Vectors
Scalars and Vectors
A vector is a quantity that has a magnitude and a direction. One example of a vector is
velocity. The velocity of an object is determined by the magnitude(speed) and direction of
travel. Other examples of vectors are force, displacement and acceleration.
A scalar is a quantity that has magnitude only. Mass, time and volume are all examples of
scalar quantities.
Example 1.
If a car travels from point O to point A, which is 50km. away in a north-easterly direction,
then the displacement of the car from O is 50km.NE. The displacement of the car is
specified by the distance travelled (50km.) and the direction of travel (NE.) from O.
Displacement is therefore a vector, and the magnitude of the displacement (distance), is a
scalar.
On the diagram below the displacement is represented by the directed line segment OA .
The length of the line represents the magnitude of the displacement and is written OA .
The arrowhead represents the direction of the displacement.
N
A
OA =50km.
450
O
E
Example 2.
A force of 50 Newton at an angle of 200 to the horizontal downward, is applied to a
wheelbarrow. The diagram below shows a vector representing this force.
A
AB = 50N
200
50N
B
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Geometric Vectors
We will be considering vectors in three-dimensional space defined by three mutually
perpendicular directions.
Definitions and conventions.
Vectors will be denoted by lower case bold letters such as a, b, c.
Unit vectors i, j, k
Vectors with a magnitude of one in the direction of the x-axis, y-axis and z-axis will be
denoted by i, j, and k respectively.
The notation (a, b, c) will be used to denote the vector (ai + bj + ck) as well as the coordinates of a point P (a, b, c). The context will determine which meaning is correct.
Example
z
• P (3,4,5)
k
j
y
O
i
x
i = j = k =1
In the diagram above the point P has coordinates (3,4,5).
The vector OP is the vector 3i + 4j + 5k. This may also be written (3,4,5).
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Directed Line Segment.
The directed line segment, or geometric vector, PQ , from P( x1 , y1 , z1 ) to Q( x2 , y2 , z2 ) is
found by subtracting the co-ordinates of P (the initial point) from the co-ordinates of Q (the
final point).
PQ =
cb x − x gi + b y − y gj + bz − z gkh
2
1
2
1
2
1
Example.
b g b g b
z
g
PQ = 5 − 3 i + 6 − 4 j + −1 − 1 k
PQ = 2i + 2 j − 2k
b g
P 3,4,1
y
b
g
Q 5,6,−1
x
The directed line segment PQ is represented by the vector 2i + 2j − 2k, or (2, 2, −2). Any
other directed line segment with the same length and same directon as PQ is also
represented by 2i + 2j − 2k or (2, 2, −2).
The directed line segment QP has the same length as PQ but is in the opposite direction.
QP = − PQ = − (2i + 2j − 2k) = − 2i − 2j + 2k or (−2,−2, 2)
Position Vector.
The position vector of any point is the directed line segment from the origin O (0,0,0) to the
point and is given by the co-ordinates of of the point.
The position vector of P(3, 4, 1) is 3i + 4j + k, or (3, 4, 1).
Exercise 1. Given the points A(3, 0,4)
(a) AB
(b) AC
(d) BC
(e) CA
B(−2, 4, 3) and
C(1,−5,0), find:
(c) CB
(f) The position vectors of A, B and C.
Compare your answers 1(b)and 1(e), and 1(c) and 1(d). What do you notice?
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Length or Magnitude of a Vector.
The length of a vector a = a1i + a2j + a3k is written a and is evaluated by:
a = a 12 + a 22 + a 23
For example the length of the vector 2i + 2j - 2k equals
b g
2 2 + 2 2 + −2
2
= 12 = 2 3
a1, a2 , a3 are often referred to as the components of vector a.
Unit Vector
A vector with a magnitude of one is called a unit vector. If a is any vector then a unit vector
parallel to a is written aɵ (a “hat”). The “hat” symbolises a unit vector.
Vector a can then be written
therefore
a = a aɵ
aɵ =
a
a
Example
A unit vector parallel to
is the vector
a = (1, 2, 3)
aɵ =
b
g
b
g
1, 2, 3
1
a
1, 2, 3
=
=
a
14
12 + 2 2 + 32
Adding and Subtracting Vectors.
Vectors are added or subtracted by
•
adding or subtracting their corresponding components
•
using the triangle rule
•
by using the parallogram rule.
Example
If a = (-3, 4, 2) and b = (−1 −2, 3), find:
(i) a + b
( ii) a −b.
Adding or subtracting components
(i) a + b = (−3, 4, 2) + (−1, −2, 3) = (−4, 2, 5)
Similarly
(ii) a −b = (−3, 4, 2) − (−1, −2, 3) = (−2, 6, −1)
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Triangle Rule
R
(i) a + b
a+b
b
P
a
Q
Place the tail of vector b at the head of vector a (point Q). The directed line segment PR
from the tail of vector a to the head of vector b is the vector a + b.
(ii) To subtract b from a, reverse the direction of b to give −b then add a and −b.
a −b = a + (−b)
Parallelogram Rule
S
(i) a + b
R
b
a+b
P
a
Q
a and b are placed “tail-to-tail”( point P) and the parallelogram (PQRS) completed. The
diagonal PR is the sum a + b.
(ii) To find (a − b), reverse the direction of b to give −b then add a and −b.
Exercise 2.
For vectors p (3, 6, 5), q (−4, 1, 0) and r (1, −3, 5) find:
(a) p + q
(b) r + p
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(c) p − q
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Multiplication of a vector by a scalar.
To multiply vector a = a1i + a2 j + a3k by a scalar, m, multiply each component of a by m.
ma = ma1i + ma2 j + ma3k
The result is a vector of length |m|×|a|
If m > 0 the resultant vector is in the same direction as a
If m < 0 the resultant vector is in the opposite direction from a.
Two vectors a and b are said to be parallel if and only if a = kb where k is a real constant.
Example 1
a = (3i + j − 2k) is multiplied by 7
7a = 7(3i + j − 2k) = 21i + 7j − 14k.
The magnitude of a is
a =
7a =
2
32 + 12 + ( −2 ) = 14
2
212 + 7 2 + ( −14 ) = 686 = 7 14 = 7 a
Example 2
Find the value of m so that the vector a, ( 4, m, 8) is parallel to the vector b, (−6, 3, 12).
For a and b to be parallel a = kb
Therefore
( 4, m, 8) = k (−6, 3, −12) = (−6k, 3k, −12k)
equating “i”components
4 = −6k
k=
−2
3
equating “j”components
m = 3k
∴ m = 3×
−2
3
m = −2
Exercise 3
Find the following
(a) 3 ×(i + 3j −5k)
(b) 8 × (7i + 2j + 4k)
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(c) − 4 × ( j − 3k)
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Multiplication of a vector by a vector
(1) Dot product or scalar product
The dot product of two vectors a ( a1 , a2 , a3 ) and b ( b1 , b2 , b3 )
is a scalar, defined by
a
a • b = a b cos θ , where θ is the angle between a and b
θ
b
and a • b = b • a
If a is perpendicular to b then a • b = 0 (cos(π/2)= 0).
In particular i • j = j • k = k • i = 0
If a is parallel to b then a • b = a b (cos(0)=1)
In particular i • i = j • j = k • k = 1
Also ( a1i + a2 j + a3k ) • ( b1i + b2 j + b3k ) =
( a1b1i • i + a1b2i • j + a1b3i • k + a2b1 j • i + a2b2 j • j + a2b3 j • k + a3b1k • i + a3b2k • j + a3b3k • k )
Thus a • b can be defined by
a • b = a1b1 + a2b2 + a3b3
Example 1
Find the dot product of (2i + 3j + 4k) and (− i − 2j + k)
(2i + 3j + 4k) • (− i − 2j + k) = 2(−1) + 3(−2) + (4)(1)
= −2 − 6 + 4
=−4
(2i + 3j + 4k) • (− i − 2j + k) = − 4
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Example 2
Find the scalar product of a and b, as drawn, below where
a = 14 and b = 6
a • b = a b cosθ
= 14 × 6 × cos 300
a
300
= 14 × 6 ×
b
3
2
=3 7
a•b = 3 7
Exercise 4 Find the dot product of the following vectors:
(a) 3i and 5j
(b) 2i + 3k and 7i + 2j + 4k
(d) (2, 0, 4) and (− 3, 1, 3)
(e) (0, 5, 1) and (4, 0, 0)
(f)
(c) 5k and j − 2k
(g)
5
3
450
4
4
(2) Cross product or vector product
The cross product of two vectors a and b is the vector a × b, which is perpendicular to both a
and b and is given by
i
j k
a × b = a1 a2 a3
b1 b2 b3
The magnitude of a × b is given by a × b = a b sin θ where θ is the angle between a and b.
The direction of a × b is that in which your thumb would point if the fingers of your right are
curled from a to b.
In particular
i × j = k,
j × k = i,
k×i=j
i×k=−j
k× j = − i
j×i=−k
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If a is parallel to b then a × b = 0. (sin00 = 0)
In particular
i×i=j×j=k×k=0
If a is perpendicular to b then a × b = a b ( sin900 =1)
a × b = − b × a ( the cross product is not commutative.)
Example 1
Find a × b if a = 2i +3j + k and b = 5j +3k
i
j k
a × b = 2 3 1 = (9 − 5)i − (6 − 0)j + (10 − 0)k
0 5 3
a × b = 4i − 6j + 10k
Example 2
Find a × b if a = (2,1,1) and b = (−2,4,0)
i
j
k
a × b = 2 1 1 = (0 − 4)i − (0 + 2)j + (8 + 2)k
−2 4 0
a × b =− 4i − 2j + 10k
Example 3
Find a × b if a = (2,1,1) and b = (8,4,4)
Because a = 4b, a is parallel to b therefore a × b = 0
Exercise 5 Find the cross product of the following vectors:
(a) j × k
(b) i × 4i
(d) 3j × 5i
(e) (i − 3j + k) × (2i + j − k)
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(c) (2i + 3j − k) × (3j + 2k)
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Projection of vectors.
Consider the diagram below:
Q
a
θ
P
S
b
R
Let PQ = a and PS = b.
Scalar projection
The scalar projection of vector a in the direction of vector b is the length of the straight line
PR or PR .
PR = a cosθ .
Also
cosθ =
a •b
ab
(because a • b = a b cosθ )
Therefore
 a •b  a •b
b ˆ
PR = ( a ) 
=
= a • bɵ (cancel a , and use
=b)
 a b 
b
b


The scalar projection of a vector a in the direction of vector b is given by
a •b
= a • bɵ
b
or
a cosθ
Vector projection
The vector projection of vector a in the direction of vector b is a vector in the direction of b
with a magnitude equal to the length of the straight line PR or PR .
Therefore the vector projection of a in the direction of b is the scalar projection multiplied by
a unit vector in the direction of b.
The vector projection of vector a in the direction of vector b is given by
(a • b ) b
( a • bɵ ) b̂ =
2
b
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Angle between two vectors
The angle, θ between two vectors can be found from the definition of the dot product
a • b = a b cos θ
therefore cosθ =
a •b
ab
θ can also be found from cosθ =
a • bɵ
a
Example
Find (a) the scalar projection of vector a = (2, 3, 1) in the direction of vector b = (5, −2, 2).
(b) the angle between a and b.
(c) the vector projection of a in the direction of b.
(a) Scalar projection
b = 25 + 4 + 4 = 33
therefore
(5, − 2, 2)
bɵ =
33
b g
(5, − 2, 2) 10 + −6 + 2
6
=
a • bɵ = (2, 3, 1)•
=
33
33
33
The scalar projection of a in the direction of b is
6
33
(b) Angle between a and b
The scalar projection of a in the direction of b is also equal to a cosθ , where
θ is the angle between a and b.
6
= a cosθ .
33
Therefore
∴
6
=
33
∴ cosθ =
a =
2 2 + 32 + 12 = 14
14 cosθ
6
= 0.2791
33 × 14
∴ θ = 74°
The angle between a and b is 740.
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(c) Vector projection
The vector projection a in the direction of b equals:
(scalar projection a in the direction of b) b̂
=
6
(5, − 2, 2)
6(5, − 2, 2)
×
=
33
33
33
The vector projection of a in the direction of b is
6(5, − 2, 2)
33
Exercise 6 For the following pairs of vectors find:
(i) the scalar projection of a on b
(ii) the angle between a and b
(iii) the vector projection of a on b
(a) a = (2, 3 ,1)
b = (5, 0, 3)
(b) a = (0, 0, 3)
(c) a = (5, 0, 0)
b = (0, 3, 0)
(d) a = (−3, 2, −1) b = (2, 1, 2)
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b = (0, 0, 7)
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Answers
1.(a) (−5, 4, −1)
(b) (−2, −5, −4)
(d) (3, −9, −3)
(f) OA = 3i + 4k ,
(c) (−3, 9, 3)
(e) (2, 5, 4)
OB = −2i + 4 j + 3k , OC = i − 5 j
2. (a) (−1, 7, 5)
(b) (4, 3, 10)
(c) (7, 5, 5)
3. (a) 3i + 9 j − 15k
(b) 56i + 16 j + 32k
(c) −4 j + 12k
4. (a) 0
(b) 26
(b) 0
5. (a) i
6.(a)(i)
13
34
(c) 9i − 4j + 6k
(ii) 530
(b)(i) 3
(ii) 00
(c)(i)
(ii) 900
0
(d)(i) −2
(d) 6
(c) −10
(ii) 122
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(e) 0
(f) 10 2
(d) −15k
(e) −i + 9j +7 k
(iii)
13
(5, 0, 3)
34
(iii)
3
0, 0, 7
7
b
(g) 0
g
(iii) 0
0
(iii)
b
−2
2, 1, 2
3
g
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