Addis Ababa Science and
Technology University
DEPARTMENT OF MPS
APPLIED MATHEMATICS IB(MATH 1041)
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Chapter One :
VECTORS AND VECTOR SPACES
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1. Scalar and Vectors
Definitions:
• A scalar is a physical quantity that is described by its magnitude alone.
For example, temperature, mass, length, time , work done and speed are
scalars because they are completely described by a number that tells
"how much"-say a temperature of 20°C, a length of 5 cm, or a speed of 10
m/s.
• A vector is a physical quantity that is described using both magnitude and
direction.
For instance, velocity, displacement, force and Torque are some examples
of vectors.
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(2)
Definitions (Cartesian product of two sets)
• Let A and B be two non empty set, then the Cartesian products of A and B defined as
Remarks:
1. Thus we have
• The set of ordered pairs of real numbers called Cartesian coordinate plane (Real plane) .
2. A vector in
(plane) represented by a point vector
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Cont.
3. Vectors in
 Thus, the vector in
(Space) represented by a point vector
Example 1. plot the following vectors in space P(1,3,2) and Q(-1,3,-2)
Solution :4. The distance b/n P(x,y,z)
(a) from xy-plane is
(b) from yz-plane is
(c) from xz-plane is
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5. The distance b/n the point
P(x,y,z) to
(a) X-axis is
(b) Y-axis is
(c) Z-axis is
6. The distance b/n the point P(x,y,z) to
is
6. The distance b/n the point P(x,y,z) to the
origin is
Example 2: Find the distance of the point
vector P(-1,2,4) from
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7) There is a one-to-one correspondence b/n points in a plane/space to
a vector in plane/space.(i.e Every point is a vector in Rectangular
representation)
8) Vector in n-dimensional space represented in rectangular form as
or
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Geometrical Vector representation
i) Position Vector: A vector in
(space ) is positioned so its initial
point is at the origin and terminal at a point P(x,y,z).i.e V = OP
z
P
V
O
y
x
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Definitions:
1. A located vector is a vector 𝐴𝐵 defined as an arrow whose initial point is at point A
and whose terminal point is at B.
Y
b2
B(b1,b2)
)
a2
A(a1,a2)
a1
b1
X
(b1 , b2 )  (a1  (b1  a1 )), (a 2  (b 2 a 2 ))
𝑏1 = 𝑎1 + 𝑏1 − 𝑎1
 (a1 , a 2 )  (b1  a1 ,b 2 a 2 )
 (a1 , a 2 )  [(b1 ,b 2 )  (a1 , a 2 )]
B  A  ( B  A)
𝑏2 = 𝑎2 + 𝑏2 − 𝑎2
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iii) EQUAL (OR EQUIVALENT) VECTORS
Example 3: Find the located vector
and B(6,-2,4).
Solution :-
where A(4,0,-2)
 Two vectors are said to be Equal/ Equivalent iff they have
the same magnitude and direction.
Let
, then
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3. Vector addition and Scalar multiplication
Definition: For any two vectors 𝑢 = 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑣 = 𝑣1 , 𝑣2 in ℝ2 , we define their sum to
be
𝑢 + 𝑣 = 𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 .
Geometrically, if we represent the two vectors 𝑢 𝑎𝑛𝑑 𝑣 by 𝐴𝐵 𝑎𝑛𝑑 𝐵𝐶
respectively, then 𝑢 + 𝑣 is represented by 𝐴𝐶 , as shown in the diagram below:
C
U+V
V
A
B
U
AB  BC  AC
a) The Triangular Law
D
C
U+V = V+U
V
V
A
B
U
b) The Parallelogram Law
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Scalar Multiple of a Vector
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Addition of vectors on the coordinate plane
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Exercise:- Let A and B be points with position vectors a and b
respectively relative to a fixed origin “o”.
a) Find
in terms of a and b.
b) If P is the mid-point of AB, then find
c) Show that
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Properties of Vector addition & Scalar Multiplication
Let 𝑢, 𝑣 and 𝑤 be vectors in ℝ2 and 𝑐 & 𝑚 are scalars. Then:
a) 𝑢 + 𝑣 ∈ ℝ2
b) 𝑢 + 𝑣 = 𝑣 + 𝑢
c) 𝑢 + 0 = 0 + 𝑢 = 𝑢, 𝑤ℎ𝑒𝑟𝑒 0 = 0,0 ∈ ℝ2 .
d) There exists 𝑤 ∈ ℝ2 such that 𝑢 + 𝑤 = 0 for every 𝑢 ∈ ℝ2 .
e) 𝑢 + 𝑣 + 𝑤 = 𝑢 + 𝑣 + 𝑤
f) 𝑐 𝑚 𝑢 = 𝑐𝑚 𝑢
g) 𝑐 + 𝑚 𝑢 = 𝑐𝑢 + 𝑚𝑢
h) 1. 𝑢 = 𝑢
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Remark: The properties described above also hold true for vectors in ℝ3 , where
0 = 0,0,0 ∈ ℝ3 , replaces the zero vector 0 in ℝ2 .
SUMS OF THREE OR MORE VECTORS
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𝑣 2 = 𝑂𝑅 2 + 𝑃𝑅 2 = (𝑂𝑄)2 + (𝑄𝑅)2 + (𝑃𝑅)2
𝑣 = 𝑣12 + 𝑣22 + 𝑣32
Similarly, for a vector 𝑣 = 𝑣1 , 𝑣2 ∈ ℝ2 , its norm is given by
𝑣 = 𝑣12 + 𝑣22
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Examples: a) If 𝑣 = −1,4,3 , then find 𝑣 .
b) If 𝑢 = 6, 𝑓𝑖𝑛𝑑 𝑥 such that𝑢 = −1, 𝑥, 5 .
Remarks: (i) 𝑣 ≠ 0 𝑖𝑓 𝑣 ≠ 0
(ii) 𝑣 = −𝑣 .
Theorem: If 𝑐 ∈ ℝ, then 𝑐 𝑣 = 𝑐
𝑣 .
Proof: suppose that 𝑣 ∈ 𝑅 𝑛 then
𝑐𝑣 =
𝑐𝑣1 2 + 𝑐𝑣2 2 + ⋯ + 𝑐𝑣𝑘 2
=
𝑐 2 [ 𝑣1 2 + 𝑣2 2 + ⋯ + 𝑣𝑘 2 ]
= 𝑐
𝑣
Definition (Unit Vector)
Any vector 𝑢 satisfying 𝑢 = 1 is called a unit vector.
Examples: The vectors 0,1 , −1,0 ,
1
,
2
−1
2
, 1,0,0 are examples of unit vectors.
N.B: 1. All unit vectors in ℝ2 are of the form cos 𝜃 , sin 𝜃 , 𝑤ℎ𝑒𝑟𝑒 𝜃 ∈ ℝ.
2. For any non-zero vector 𝑣 , the unit vector 𝑢 corresponding to 𝑣 in the
direction of 𝑣 can be obtained as:
𝑢=
𝑉
𝑉
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Example:- Find the unit vector in the direction of u=2i-j-2k.
Solution:-
Example l:- let u and v be parallel vectors such that u=2i-j-2k and
IIvII=12, then find v.
Solution:-
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3. For two points 𝑃 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑄 𝑣1 , 𝑣2 on the plane ℝ2 , we
calculate the distance 𝑑 Ρ, 𝑄 between the two points, as :
𝑑 Ρ, 𝑄 = Ρ𝑄 =
𝑣1 − 𝑢1 2 + 𝑣2 − 𝑢2 2 , 𝑤ℎ𝑒𝑟𝑒 Ρ𝑄 is the vector with initial
point P and terminal point Q,
Ρ𝑄 = 𝑣1 − 𝑢1 , 𝑣2 − 𝑢2 .
4. Distance between two vectors can be viewed as the length of U -V where
𝑈 = 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑉 = 𝑣1 , 𝑣2
.
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The Dot Product (or Scalar Product)
Definition: Suppose that 𝑢 = 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑣 = 𝑣1 , 𝑣2 are vectors in ℝ2 , and that 𝜃 ∈ 0, 𝜋
represents the angle between them. We define the dot product of 𝑢 𝑎𝑛𝑑 𝑉 denoted by 𝑢 ∙ 𝑣
by:
𝑢 ∙ 𝑣 = 𝑢1 𝑣1 + 𝑢2 𝑣2
Or alternatively, we write
𝑢∙𝑣 =
Thus 𝑢
𝑢
𝑣 cos 𝜃, 𝑖𝑓 𝑢 ≠ 0 & 𝑣 ≠ 0
0
𝑖𝑓 𝑢 = 0 𝑜𝑟 𝑣 = 0
𝑣 cos 𝜃 = 𝑢1 𝑣1 + 𝑢2 𝑣2 , for non – zero vector 𝑢 𝑎𝑛𝑑 𝑣 𝑖𝑛 ℝ2 .
Similarly, if 𝑢 = 𝑢1 , 𝑢2 , 𝑢3 𝑎𝑛𝑑 𝑣 = 𝑣1 , 𝑣2 , 𝑣3 are vectors in ℝ3 , we define 𝑢 ∙ 𝑣 as bellow:
𝑢 ∙ 𝑣 = 𝑢1 𝑣1 + 𝑢2 𝑣2 + 𝑢3 𝑣3
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Remark: The dot product of two vectors is a scalar quantity,
and its value is maximum when 𝜃 = 0° and minimum if 𝜃 = 180° 𝑜𝑟 𝜋 radians.
Example: If 𝑢 = 𝑖 − 2𝑗 + 3𝑘 𝑎𝑛𝑑 𝑣 = 0,1,−5 , then find:
a) 𝑢 ∙ 𝑣
b) 𝑢 ∙ 𝑢
c) 𝑢 + 𝑣 ∙ 𝑣
Properties of the dot product
If 𝑢,𝑣,𝑎𝑛𝑑 𝑤 are vectors in the same dimension, and 𝑐 ∈ ℜ, then:
1. 𝑢 ∙ 𝑢 = 𝑢 2
4. 0 ∙ 𝑢 = 0
2. 𝑢 ∙ 𝑣 = 𝑣 ∙ 𝑢
5. 𝑐 𝑢 ∙ 𝑣 = 𝑐 𝑢 ∙ 𝑣 = 𝑣 ∙ 𝑐 𝑢
3. 𝑢 ∙ 𝑣 + 𝑤 = 𝑢 ∙ 𝑣 + 𝑢 ∙ 𝑤
6. 𝑢 ∙ 𝑢 ≥ 0 𝑎𝑛𝑑 𝑢 ∙ 𝑢 = 0 𝑖𝑓𝑓 𝑢 = 0.
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Angle between two vectors
If 𝜃 is the angle between two non – zero vectors 𝑢 𝑎𝑛𝑑 𝑣 , then,
the angle between the two vectors can be obtained by:
𝑢 ∙𝑣
cos 𝜃 = 𝑢
⇒ 𝜃 = 𝑐𝑜𝑠 −1
𝑣
𝑢 ∙𝑣
𝑢
𝑣
, 𝑎𝑛𝑑 𝜃 ∈ 0, 𝜋 .
Example: Find the angle between the vectors 𝑢 = 1,0, −1 𝑎𝑛𝑑 𝑣 = 1,1,0 .
Solution: Let 𝜃 be the angle between the two vectors:
𝑢 ∙𝑣
Then cos 𝜃 = 𝑢
𝑣
=
1
1
−1 1
= 2 ⇒ 𝜃 = 𝑐𝑜𝑠
2∙ 2
2
𝜋
=3
Definition: Two non – zero 𝑢 𝑎𝑛𝑑 𝑣 are said to be orthogonal
𝜋
(perpendicular) iff 𝑢 ∙ 𝑣 = 0, i.e, if 𝜃 = .
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Example: Find the value (s) of x such that the vectors 𝐴 = 1 ,4, 3 𝑎𝑛𝑑
Β = 𝑥, −1, 2 are orthogonal.
Definition: If P and Q are points in 2 or 3 spaces, the distance between P
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and Q, denoted by Ρ − 𝑄 is given by
Ρ−𝑄 =
Ρ−𝑄 . Ρ−𝑄
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Theorem: Given two vectors 𝑢 & 𝑣 in space, 𝑢 + 𝑣
=
𝑢−𝑣
iff
𝑢 𝑎𝑛𝑑 𝑣 are orthogonal vectors.
Proof : (⤏) Given
WTS
2
𝑢+𝑣
2
2
𝑢−𝑣
𝑢 𝑎𝑛𝑑 𝑣 are orthogonal
+ 2𝑢. 𝑣 +
2
𝑣
= 𝑢 − 𝑣 ∙ (𝑢 − 𝑣)
= 𝑢
By hypothesis
𝑢
=
= 𝑢 + 𝑣 ∙ (𝑢 + 𝑣)
= 𝑢
𝑢−𝑣
𝑢+𝑣
2
2
− 2𝑢. 𝑣 +
𝑢+𝑣
+ 2𝑢. 𝑣 +
=
𝑣
2
𝑣
2
𝑢−𝑣
=
𝑢
2
− 2𝑢. 𝑣 +
𝑣
2
4𝑢. 𝑣 = 0
𝑢. 𝑣 = 0
Hence , 𝑢 𝑎𝑛𝑑 𝑣 are orthogonal
Proof :(⤌) Given 𝑢 𝑎𝑛𝑑 𝑣 are orthogonal
WTS :
𝑢+𝑣
𝑢+𝑣
2
=
=
𝑢
= 𝑢
2
2
𝑢−𝑣
− 2𝑢. 𝑣 +
𝑣
2
And
𝑢−𝑣
2
− 2𝑢. 𝑣 +
𝑣
2
Since 𝑢 𝑎𝑛𝑑 𝑣 are orthogonal 𝑢 . 𝑣 = 0
Therefore,
𝑢+𝑣
=
𝑢−𝑣
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Example:- let u,v and w are vectors in space such that u+v+w=0, IIuII=3,
IIvII=5 and IIwII=7, then find the angle b/n u and v.
Solution:Example:- If u and v are unit vectors make an angle of
b/n them,
then find II3u+vII.
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Properties of NORMED Space
1.
2.
3.
4.
5.
6.
7.
…… Parallelogram Law
……………… Polarization identity
…….. Cauchy-Schwarz Inequality
…….. Triangle inequality
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Pythagoras Theorem:
If A and B are orthogonal vectors, then 𝐴 + 𝐵 2 = 𝐴 2 + 𝐵 2
Proof: 𝐴 + 𝐵 2 = 𝐴 + 𝐵 ∙ 𝐴 + 𝐵
= 𝐴 2 + 2𝐴 ∙ 𝐵 + 𝐵 2
= 𝐴 2 + 𝐵 2 𝑠𝑖𝑛𝑐𝑒 𝐴 ∙ 𝐵 = 0.
Note: If A is perpendicular to B, then it is also perpendicular to
any scalar multiple of B.
Orthogonal Projection
Definition: Suppose S is the foot of the perpendicular from R to
the line containing 𝑃𝑄 , then the vector with representation
𝑃𝑆 is called the vector projection of B on to A, and is denoted
by project𝐴𝐵 .
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R
R
B
B
A
S
A
Q
P
P
Proj AB
S
Q
projAB
𝑝𝑟𝑜𝑗𝐴𝐵 = 𝑃𝑆 = 𝑡𝐴
𝐵 − 𝑝𝑟𝑜𝑗𝐴𝐵 = 𝑆𝑅
(𝐵 − 𝑝𝑟𝑜𝑗𝐴𝐵 ).A=0
(B- t A).A =0
B.A –t A.A =0
𝑨∙𝑩
𝒕= 𝑨 𝟐
𝑨∙𝑩
𝑝𝑟𝑜𝑗𝐴𝐵 = 𝑃𝑆 = 𝒕. 𝑨 = 𝑨 𝟐 𝐴
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The scalar projection of B onto A (also called the component of B a long A)
is defined to be the length of project𝐴𝐵 , which is equal to 𝑩 𝐜𝐨𝐬 𝜽 and is
denoted by comp𝐴𝐵 . Thus:
𝒑𝒓𝒐𝒋𝑩𝑨 =
𝒄𝒐𝒎𝒑𝑩𝑨 = 𝒑𝒓𝒐𝒋𝑩𝑨 =
𝑨∙𝑩
𝑨
𝑨
=
𝑨
𝑨∙𝑩
𝑨 =
𝑨 𝟐
𝑨∙𝑩
𝑨 𝟐
𝑨∙𝑩
𝑨 𝟐
𝑨
𝑨∙𝑩
𝑨 = 𝑨
Example: Let 𝐴 = −1,3,1 = −𝑖 + 3𝑗 + 𝑘 and 𝐵 = 2,4,3 = 2𝑖 + 4𝑗 + 3𝑘
13
Then find (i) 𝒑𝒓𝒐𝒋𝑨𝑩 [𝑎𝑛𝑠. 29 2,4,3 ] ,
13
(ii) 𝒑𝒓𝒐𝒋𝑩𝑨 [𝑎𝑛𝑠. 11 −1,3,1 ]
(iii) 𝒄𝒐𝒎𝒑𝑩𝑨
Directional angles and cosines
Let 𝐴 = 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 be a vector positioned at the origin in ℝ3 , making an
angle of 𝛼 , 𝛽 𝑎𝑛𝑑 𝛾 with the positive 𝑥 , 𝑦 𝑎𝑛𝑑 𝑧 axes respectively. Then
the angles 𝛼 , 𝛽 𝑎𝑛𝑑 𝛾 are called the directional angles of A, and the quantities
cos 𝛼 , 𝑐𝑜𝑠 𝛽 𝑎𝑛𝑑 cos 𝛾 are called the directional cosines of A, which can
be computed as follows:
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𝐴.𝑖
cos 𝛼 = 𝐴
𝑎
𝐴.𝑗
1
=
, cos 𝛽 = 𝐴
𝑖
𝐴
𝑎
𝑗
𝐴.𝑘
= 𝐴2 and cos 𝛾 = 𝐴
𝑎
3
=
𝑘
𝐴
From this relation we can deduce a unit vector
𝐴 = 𝐴 (𝑐𝑜𝑠𝛼, 𝑐𝑜𝑠𝛽, 𝑐𝑜𝑠𝛾)
𝐴
𝐴
= (𝑐𝑜𝑠𝛼, 𝑐𝑜𝑠𝛽, 𝑐𝑜𝑠𝛾) is a unit vector.
𝑎
cos 𝛼 = 𝐴1 , 𝛼 ∈ 0, 𝜋
𝑎
𝑎
, 𝑐𝑜𝑠 𝛽 = 𝐴2 , 𝛽 ∈ 0, 𝜋 , cos 𝛾 = 𝐴3 , 𝛾 ∈ 0, 𝜋
𝑍
A = a1 i + a2 j +a3k
𝛾
𝛼
𝛽
𝑌
𝑋
Remark: 𝑐𝑜𝑠 2 𝛼 + 𝑐𝑜𝑠 2 𝛽 + 𝑐𝑜𝑠 2 𝛾 = 1. (Verify!)
Solution
𝑎1
𝐴
2
+
𝑎2
𝐴
2
+
𝑎3
𝐴
2
=
𝑎1 2 + 𝑎2 2 + 𝑎3 2
𝐴 2
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𝐴 2
= 𝐴 2=1
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Solution:Solution:-
Solution:Exercise:- Use vector, Show that the altitudes of the triangle are concurrent.
Solution:-
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Definition: Suppose that 𝐴 =
𝐵 =
𝑏1 , 𝑏2 , 𝑏3
𝑎1 , 𝑎2 , 𝑎3
= 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 and
= 𝑏1 𝑖 + 𝑏2 𝑗 + 𝑏3 𝑘 be two vectors inℝ3 . Then the cross
product 𝐴 × 𝐵 of the two vectors is defined as:
𝑨×𝑩 =
Or
𝒂𝟐 𝒃𝟑 − 𝒂𝟑 𝒃𝟐 𝒊 + 𝒂𝟑 𝒃𝟏 − 𝒂𝟏 𝒃𝟑 𝒋 + 𝒂𝟏 𝒃𝟐 − 𝒂𝟐 𝒃𝟏 𝒌
𝐴𝑋𝐵 = 𝐷𝑒𝑡
Example: Let
𝑖
𝑗
𝑘
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝐴 = 4𝑖 − 3𝑗 + 2𝑘
𝐵 = 2𝑖 − 5𝑗 − 𝑘
Then find a) 𝐴 × 𝐵
b) 𝐵 × 𝐴
Remarks: For two non – zero vectors A & B,
1.
𝐴 × 𝐵 is a vector which is orthogonal to both A and B.
2.
𝐴 × 𝐵 is not defined for 𝐴, 𝐵 ∈ ℝ2 .
3.
𝑖 × 𝑗 = −𝑗 × 𝑖 = 𝑘
𝑗×𝑘 = − 𝑘×𝑗
= 𝑖
𝑘×𝑖 = − 𝑖×𝑘
= 𝑗
A XB
B
A
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-A XB = B XA
37
Properties of Cross Product
Let A, B and C be vectors in ℝ3 and 𝛼 be any scalar. Then:
(1) 𝐴 × 0 = 0 × 𝐴 = 0, 𝑤ℎ𝑒𝑟𝑒 0 = 0,0,0
(2) 𝐴 × 𝐵 = − 𝐵 × 𝐴
(3) 𝐴 × 𝐵 × 𝐶
≠
𝐴×𝐵
(4) 𝛼𝐴 × 𝐵 = 𝐴 × 𝛼𝐵
(5) 𝐴 ×
𝐵+𝐶
(6) 𝐴 ∙ 𝐴 × 𝐵
×𝐶
= 𝛼
𝐴×𝐵
= 𝐴×𝐵+ 𝐴×𝐶
= Β∙ 𝐴×𝐵
= 0
(7) If A and B are parallel, then 𝐴 × 𝐵 = 0
(8) 𝐴 × Β
2
=
𝐴
(9) 𝐴 × Β
=
𝐴
(10)
2
𝐵
Β
𝐴×𝐵 = 𝑛
2
− 𝐴∙Β 2
sin 𝜃 , 𝜃 ∈ 0, 𝜋 .
𝐴
𝐵
sin 𝜃, 𝑤ℎ𝑒𝑟𝑒 𝑛 is the unit
vector in the direction of 𝐴 × 𝐵 𝑎𝑛𝑑 𝜃 ∈ 0, 𝜋 is the
angle between A and B.
Example: If
𝐴
= 2,
A and B then find
Β
= 4 𝑎𝑛𝑑 𝜃 = 𝜋 4 for two vectors
𝐴×Β .
Note: The angle 𝜃 between A and B can be obtained by
sin 𝜃 =
𝐴×Β
𝐴
Β
, for two non – zero vectors A and B.
Melak Mesfin
38
Applications of cross Product:
(i)Area: The area of a parallelogram whose adjacent sides
coincides with the vectors A and B is given by 𝐴 × Β = 𝐴
d
B
Β
sin 𝜃
b
h
O
a
A
So, 𝑎 𝑜𝑎𝑏𝑐 = 𝐵𝑎𝑠𝑒 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡
= 𝐴
Β
sin 𝜃
N.B: The area of the triangle formed by A and B as its adjacent sides is given by
1
𝑎𝑟𝑒𝑎 = 2 𝐴 × Β .
ii.Volume
The volume V of a parallelepiped with the three vectors A, B and C in ℝ3 as three of its
adjacent edges is given by:
𝑽= 𝑨∙ 𝑩×𝑪
𝒂𝟏
= 𝐝𝐞𝐭 𝒃𝟏
𝒄𝟏
𝒂𝟐
𝒃𝟐
𝒄𝟐
𝒂𝟑
𝒃𝟑
𝒄𝟑
Melak Mesfin
39
BXC
A
B
C
ℎ=
𝑝𝑟𝑜𝑗
Hence,
𝐴
𝐵×𝐶
=
𝐴∙ 𝐵×𝐶
𝐵×𝐶 2
𝐵×𝐶
=
𝐴∙ 𝐵×𝐶
𝐵×𝐶
V =𝐵𝑎 ℎ = 𝐴 ∙ 𝐵 × 𝐶
Examples:
1. Find the area of a triangle whose vertices are 𝐴 1, −1,0 , 𝐵 2,1, −1
𝑎𝑛𝑑 𝐶 −1,1,2 .
Solution: The vectors on the sides of the
triangle ∆ 𝐴Β𝐶 are 𝐴𝐵 = 𝐵 − 𝐴 = 1,2, −1 𝑎𝑛𝑑 𝐴𝐶 = −2,2,2 = 𝐶 − 𝐴.
So, 𝑎 ∆ 𝐴Β𝐶 =
1
2
𝐴𝐵 × 𝐴𝐶
= 3 2 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠.
1. Find the volume of the parallelepiped with edges
𝑢 = 𝑖 + 𝑘 , 𝑣 = 2𝑖 + 𝑗 + 4𝑘 𝑎𝑛𝑑 𝑤 = 𝑗 + 𝑘.
Solution: V= 𝑢 ∙ 𝑣 × 𝑤
= 1 𝑢3
Melak Mesfin
N.B: Three vectors 𝐴, 𝐵 𝑎𝑛𝑑 𝐶 are coplanar iff 𝐴 ∙ 𝐵 × 𝐶 = 0.
40
Lines and Planes in ℝ𝟑
Definition: A vector 𝑣 = 𝑎, 𝑏, 𝑐 is said to be parallel to a line ℓ 𝑖𝑓 𝑣
is parallel to Ρ0 Ρ1 for any two distinct points Ρ0 𝑎𝑛𝑑 Ρ1 𝑜𝑛 ℓ.
A line ℓ 𝑖𝑛 ℝ3 is determined by a given point Ρ0 𝑥0 , 𝑦0 , 𝑧0 on ℓ
and a parallel vector 𝑣 (directional vector) 𝑣 = 𝑎, 𝑏, 𝑐 𝑡𝑜 ℓ.
Equations of a line in space:
Let Ρ0 𝑥0 , 𝑦0 , 𝑧0 be a given point on a line ℓ and 𝑃 𝑥, 𝑦, 𝑧 be any
arbitrary point on ℓ. If 𝑉 = 𝑎, 𝑏, 𝑐 is the parallel vector to ℓ, then
1) The parametric equation of ℓ is given by 𝑥 = 𝑥0 + 𝑎𝑡,
𝑦 = 𝑦0 + 𝑏𝑡 ,
𝑧 = 𝑧0 + 𝑐𝑡, 𝑡 ∈ ℜ, where t is called
the parameter.
2) The symmetric form of equation of ℓ is given by:
𝒙 − 𝒙 𝟎 𝒚 − 𝒚𝟎 𝒛 − 𝒛 𝟎
=
=
, 𝒇𝒐𝒓 𝒂, 𝒃, 𝒄 ≠ 𝟎.
𝒂
𝒃
𝒄
3) The vector equation of ℓ is written as 𝒓 − 𝒓𝟎 = 𝒕𝒗, 𝑤ℎ𝑒𝑟𝑒 𝑡 ∈ ℜ.
Melak Mesfin
41
Remarks:
1. The above equations of the line can be derived using vector
algebra as follows:
Z
p
P0
r0
r
Y
X
From vector addition, we have 𝑟 − 𝑟0 = Ρ0 Ρ & 𝑟𝑜 = 𝑥0 𝑖 + 𝑦0 𝑗 + 𝑧0 𝑘,
𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘. Since Ρ0 Ρ 𝑣 , there exists 𝑡 ∈ ℜ such that
𝑟 − 𝑟0 = 𝑡𝑣 = Ρ0 Ρ

𝑥 − 𝑥0 , 𝑦 − 𝑦0 , 𝑧 − 𝑧0 = 𝑡 𝑎, 𝑏, 𝑐
 𝑥 = 𝑥0 + 𝑎𝑡 , 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 + 𝑐𝑡.
2. If one of 𝑎, 𝑏 𝑜𝑟 𝑐 𝑖𝑠 0, (say for instance 𝑏 = 0),
the symmetric equation of ℓ is given as:
𝑥 − 𝑥0
𝑧 − 𝑧0
Melak
= Mesfin
, 𝑦 = 𝑦0 .
𝑎
𝑐
42
Note that:- For two distinct lines
1)
Are intersecting lines iff
such that
2) Thus
3) Non-intersecting and non-parallel lines called skew line.
4)
Example: Find parametric equations and symmetric equations of the line l
containing the points P1 =(4, – 6, 5) and P2 = (2, –3, 0).
Solution:-
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43
Exercise:-1 Show that the lines and with parametric equations
are skew lines; that is, they do not intersect and are not parallel (and
therefore do not lie in the same plane).
Exercise:- 2 Find the point of intersection of the two lines
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44
Equation of a plane:
A plane in space is determined by a point Ρ0 𝑥0 , 𝑦0 , 𝑧0 in the plane
and a vector 𝑛 that is orthogonal to the plane. This orthogonal
vector 𝑛 is called a normal vector. Suppose that 𝑃 𝑥, 𝑦, 𝑧 be any
arbitrary point in the plane, and let 𝑟 𝑎𝑛𝑑 𝑟0 be the position
vectors of 𝑃 𝑥, 𝑦, 𝑧 and Ρ0 𝑥0 , 𝑦0 , 𝑧0 . Then we have
𝑛 ∙ 𝑟 − 𝑟0 = 0 (Since 𝑛 perpendicular to any vector in the plane).
 𝑛 ∙ 𝑟 = 𝑛 ∙ 𝑟𝑜 ,which is called the vector equation of the plane.
If we let 𝑛 = 𝑎, 𝑏, 𝑐 = 𝑎𝑖 + 𝑏𝑗 + 𝑐𝑘, we get
𝑎, 𝑏, 𝑐 ∙ 𝑥 − 𝑥0 , 𝑦 − 𝑦0 , 𝑧 − 𝑧0 = 0
 𝑎 𝑥 − 𝑥0 + 𝑏 𝑦 − 𝑦0 + 𝑐 𝑧 − 𝑧0 = 0
(point - normal form of equation of a plane)
⟺ 𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎, 𝒘𝒉𝒆𝒓𝒆 𝒅 = − 𝒂𝒙𝟎 + 𝒃𝒚𝟎 + 𝒄𝒛𝟎
Melak Mesfin
(general or standard form of the
equation of a plane).
45
Examples:
1. Find the equations of a line that contains the point 1,4, −1
and parallel to 𝑣 = −2𝑖 + 3𝑗.
Solution: let ΡΟ = 1 ,4, −1 = 𝑥0 , 𝑦0 , 𝑧0
Then the parametric form of the equation of the line is:
𝑥 = 1 − 2𝑡
𝑦 = 4 + 3𝑡
𝑧 = −1
𝑥−1
=
−2
𝑦−4
3
and its symmetric form is given as:
, 𝑧 = −1.
2. Find the equation of the plane through the points Ρ0 1,1,1 ,
Ρ1 2,2,0 𝑎𝑛𝑑 Ρ2 4, −6,2 .
Solution: The vectors 𝐴 = Ρ0 Ρ1 = 1,1, −1 𝑎𝑛𝑑
Β = Ρ1 Ρ2 = 3, −7,1
are parallel to the plane, and hence, their
cross product 𝑛 = 𝐴 × Β = −6, −4, −10 is normal to the plane.
Thus, the equation of the plane is given by:
−6 𝑥 − 1 − 4 𝑦 − 1 − 10 𝑧 − 1 = 0
 3 × +2𝑦 + 5𝑧 − 10 = 0
Melak Mesfin
46
OR: The equation of the plane can be obtained by computing:
𝑥−1
𝑑𝑒𝑡 𝑥 − 2
𝑥−4
𝑦−1
𝑦−2
𝑦+6
𝑧−1
𝑧−0
𝑧−2
=0
Distance in Space
a) Distance from a point to a line
The distance D from a point Ρ1 (not on ℓ) to a line ℓ in space is given by:
𝒗 × 𝚸𝚶 𝚸𝟏
𝒗
𝑫=
, where 𝑣 is the directional vector of ℓ and ΡΟ is any point on ℓ.
Proof:
Z
P1
L
D
𝜃
P0
V
Y
X
sin 𝜃 =
𝐷
ΡΟ Ρ1
⟹ 𝐷 = ΡΟ Ρ1 sin 𝜃
But since 𝑣 × ΡΟ Ρ1 = 𝑣
ΡΟ Ρ1
sinMelak
𝜃 , =Mesfin
𝑣 𝐷 , we have: 𝐷 =
𝑣 ×Ρ Ο Ρ 1
𝑣
.
47
a)
Distance from a point to a plane
The perpendicular distance D of a point Ρ0 𝑥0 , 𝑦0 , 𝑧0
in space to the
plane with the equation 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑 = 0 is given by:
𝐷 =
𝑎𝑥 0 +𝑏𝑦 0 +𝑐𝑧 0 +𝑑
𝑎 2 +𝑏 2 +𝑐 2
=
𝑛 ∙𝑜Ρ
𝑛
, where o is the foot of 𝑛 within the plane.
Proof: Consider the diagram below:
P(x0,y0,z0)

O(x1,y1,z1)
𝐷 =
𝑝𝑟𝑜𝑗𝑛0Ρ
 𝐷 =
=
𝑜Ρ ∙ 𝑛
𝑛 2
𝑎𝑥 0 +𝑏𝑦 0 +𝑐𝑧 0 +𝑑
𝑎 2 +𝑏 2 +𝑐 2
𝑛
=
𝑜Ρ ∙ 𝑛
𝑛
, 𝑤ℎ𝑒𝑟𝑒 𝑑 = − 𝑎𝑥1 + 𝑏𝑦1 + 𝑐𝑧1
P1(x1,y1,z1)
𝜃

B
O
P0(x0,y0,z0)
Melak Mesfin
48
Melak Mesfin
49
Melak Mesfin
50
Distance between two parallel planes
Given two parallel planes 𝜋1 𝑎𝑛𝑑 𝜋2 . Then we can have normal vectors
with coefficients 𝑎, 𝑏, 𝑐 to be the same such that:
𝜋1 : 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑1 and 𝜋2 : 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑2.
Then the distance between 𝜋1 𝑎𝑛𝑑 𝜋2 is the same as the distance
from any arbitrary point 𝑃 𝑥0 , 𝑦0 , 𝑧0
plane 𝜋2 . ⇒𝐷 =
Thus,
𝑎 𝑥 𝑜 +𝑏𝑦 𝑜 +𝑐𝑧𝑜 −𝑑 2
𝑎 2 +𝑏 2 +𝑐 2
has been taken from 𝜋1 to the
. But 𝑎𝑥𝑜 + 𝑏𝑦𝑜 + 𝑐𝑧𝑜 = 𝑑1
𝑑 1 −𝑑 2
𝐷=
𝑎 2 +𝑏 2 +𝑐 2
Example: Find the distance between the planes
𝜋1 : 𝑥 + 2𝑦 − 2𝑧 = 3 𝑎𝑛𝑑 𝜋2 : 2𝑥 + 4𝑦 − 4𝑧 = 7
Solution: We first rewrite the equations 𝜋1 𝑎𝑛𝑑 𝜋2 so that they
have the same 𝑛 = 𝑎, 𝑏, 𝑐 . That is:
𝜋1 : 𝑥 + 2𝑦 − 2𝑧 = 3
𝜋2 : 𝑥 + 2𝑦 − 2𝑧 = 7 2 ⟹ 𝑑1 = 3& 𝑑2 = 7 2
Thus, 𝐷 =
=
𝑑 1 −𝑑 2
𝑎 2 +𝑏 2 +𝑐 2
7
2−3
1+4+4
=
, 𝑤ℎ𝑒𝑟𝑒 𝑎 = 1, 𝑏 = 2, 𝑐 = −2
1
2
3
=1 6
Melak Mesfin
51
Remarks:1. Let
a) If two points of the line lies on the plane, the lies on .
b)
i.e.
c)
i.e.
2. Let
be two planes:
a)
i.e.
b)
i.e.
c) the angle b/n
is the angle b/n
thus, we have
Melak Mesfin
52
Example:- Find the equation of the plane passing through p(-2,3,5) and is
perpendicular to the line with parametric equation
Solution:Exercise:- 1. Find the point of intersection of the line with parametric
equations
intersect the plane
Exercise:-2 For two planes
a) Find the angle b/n
.
b) Find the symmetric equations for the line of intersection of the two
planes
.
Melak Mesfin
53
Vector Space
Definition(Vector Space)
A real vector space V is a set
of objects together with two
operations addition “+” and
Scalar multiple “.” satisfying
the following axioms:
Melak Mesfin
54
Note that :
i) R(the set of real number) is a vector space over a field R.
ii) any vector space contain at least one element(i.e. Nonempty)
Example:-1 The set of all vectors in a plane is a vector space over R
under the standard operation vector addition “+” and Scalar
Multiple.
Solution:iii) Under the standard operation
are real vector
space.
Example:- 2
together with the standard
operations Matrix addition and scalar Multiple form a real vector
spaces.
Melak Mesfin
55
Example:- 3
, the set of all polynomial of degree less than or
equal to 2 with standard operations is a real vector space.
iv) In order to show that V is not a vector space it is enough to show
that one axiom of the vector space doesn’t hold
Solution:Example:-4 Is
with standard operation is a vector space?
Solution:Example:-5
be the set of all vectors in a plane with defined
operations. Examine whether
is a vector space or not.
a)
b)
Melak Mesfin
56
Subspaces
Definition: Let V be a given vector space over a field F. Then a non-empty subset W of V is said
to be a subspace of V if W itself is a vector space over F under the operations of V.
Theorem: Suppose that V is a vector space over a field F.A non-empty subset W of V is a
subspace of V if it satisfies the following conditions:
i.
ii.
iii.
𝑢, 𝑣 ∈ 𝑊 ⇒ 𝑢 + 𝑣 ∈ 𝑊.
∀𝑢 ∈ 𝑊 𝑎𝑛𝑑 𝑐 ∈ 𝐹 ⇒ 𝑐𝑢 ∈ 𝑊.
0 ∈ 𝑊.
Examples
1. V and 0 are the trivial subspaces of any vector space V.
2. For the vector space 𝑉 = ℝ3 = { 𝑥, 𝑦, 𝑧 ; 𝑥, 𝑦, 𝑧 ∈ ℝ} over ℝ.Then the set 𝑊 =
{ 𝑥, 𝑦, 0 ; 𝑥, 𝑦 ∈ ℝ} is a subspace of V. (verify!)
Solution
i) Let 𝑈 = 𝑥1 , 𝑦1 , 0 and 𝑉 = 𝑥2 , 𝑦2 , 0
then 𝑈 + 𝑉 = 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 0 ∈ 𝑊
ii) Let 𝑈 = 𝑥1 , 𝑦1 , 0 and 𝑐𝑈 = {𝑐𝑥1 , 𝑐𝑦1 , 0) ∈ 𝑊}
iii) (0,0,0)W
Therefore W is the subspace of V Melak Mesfin
57
3. The set of all lines passing through the origin, 𝐿 = {𝑎𝑥 + 𝑏𝑦 = 0, 𝑎, 𝑏 ∈ ℝ} is a
subspace of the vector space 𝑉 = ℝ2 .
Solution
i) Let 𝐿1 = {𝑎𝑥1 + 𝑏𝑦1 = 0, 𝑎, 𝑏 ∈ ℝ} and 𝐿2 = {𝑎𝑥2 + 𝑏𝑦2 = 0, 𝑎, 𝑏 ∈ ℝ
the 𝐿1 + 𝐿2 = {𝑎 𝑥1 + 𝑥2 , 𝑏 𝑦1 + 𝑦2 = 0, 𝑎, 𝑏 ∈ ℝ} ∈ 𝐿
ii)Let 𝐿1 = {𝑎𝑥1 + 𝑏𝑦1 = 0, 𝑎, 𝑏 ∈ ℝ} and 𝑐𝐿1 = {𝑐(𝑎𝑥1 + 𝑏𝑦1 ) = 0, 𝑎, 𝑏 ∈ ℝ} ∈ 𝐿
iii) 𝐿1 = {𝑎(0) + 𝑏(0) = 0, 𝑎, 𝑏 ∈ ℝ} ∈ 𝐿 this implies that line L passes through the
origin.
Therefore L is the subspace of V
Exercise: 1.Is the set 𝑊 = {𝑥 − 4𝑦 = 1} a subspace of 𝑉 = ℝ2 ? Justify.
2. Which of the following is the subspace of 𝑉 = ℝ3 ?
a) 𝑊 = { 𝑥1 , 𝑥2 , 1 ; 𝑥1 , 𝑥2 ∈ 𝑅
b) 𝑊 = { 𝑥1 , 𝑥1 + 𝑥3 , 𝑥3 ; 𝑥1 , 𝑥Melak
3∈𝑅
Mesfin
58
Theorem2: A subset W of a vector space V is a subspace of v iff
Example:- Show that
of
.
Exercise:- Let V be a vector space
following are subspace or not.
a.
b.
c.
d.
Melak Mesfin
is the subspace
. Examine whether the
59
Definition 1.1.8 A vector w in Rn is said to be a linear combination.. of the vectors
V1, V2, … , Vk in Rn if w can be expressed in the form W = c1 v1 + c2v2 + . . . + ckvK
The scalars c1, c2, ... , ck are called the coefficients in the linear combination. In the case
where k = 1, Formula (14) becomes w = c1 v1, so to say that w is a linear combination of v1
is the same as saying that w is a scalar multiple of v1.
Example : Provided that X= (-1 , -2 , -2 ) , U = (0 ,1 ,4) , V = (-1 , 1 ,2) and W = (3 ,1 ,2) in
R3 find scalars a , b and c such that X = a U + b V + c W
SOLUTION (-1 , -2 , -2 ) = a (0, 1 ,4) + b ( -1 ,1 ,2 ) + c (3 , 1 ,2 )
= (-b +3c , a +b +c , 4a +2b +2c )
you can equate corresponding components so that they form the system of three linear
equations as shown below.
-b +3c = -1
a +b +c = - 2
,4a +2b +2c = -2
Answer a =1 ,b= - 2 and c = -1
Try using vector addition and scalar multiplication to check this result.
Melak Mesfin
60
Linear Dependence and Independence
Definition: Let 𝑣1 , 𝑣2, … , 𝑣𝑛 be elements of an arbitrary vector space V, and
𝛼1 , 𝛼2, … , 𝛼𝑛 be scalars. An expression of the form 𝛼1 𝑣1 + 𝛼2 𝑣2 + ⋯ 𝛼𝑛 𝑣𝑛
is called linear combinations of the vectors 𝑣1 , 𝑣2, … , 𝑣𝑛 .
Examples:
1) For the set of vectors in ℝ3 , 𝑆 = { 1,3,1 , 0,1,2 , 1,0,5 }
Let 𝑣1 = 1,3,1 , 𝑣2 =
0,1,2 , 𝑎𝑛𝑑𝑣3 1,0,5
V1 is a linear combinations of v2 and v3 because
V1= c1 (0,1, 2) +c2 (1,0,5),
C1 =1 and
(1, 3, 1) = c1 (0, 1, 2) +c2 (1,0,5)
2c1 -5c2 =1, C2 =3
V1 =3 V2 + V3
This implies that S is linearly independent .
Exercise:
1) Write the vector W = (1,1,1) as a linear combination of vectors in the set S.
S = { (1,2,3),(0,1,2), (-1,0,1)}.
2) If possible, write the vector W = (1,-2, 2) as a linear combinations of vectors in the set
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S ={ (1,2,3),(0,1,2), (-1,0,1)}.
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Definition: Let 𝑣1 , 𝑣2, … , 𝑣𝑛 are vectors in a vector space V over ℝ.Then vectors are called:
1. linearly dependent if there exist scalars 𝛼1 , 𝛼2, … , 𝛼𝑛 not all zero such that
𝛼1 𝑣1 + 𝛼2 𝑣2 + ⋯ 𝛼𝑛 𝑣𝑛 = 0.
2. Linearly independent if 𝛼1 𝑣1 + 𝛼2 𝑣2 + ⋯ 𝛼𝑛 𝑣𝑛 = 0 implies
𝛼1 = 𝛼2 = … = 𝛼𝑛 = 0.
Linear dependence in the vector space V ¼ R3 can be described geometrically as follows:
(a) Any two vectors u and v in R3 are linearly dependent if and only if they lie on the same line
through the origin O, as shown in Fig. 4-3(a).
(b) Any three vectors u, v, w in R3 are linearly dependent if and only if they lie on the same
plane through the origin O, as shown in Fig. 4-3(b)Later, we will be able to show that any four
or more vectors in R3 are automatically linearly dependent.
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Examples:
1. Determine whether the following set of vectors in the vector space 𝑉 = ℝ3 are linearly
dependent or independent. a) 1, 0,0 , 0 , 1, 0 , (0, 0,3) b) 2, 6,0 , 2 , 4, 1 , (1 , 1, 1) .
c) {(1,2,3),(0,1,2),(-2,0,1)}
2. Let V be the vector space of all real valued functions of the variable t. Then which of the
following set of functions are LD/LI? Justify!
a) 𝑡, 𝑡 2 , 𝑠𝑖𝑛𝑡
b) 𝑐𝑜𝑠 2 𝑡, 𝑠𝑖𝑛 2 𝑡, 1
3. Determine whether the following set of vectors in the vector space 𝑉 = ℝ2 are linearly
dependent or independent a) {(1, 2), (2,4)} b) {(1, 0), (0,1),(-2,5)}
4 . Determine whether the following set of vectors in P2 are linearly dependent or
independent . S = {1+ x -2x2 , 2+5x –x2 , x+x2 }
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Basis of a vector space
Definitions: Let V be any vector space over a field F, and let the set 𝑆 = 𝑣1, 𝑣2 , … , 𝑣𝑛 be a set of
vectors in V. Then:
i)
ii)
iii)
S is said to spans (or generates) V if each element of V is a linear combinations of
elements of S.
S is called a basis for V if S is a linearly independent set and it spans V.
The dimension of V is said to be n (dim V=n) if V has basis consisting of n-elements.
Examples:
1. Show that the set 𝑆 = 1,0,0 , 0,1,0 , (0,0,5) form a basis of the vector spaceℜ3 .
Solution
I) we need to show that S is linearly independent
𝛼 1,0,0 + 𝛽 0,1,0 , + 𝛾 0,0,5 = (0,0,0)
𝛼=𝛽=𝛾=0
Hence S is linearly independent
ii) we need to show that S spans ℜ3
Let (x, y ,z) ℜ3 then (𝑥, 𝑦, 𝑧) = 𝛼 1,0,0 + 𝛽 0,1,0 , + 𝛾 0,0,5
𝑧
𝛼 = 𝑥 , 𝛽 = 𝑦 𝑎𝑛𝑑 𝛾 =
5
𝑧
(𝑥, 𝑦, 𝑧) = 𝑥 1,0,0 + 𝑦 0,1,0 , + 5 0,0,5
Therefore , S Spans ℜ3
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1.8. Basis of a vector space.
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