) DCBD 2
j) DBBO 22
4. Points A, B and C has the following position vectors qp 2 , qp 6 and qp 412
respectively. Prove that the three points are collinear.
5. The position vector for points P, Q and R are ba~~2 , ba
~2~4 and ba
~3~6
respectively. Show that the three points are on a straight line. Then, determine the ratio
of PRPQ : .
6. Show that point L, M and N with position vectors ba~
2~ , ba~~3 and ba
~4~5 are
on a straight line. Hence, determine the ratio of LM and MN .
7. Points S, T and U has the following position vectors qp 3 , qp 22 and qp 35
respectively. Prove that the three points are collinear. Hence, find the ratio of
SUST : .
8. Points U, V and W has the position vectors ji 2 , ji 114 and ji 83
respectively. Prove that the three points are collinear. Hence, find the ratio of
UWUV : .
9. Given jia 2~ and jib 43~
, find the following. Hence, determine the
magnitude.
a) ba~~
b) ba~
2~
c) )~~(2 ba
d) ba~
4~3
e) )~2~
(3 ab
f) ba~
2~2
10. Given jia 32~ , jib 4~
and jic 53~ , calculate the following and
determine the magnitude.
a) cba ~~~
b) cab ~3~~
2
c) abc ~2~
3~2
d) bca~
)~~(2
e) )~~3(~ bac
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5.4 Apply Scalar (Dot) Product of Two Vectors
The scalar or dot product is often defined either algebraically or geometrically.
Algebraically, if given vectors jyixOA11
and jyixOB22
, then the
scalar product for the vector is calculated as:
2121yyxxOBOA (for 2D vectors) or
212121zzyyxxOBOA (for 3D vectors)
Example 1: Given that A(2, -3, 5), B(-3, 1, 4), C(-2, 7, -6). Find: a) BA
b) BC
Answers: a) BA )4(5)1)(3()3(2
2036
11 b) BC )4)(6()1)(7()3(2
2476
11 Example 2:
If jip 2~ and jiq 3~ , find;
a) qp ~~
b) qp ~3~2
Answers:
a) qp ~~ )3)(1()1(2
32
1
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b) qp ~3~2 )3)(3()2(2 jiji )93()24( jiji
9234 1812
6
Example 3:
If kjiA 32 and kjiB 24 , find;
a) BA
b) AB3
1
Answers:
a) BA 13)2)(1()4(2 328
7
b) AB3
1 kjikji 32
3
124
kjikji
3
1
3
224
11)3
1)(2()
3
2(4
1)3
2()
3
8(
3
7
Geometrically, if given vectors a~OA and b~
OB , then the scalar product for the vector is written as:
cosOBOAOBOA or
cos~~~~ baba
Theta, is the angle between vectors where a~ and b~
converges at point O or
diverges out of point O and 1800 .
O
A
B
Diverge Converge
O
A
B
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If cosOBOAOBOA then,
2
2
2
2
2
1
2
1
2121cosyxyx
yyxx
OBOA
OBOA
for 2D vector and
2
2
2
2
2
2
2
1
2
1
2
1
212121
zyxzyx
zzyyxx
for 3D vector.
i. If 0 then, baba
~~~~
Then, 2~~ aaa and 2
~~bbb
1~~ 2 iii and 1
~~ 2 jjj
ii. If 90 then, 0
~~ ba
0~~ ji and 0
~~ ij
Example:
Based on figure below, kjiOA 432 and kjiOB 34 . Find:
a) OBOA
b) The value of .
Answers:
a) OBOA )1(4)3)(3()4(2
498
13
B
A O
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b) The value of .
OBOA
OBOAcos
222222 134432
13
2629
13cos 1
74.61
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1. If jia 2~ and jib 3~
, find;
a) ba~~
b) ba~~2
c) )~~(
~bab
d) )~~(2)~
~2( baab
e) abab ~2~~2~
2. Given kjiAB 32 and kjiBC 24 , find;
a) BCAB2
12
b) )(2
1ABBCAB
c) )( ABBCAB
d) ))(( ABBCBCAB
e) )2)(( ABBCBCAB
EXERCISE
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5.5 Apply Vector (Cross) Product of Two Vectors Vector Product just like scalar product, is another way of multiplying vectors which see the most applications in physics and astronomy. In Mathematical context, vector product is used to calculate the area of triangle and parallelogram.
a) If kzjyixOA111
and kzjyixOB 222 are given, then
222
111
zyx
zyx
kji
OBOA k
y
y
j
x
x
i
j
z
z
k
x
x
i
i
z
z
k
y
y
j
2
1
2
1
2
1
2
1
2
1
2
1
kyxyxjzxzxizyzy122112211221
b) To find the Area of Triangle OAB = OBOA2
1
c) The Area of Parallelogram OACB = sinOBOA = OBOA
Example 1:
Find the cross product for kjiP 32
and kjiR 42
.
Answers:
421
312
kji
RP
kji 112231423241 kji 5112 kji 5112
O
A
B
C
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apter 5: Vector &
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Example 2:
Find the vector product for
A and
B if given kji 2 and kji 22 respectively.
Answers:
122
211
kji
BA
kji 122122112211
kji 055
ji 55
Example 3:
Find the area of triangle OAB if given
OA kji 34 and
OB kji 33
respectively.
Answers:
Area of triangle,
313
1342
1
2
1
kji
OBOA
kji )94()312()19(2
1
kji 59102
1
222 59102
1
2062
1
18.7 units2
O
A
B
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Example 4:
Find the area of parallelogram spanned by vectors kjiOS 32 and
kjiOT 54 .
Answer:
Area of parallelogram,
541
132
kji
OTOS
kji )38()110()415(
kji 51119
222 51119
507
52.22 units2
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apter 5: Vector &
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1. Find the cross product for kjiP 32
and kjiR 42
.
2. Find the vector product for
A and
B if given kji 2 and kji 22 respectively.
3. Given kjiS
32 and kjiT2
123
, find the cross product of the vectors.
4. Solve
NM if kjiM 223
and kjiN 23
1
.
5. Given kjOA 2 and jiOB 4 . Find the area of triangle OAB.
6. Find the area of parallelogram spanned by vectors kjiOS 32 and
kjiOT 54 .
7. Find the area of triangle defined by A(-1, 2, 3) and B(4, -3, 5).
8. Given the vertices P(-2, -3, 0), Q(4, 5, -3) and R(6, 2, 1). Find the area of triangle PQR.
9. Given vector kjiOA 32 and kjiOB 22 and kjiOC 43 . Determine:
a) AB
b) BC
c) CA
d) AC
e) BA
f) CB
g) BCAB
h) CAAC
i) CBBACA j) BABC
k) BAACBC l) CACBABBC
10. A triangle PQR has the corners P(1,-2,4), Q(3,1,-2) and R(-2,3,1). Find PQ , QR and RP .
Then find the length of all the sides.
11. If jiOA 34 , jiOB 26 and jiOC 2 . Find AB , BC and CA . Hence,
determine the length of each side.
12. Determine the angles between the following vectors.
a) kjiA 3 and kjiB 253
b) jia 2~ and jib 43~
c) jia 2~ and jib 3~
d) kjiPQ 32 and kjiPR 24
e) kjiOA 32
and kjiOB 42
EXERCISE
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13. Given kjiOA 32 , kjiOB 22 , kjiOC 43 and kjiOD 25 .
Find the angles of the following vectors.
a) AB and AC
b) BA and CA
c) BA and BC
d) DA and DB
e) CD and CB
f) BD and BD